Number Play ⇒⇒ Exercise 5

Question 1

Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and ‘–’ signs in between the numbers. How many different possibilities exist? Write all of them.
Eight such expressions are possible. You can use the diagram below to systematically list all the possibilities.
Evaluate each expression and write the result next to it. Do you notice anything interesting?

Solution :

Systematic List of All 8 Possibilities

$3 + 4 + 5 + 6 $ = 18

$3 + 4 + 5 - 6 $ = 6

$3 + 4 - 5 + 6 $ = 8

$3 + 4 - 5 - 6$ = -4

$3 - 4 + 5 + 6$ = 10

$3 - 4 + 5 - 6$ = -2

$3 - 4 - 5 + 6$ = 0

$3 - 4 - 5 - 6$ = -12

Interesting Observations

Notice that all the results are even. If you start with any four consecutive numbers, the sum or difference will always result in an even integer .

The Zero Result:

For any four consecutive numbers $n, n+1, n+2, n+3$, the pattern $n - (n+1) - (n+2) + (n+3)$ will always equal 0.

Question 2

Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs as you have done before. Find out the results of each expression. What do you observe?

Solution :

Let’s choose four different consecutive numbers: 10, 11, 12, and 13.

Just like before, there are 8 possible combinations of '+' and '–' signs.

Systematic List of All 8 Possibilities

$10 + 11 + 12 + 13$ = 46

$10 + 11 + 12 - 13$ = 20

$10 + 11 - 12 + 13$ = 22

$10 + 11 - 12 - 13$ - 4

$10 - 11 + 12 + 13$ = 24

$10 - 11 + 12 - 13$ = -2

$10 - 11 - 12 + 13$ = 0

$10 - 11 - 12 - 13$ = -26

Interesting Observations

Always Even: If you start with any four consecutive numbers, the sum or difference will always result in an even integer .

The Zero Result:

For any four consecutive numbers $n, n+1, n+2, n+3$, the pattern $n - (n+1) - (n+2) + (n+3)$ will always equal 0.

The "Fixed" Results: Notice that three of the results are exactly the same as in the first example:

(+ – –) always equals -4.

(– + –) always equals -2.

(– – +) always equals 0.

Range and Symmetry:

The results spread out quite significantly, but the patterns of how they increase or decrease remain consistent between different sets of numbers.

Question 3

Repeat this for one more set of 4 consecutive numbers. Share your findings

Solution :

Let’s choose four different consecutive numbers: 20, 21, 22, and 23

Just like before, there are 8 possible combinations of '+' and '–' signs.

Systematic List of All 8 Possibilities

$20 + 21 + 22 + 23$ = 86

$20 + 21 + 22 - 23$ = 40

$20 + 21 - 22 + 23$ = 42

$20 + 21 - 22 - 23$ = -4

$20 - 21 + 22 + 23$ = 44

$20 - 21 + 22 - 23$ = -2

$20 - 21 - 22 + 23$ = 0

$20 - 21 - 22 - 23$ = -46

This exercise demonstrates that while the specific numbers might change, the underlying structure of mathematics ensures that certain relationships and patterns remain perfectly constant!

Question 1

We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even.

Solution :

To determine if an expression is even without calculating the final answer, we can use the rules of parity (how even and odd numbers interact).

Addition and Subtraction

The rules are simple:

Even ± Even = Even

Odd ± Odd = Even

Even ± Odd = Odd

1: $43 + 37$

Analysis : Odd + Odd, Is it Even? : Yes Even

2: $672 - 348$

Analysis : Even - Even, Is it Even? : Yes Even

3: $708 - 477$

Analysis : Even - Odd, Is it Even? : No (Odd)

4: $809 + 214$

Analysis : Odd + Even, Is it Even? : No (Odd)

5: $543 - 479$

Analysis : Odd - Odd, Is it Even? : Yes Even

Multiplication and Powers

The rules are simple:

Even $\times$ Anything = Even

Odd $\times$ Odd = Odd

1: $4 \times 347 \times 3$

Analysis : Contains an even number (4) , Is it Even? : YES, Even

2: $119 \times 303$

Analysis : Odd $\times$ Odd, Is it Even? : No (Odd)

3: $513^3$

Analysis : Odd $\times$ Odd $\times$ Odd, Is it Even? : No (Odd)

Question 2

Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.

Solution :

To determine if an algebraic expression is even without calculating the final answer, we can use the rules of parity:

Any integer multiplied by an even number (like 2, 4, or 6) is always even, and the sum of two even numbers is always even.

Always Even

These expressions will result in an even number regardless of whether the variables ($a, b, m, n, u, v, k, j$) are even or odd integers:

1: $2a + 2b$

This is the sum of two even terms ($Even + Even$).

2: $4m + 2n$

Since both 4 and 2 are even, both terms are always even ($Even + Even$).

3: $2u - 4v$

Similar to the above, both terms are multiples of 2, so the result is always even ($Even - Even$).

4: $13k - 5k$

This simplifies to $8k$. Since 8 is even, the product is always even.

5: $4k \times 3j$

This simplifies to $12kj$. Since 12 is even, the entire product is always even.

Not Necessarily Even

These expressions depend on the specific values of the variables and will not be even for all integers:

1: $3g + 5h$

Analysis : If $g=1$ and $h=0$, the result is 3 (odd).

2: $6m - 3n$

Analysis : While $6m$ is always even, $3n$ can be odd. If $n$ is odd, the result is odd ($Even - Odd = Odd$).

3: $x^2 + 2$

Analysis : If $x$ is odd, $x^2$ is odd. Then $Odd + 2$ results in an odd number.

If $x$ is even, $x^2$ is even. Then $Odd + 2$ results in an even number.

4: $b^2 + 1$

Analysis : If $b$ is even, $b^2$ is even. Then $Even + 1$ results in an odd number.

If $b$ is odd, $b^2$ is odd. Then $odd + 1$ results in an even number.

Question 3

Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.

Solution :

To determine if an algebraic expression is even without calculating the final answer, we can use the rules of parity:

Any integer multiplied by an even number (like 2, 4, or 6) is always even, and the sum of two even numbers is always even.

Expressions that are Always Even

These expressions are guaranteed to be even because they can be simplified into a form that is a multiple of 2.

(i): $4k \times 3j$

Reason: This simplifies to $12kj$. Since 12 is an even number, the product of 12 and any other integers will always be even.

Example: If $k = 1, j = 1$, $12 \times 1 \times 1 = 12$ (Even).

Example: If $k = 2, j = 3$, $12 \times 2 \times 3 = 72$ (Even).

Expressions that are NOT Always Even

These expressions depend on the specific values of the variables and will not be even for all integers:

(i): $3g + 5h$

Reason: This is a sum of two terms that can both be odd. If the total sum is $Odd + Even$, the result is odd.

Non-Example: If $g = 1$ and $h = 2$, $3(1) + 5(2) = 3 + 10 = 13$ (Odd).

Example: If $g = 2$ and $h = 2$, $3(2) + 5(2) = 6 + 10 = 16$ (Even).

Question 4

Write a few algebraic expressions which always give an even number.

Solution :

To create algebraic expressions that are always even, the fundamental rule is that the expression must be reducible to a form that is a multiple of 2 (i.e., $2 \times \text{integer}$).

1. Simple Multiples of Even Numbers

Any integer multiplied by an even number results in an even number.

$10n$: Since 10 is even, any integer $n$ multiplied by it remains even.

$6k$: This is always even because 6 is a multiple of 2.

2. Sums of Even Terms

If every term in an addition or subtraction expression is even, the final result is guaranteed to be even.

$2a + 4b$: Both $2a$ and $4b$ are even, so their sum is even.

$8x - 2y$: Similar to the sum, subtracting one even number from another always yields an even result.

3. Combining Odds to make Evens

We know that $Odd + Odd = Even$ and $Odd - Odd = Even$. We can use this to build expressions that "force" an even result.

$n^2 + n$: If $n$ is even, $Even + Even = Even$. If $n$ is odd, $Odd + Odd = Even$. Therefore, this expression is always even.

$2k + 2$: Since $2k$ is even and 2 is even, the result is always even.

4. Expressions from the Exercise

Based on our previous analysis, these specific forms are always even regardless of the integer values used for the letters:

$4m + 2n$

$4k \times 3j$ (which simplifies to $12kj$)

Question 1

When will two even numbers add up to give a multiple of 4?

Solution :

Two even numbers add up to a multiple of 4 when both numbers share the same "type" of evenness (either both are multiples of 4, or both are even numbers that are not multiples of 4).

The Two Conditions for a Multiple of 4

1. When both numbers are multiples of 4

add two numbers that are already divisible by 4, their sum will naturally be divisible by 4.

Example: $4 + 8 = 12$ (a multiple of 4)

2. When both numbers are NOT multiples of 4

If you add two even numbers that each have a remainder of 2 when divided by 4, those two "remainders" add up to 4, completing a new multiple.

Example: $2 + 6 = 8$ (a multiple of 4)

Question 2

What happens when we add a multiple of 4 to an even number that is not a multiple of 4?
Is it similar to the case of the parity of the sum of an even and an odd number?

Solution :

When you add a multiple of 4 to an even number that is not a multiple of 4, the result will never be a multiple of 4.

Why

We can look at the structure of these numbers

Multiple of 4: Can be written as $4n$ (e.g., 4, 8, 12).

Even (not multiple of 4): Can be written as $4m + 2$ (e.g., 2, 6, 10).

When you add them together:

$4n + (4m + 2) = 4(n + m) + 2$

is not a multiple of 4

2. Comparison to Even and Odd Parity

Yes, it is very similar! You can think of this as a "higher level" of parity.

Question 1

If 8 exactly divides two numbers separately, it must exactly divide their sum.
Statement 1 is always true. Determine if it is true with subtraction.

Solution :

Yes, Statement 1 is also always true for subtraction. If 8 exactly divides two numbers separately, it must exactly divide their difference.

The Mathematical Reason

This works because of the distributive property of multiplication and division. If 8 divides two numbers, we can write them as multiples of 8:

Let the first number be $8x$ (where $x$ is an integer).

Let the second number be $8y$ (where $y$ is an integer).

When we subtract them

$$8x - 8y = 8(x - y)$$

Since $(x - y)$ is still an integer, the resulting number is clearly a multiple of 8, meaning 8 divides it exactly.

Example :

Take 80 and 24. Both are divisible by 8.

$80 - 24 = 56$. (56 is divisible by 8).

Question 2

If a number is divisible by both 9 and 4, it must be divisible by 36.

Solution :

This statement is always true. When a number is divisible by two different numbers that are coprime (meaning they share no common factors other than 1), it must also be divisible by their product.

Why it works with 9 and 4

To be divisible by 36, a number must contain all the "prime ingredients" of 36.

Let's look at the prime factorization:

$36 = 2 \times 2 \times 3 \times 3$ (or $2^2 \times 3^2$)

Now let's look at your two numbers:

Divisible by 4: This means the number contains $2 \times 2$.

Divisible by 9: This means the number contains $3 \times 3$.

If a number satisfies both conditions, it must contain $2 \times 2$ AND $3 \times 3$. When you combine those factors, you get $36$.

The "Coprime" Rule

This logic only works if the two numbers don't "overlap" in their factors.

Example:

Now, consider the number 144, which is divisible by 9 and 4, and it is also divisible by LCM (9, 4), i.e., 36.

Question 3

If a number is divisible by both 6 and 4, it must be divisible by 24.

Solution :

This statement is not always true. When a number is divisible by two different numbers that are coprime (meaning they share no common factors other than 1), it must also be divisible by their product.

Why it fails for 6 and 4

Now let's look at your two numbers:

Divisible by 4: This means the number contains $2 \times 2$.

Divisible by 6: This means the number contains $2 \times 3$.

Because they both numbers 6 and 4 share a common factor: 2 , a number doesn't need to be as large as 24 ($6 \times 4$) to be divisible by both. It only needs to be divisible by their Least Common Multiple (LCM).

The "Coprime" Rule

This logic only works if the two numbers don't "overlap" in their factors.If a number is divisible by both 6 and 4, it must be divisible by their LCM, which is 12.

Example:

The smallest number divisible by both 6 and 4 is 12.

Since 12 is divisible by both but not by 24, the statement is false. Other examples include 36, 60, and 84.

Question 4

Find a number that has a remainder of 3 when divided by 5. Write more such numbers.

Solution :

To find numbers that leave a remainder of 3 when divided by 5, you can use the general formula $5n + 3$, where $n$ is any whole number (0, 1, 2, 3...).

Examples :

8: $5 \times 1 = 5$, and $5 + 3 = 8$

13: $5 \times 2 = 10$, and $10 + 3 = 13$

18: $5 \times 3 = 15$, and $15 + 3 = 18$

$$3, 8, 13, 18, 23, 28, 33, 38, 43, \dots$$

Connection to Parity and Evenness

They always end in either 3 or 8.

Numbers ending in 8 are always even, while those ending in 3 are always odd.

Question 1

The sum of four consecutive numbers is 34. What are these numbers?

Solution :

Let the first number be $x$. Since the numbers are consecutive, they are:

x, (x + 1), (x + 2) and (x + 3).

The sum of these numbers is 34:

$x + (x + 1) + (x + 2) + (x + 3) = 34$

$4x + 6 = 34$

$4x = 28$

$x = 7$

Now that we know $x = 7$, we can list the four numbers:

The Numbers

7, 8, 9, and 10

Observations

The sum of any four consecutive numbers is always an even number

Note that in our sequence (7, 8, 9, 10), we have two odd numbers (7 and 9) and two even numbers (8 and 10).

As we discussed, Odd + Odd = Even and Even + Even = Even, so the total must be even.

Question 2

Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

Solution :

If $p$ is the greatest of five consecutive numbers, the sequence of numbers is built by counting backward from $p$

The Five Numbers in terms of $p$

Since $p$ is the largest, each preceding number is 1 less than the one following it:

The Greatest: $p$

The 2nd Number: $p - 1$

The 3rd Number: $p - 2$

The 4th Number: $p - 3$

The 5th (Smallest): $p - 4$

∴ p > (p -1) > (p – 2) > (p – 3) > (p – 4).

Question 3 (i)

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(i) The sum of two even numbers is a multiple of 3.

Solution :

(i) The sum of two even numbers is a multiple of 3.

Conclusion: Sometimes true.

Explanation:

The sum of two even numbers is always even, but being even does not guarantee divisibility by 3.

A number must be a multiple of both 2 and 3 (a multiple of 6) for this to work.

Example: $2 + 4 = 6$. (6 is a multiple of 3).

Non-example: $2 + 2 = 4$. (4 is not a multiple of 3).

Algebraic Justification:

Let the numbers be $2m$ and $2n$. The sum is $2(m + n)$.

For this to be a multiple of 3, the term $(m + n)$ must itself be a multiple of 3.

This only happens for certain values of $m$ and $n$.

Question 3 (ii)

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(ii) If a number is not divisible by 18, then it is also not divisible by 9.

Solution :

(ii) If a number is not divisible by 18, then it is also not divisible by 9.

Conclusion: Never True (as a general rule) / False

Explanation:

A number can be a multiple of 9 without reaching the "size" of 18 or being a multiple of 18 (specifically, the odd multiples of 9).

Example: 9 is not divisible by 18, but it is divisible by 9.

Non-example: 7 is not divisible by 18 and also not divisible by 9.

Algebraic Justification:

A number $x$ is divisible by 9 if $x = 9k$.

A number is divisible by 18 if $x = 18k = 9(2k)$.

Every number divisible by 18 must be divisible by 9, but a number divisible by 9 need not be divisible by 18.

Question 3 (iii)

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.

Solution :

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.

Conclusion: The statement is not always true.

Explanation:

When a number is not divisible by 6, it must leave a remainder of 1, 2, 3, 4, or 5.

It fails whenever the remainders of the two numbers add up to 6.
This is similar to the parity logic where adding two odd numbers (remainders of 1 when divided by 2) results in an even number.

Example: 1 and 2 are not divisible by 6.and their sum $1 + 2 = 3$ (not divisible by 6).

Non-example: 2 and 4 are not divisible by 6. and their sum $2 + 4 = 6$ (is divisible by 6).

Algebraic Justification:

A number divisible by 6 must be divisible by both 2 and 3.

A number not divisible by 6 may still be even or divisible by 3.

Question 3 (iv)

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

Solution :

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

Conclusion: Always True

Explanation:

Both 6 and 9 are themselves multiples of 3. Any combination of them will still be built out of "blocks" of 3.

Example: $6 + 9 = 15$. (15 is a multiple of 3).

Non-example: $12 + 18 = 30$. (30 is a multiple of 3).

Algebraic Justification:

A multiple of 6 is $6n$ and a multiple of 9 is $9m$.

$$6n + 9m = 3(2n + 3m)$$

Because the entire expression can be factored by 3, the sum is always a multiple of 3.

Question 3 (v)

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Solution :

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Conclusion: Sometimes true.

Explanation:

While both are multiples of 3, they don't always "add up" to a 9.

Example: $6 + 12 = 18$. (18 is a multiple of 9).

Non-example: Let's try $12 + 3 = 15$. (15 is not a multiple of 9).

Algebraic Justification:

Let the sum be $6n + 3m$.

This simplifies to $3(2n + m)$. For this to be a multiple of 9, the term $(2n + m)$ must be a multiple of 3.

This is not true for all integers $n$ and $m$.

Question 4

Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Solution :

(I) Here, Remainder = 2, Dividend = 3

∴ Number = (Quotient × Dividend) + Remainder = (n × 3) + 2

where, n = 1, 2, 3,…..

We can list the numbers that satisfy each condition

Remainder of 2 when divided by 3 ($3n + 2$):

2, 5, 8, 11, 14, 17, 20, 23, 26, 29...

(II) Here, Remainder = 2, Dividend = 4

∴ Number = (Quotient × Dividend) + Remainder = (m × 4) + 2

where, m = 1, 2, 3,…..

We can list the numbers that satisfy each condition

Remainder of 2 when divided by 4 ($4m + 2$):

2, 6, 10, 14, 18, 22, 26, 30...

The first few numbers that satisfy both conditions are 2, 14, and 26.

Notice that the numbers (2, 14, 26) increase by 12 each time.
This is because 12 is the Least Common Multiple (LCM) of 3 and 4.

The algebraic expression for all such numbers $x$ is:

$$x = 12k + 2$$

if a number is divisible by both 3 and 4, it must be divisible by 12.
Therefore, if a number is "2 more" than a multiple of both 3 and 4, it must be "2 more" than a multiple of 12.

Question 5

“I hold some pebbles, not too many, When I group them in 3’s, one stays with me.
Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around,

But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Solution :

The Conditions

let the number of pebbles be $x$:

Group by 3, one remains: $x \div 3$ leaves a remainder of 1.

Pairing them up leaves one: $x$ is an odd number ($x \div 2$ leaves a remainder of 1).

Group by 5, one remains: $x \div 5$ leaves a remainder of 1.

Group by 7, perfection: $x$ is a multiple of 7 (no remainder).

Not too many / Less than 100: $x < 100$.

We are looking for a number that leaves a remainder of 1 when divided by both 3 and 5.

this means the number must be 1 more than a multiple of their Least Common Multiple (LCM).

The LCM of 3 and 5 is 15.

The possible numbers are: $15k + 1$.

List: 16, 31, 46, 61, 76, 91...

From the list above, we must keep only the odd numbers (because "pairing them up simply won't do"):

List: 31, 61, 91...

Check for Divisibility by 7 ("grouping by seven, perfection is found"):

$31 \div 7 = 4$ remainder 3 (No)

$61 \div 7 = 8$ remainder 5 (No)

$91 \div 7 = 13$ remainder 0 (Yes!)

Answer : The number of pebbles you hold is 91.

Question 6

Tathagat has written several numbers that leave a remainder of 2 when divided by 6.
He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?

Solution :

Our algebraic breakdown

A number leaving a remainder of 2 when divided by 6 is written as $6k + 2$.

When you add three such numbers ($6a + 2$), ($6b + 2$), and ($6c + 2$), the constants ($2 + 2 + 2$) combine to make 6.

= ($6a + 2$)+($6b + 2$)+ ($6c + 2$)

= $6a + 6b + 6c + 6$

= $6(a + b + c + 1)$

Since the entire expression can be factored as $6(a + b + c + 1)$, it is always a multiple of 6.

Such numbers are of the form 6k + 2: 8, 14, 20, 26,.....

Example:

Take 2, 8, 14 → sum = 24 → divisible by 6.

Take 20, 26, 32 → sum = 78 → divisible by 6.

So, Tathagat’s claim is true.: the sum of any three numbers of the form 6k+2 is divisible by 6.

Question 7

When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.

(i) 4779 + 661 (ii) 4779 – 661

Solution :

Yes, we can determine the remainders without calculating the full values by using the properties of remainders

The core principle is that the remainder of a sum or difference is simply the sum or difference of the individual remainders.

(i) $4779 + 661$

Algebraic Solution

Let Numbers 4779 and 661 can be express based on their quotients ($a, b$) and remainders ($r$) when divided by 7:

$ 4779$ = $ = 7a + 5$

$ 661$ = $ = 7b + 3$

Now, add them together:

$$= 7(a + b) + 8$$

Since 8 is larger than the divisor 7, we divide 8 by 7:

$8 = 7(1) + 1$.

The total expression becomes: $7(a + b + 1) + 1$. The remainder is 1.

(i) Visualization Method:

4779 + 661 = (682 × 7) + 5 + (94 × 7) + 3

= 7 × (682 + 94) + 5 + 3

= 7 × 776 + 8

= Divisible by 7 + 8

= 1, Remainder

(ii) 4779 – 661

Algebraic Solution

Let Numbers 4779 and 661 can be express based on their quotients ($a, b$) and remainders ($r$) when divided by 7:

$ 4779$ = $ = 7a + 5$

$ 661$ = $ = 7b + 3$

Now, subtract them together:

$$= 7(a - b) + 2$$

Since 2 is smaller than 7 and positive, it is our final remainder. The remainder is 2.

(ii) Visualization Method:

4779 - 661 = (682 × 7) + 5 - (94 × 7) + 3

= 7 × (682 - 94) + 5 - 3

= 7 × 588 + 2

= Divisible by 7 + 2

= 2, Remainder

Question 8

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Solution :

To solve this, we can look for a pattern in the relationship between the divisors and the remainders.

The expression of a number that leaves a remainder of 2 when divided by 3.

Number = 3K + 2 (This is $3 - 1$)

The expression of a number that leaves a remainder of 3 when divided by 4.

Number = 4K + 3 (This is $4 - 1$)

The expression of a number that leaves a remainder of 4 when divided by 5.

Number = 5K + 4 (This is $5 - 1$)

In every single case, the number is exactly 1 short of being a multiple of the divisor.

Since the number is "1 short" of being a multiple of 3, 4, and 5 simultaneously, we first find the Least Common Multiple (LCM) of 3, 4, and 5.

$$LCM(3, 4, 5) = 3 \times 4 \times 5 = 60$$

The number 60 is the smallest number that is exactly divisible by 3, 4, and 5.

To satisfy our "remainder" conditions, we simply subtract that 1 "stubborn" missing piece:

$$60 - 1 = 59$$

The smallest such number is 59.

59 is the smallest number that leaves a remainder of 2 when divided by 3, a remainder of 3, when divided by 4, and a remainder of 4 when divided by 5.

Question 1

Find, without dividing, whether the following numbers are divisible by 9.

(i) 123
(ii) 405
(iii) 8888
(iv) 93547
(v) 358095

Solution :

To determine if a number is divisible by 9 without performing division, we use the Sum of Digits Rule.

A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

(i) 123

Sum of digits: $1 + 2 + 3 = 6$

Since 6 is not divisible by 9.

Therefore, 123 is not divisible by 9.

(ii) 405

Sum of digits: $4 + 0 + 5 = 9$

Since 9 is divisible by 9.

Therefore, 405 is divisible by 9.

(iii) 8888

Sum of digits: $8 + 8 + 8 + 8 = 32$

$3 + 2 = 5$ , Since 5 is not divisible by 9

Therefore, 8888 is not divisible by 9.

(iv) 93547

Sum of digits: $9 + 3 + 5 + 4 + 7 = 28$

$2 + 8 = 10 \rightarrow 1 + 0 = 1$ , Since 1 is not divisible by 9.

Therefore, 93547 is not divisible by 9.

(v) 358095

Sum of digits: $3 + 5 + 8 + 0 + 9 + 5 = 30$

$3 + 0 = 3$ , Since 3 is not divisible by 9.

Therefore, 358095 is not divisible by 9.

Question 2

Find the smallest multiple of 9 with no odd digits.

Solution :

To find the smallest multiple of 9 with no odd digits, we need to look for a number where every digit is from the set $\{0, 2, 4, 6, 8\}$

and the sum of those digits is a multiple of 9.

Multiples of 9 = 9, 18, 27, 36,…, 288,………

The smallest multiple of 9 with an odd digit is 9.

The smallest multiple of 9 that can be formed by summing even digits is 18 (since 9 is odd).

Here, 18 is even, so it is possible to reach this sum using only even digits.

Determine the Number of Digits

To find the smallest number, we want the fewest number of digits possible.

1-digit numbers:

The largest even digit is 8. (Sum cannot be 18).

2-digit numbers:

The largest possible sum is $8 + 8 = 16$. (Still less than 18).

3-digit numbers:

We can reach a sum of 18 with three digits.

To keep the number small, we want the hundreds digit to be as small as possible.

If the first digit is 2, the remaining two digits must sum to $16$ ($18 - 2 = 16$).

The only even digits that sum to 16 are $8 + 8$.

This gives us the number 288.

Therefore, The smallest multiple of 9 with no odd digits is 288.

Question 3

Find the multiple of 9 that is closest to the number 6000.

Solution :

For a number to be a multiple of 9, its digits must sum to a multiple of 9.

Given, 6000

Sum of the digits = 6 + 0 + 0 + 0 = 6

So it’s not a multiple of 9. Adding 3 gives 6003, which is a multiple of 9.

For 6003, the sum is $6 + 0 + 0 + 3 = 9$.

So, 6003 is the multiple of 9 that is closest to the number 6000.

Question 4

How many multiples of 9 are there between the numbers 4300 and 4400?

Solution :

The "Gap" Method

Find the total range: Subtract the smaller number from the larger number to find the distance between them.

$$4400 - 4300 = 100$$

Divide the gap by the divisor:

100 = (9 × 11) + 1

This calculation shows that there are 11 complete multiples of 9 within a span of 100 units.

There are 11 multiples of 9 between 4300 and 4400.

Question 1

Between the numbers 600 and 700, which numbers have the digital root:

(i) 5
(ii) 7
(iii) 3

Solution :

We first need to understand that the digital root of a number is the same as the remainder when that number is divided by 9. (If the remainder is 0, the digital root is 9).

(i) Digital Root: 5

If 600 has a digital root of 6 ( $6 + 0 + 0 = 6$), then 599 has a digital root of 5 (too small). The next one is $599 + 9 = \mathbf{608}$.

608 = ($6 + 0 + 8 = 1 + 4 \rightarrow 5$)

617 = ($6 + 1 + 7 = 1 + 4 \rightarrow 5$)

Numbers: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698.

(ii) Digital Root: 7

If 600 has a digital root of 6, then the very next number, 601, has a digital root of 7.

601 = ($6 + 0 + 1 \rightarrow 7$)

610 = ($6 + 1 + 0 \rightarrow 7$)

Numbers: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691, 700.

(iii) Digital Root: 3

If 600 has a digital root of 6 ( $6 + 0 + 0 = 6$), then 606 has a digital root of 3 (too small).

606 = ($6 + 0 + 6 = 1 + 2 \rightarrow 3$)

615 = ($6 + 1 + 5 = 1 + 2 \rightarrow 3$)

Numbers: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696.

Question 2

Write the digital roots of any 12 consecutive numbers. What do you observe?

Solution :

Let's look at the digital roots for any 12 consecutive numbers. We can start from any point, for example, starting with the number 15.

Digital Roots of 12 Consecutive Numbers

15 = $1 + 5 = 6$

16 = $1 + 6 = 7$

17 = $1 + 7 = 8$

18 = $1 + 8 = 9$

19 = $1 + 9 = 10 \rightarrow 1+0$ = 1

20 = $2 + 0 = 2$

21 = $2 + 1 = 3$

22 = $2 + 2 = 4$

23 = $2 + 3 = 5$

24 = $2 + 4 = 6$

25 = $2 + 5 = 7$

26 = $2 + 6 = 8$

Observations :

1. Repeating Cycle:

The digital roots follow a repeating cycle from 1 to 9. Once the root hits 9, it resets to 1.

6 → 7 → 8 → 9 → 1 → 2 → 3 → 4 → 5 → 6 → 7 → 8

2. The "Plus One" Pattern:

Adding 1 to a number also adds 1 to its digital root (except when the root is 9, where it loops back to 1).

3. Connection to 9

This sequence confirms that digital roots behave exactly like remainders when divided by 9.

Question 3

We saw that the digital root of multiples by 9 is always 9. Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.

Solution :

When we find the digital roots of multiples of other numbers, we see that they also form predictable, repeating patterns.

(i) Multiples of 3

3 = 3

6 = 6

9 = 9

12 = $1 + 2 = 3$

15 = $1 + 5 = 6$

18 = $1 + 8 = 9$

21 = $2 + 1 = 3$

24 = $2 + 4 = 6$

Observations :

The digital roots follow a simple 3-step cycle: 3, 6, 9. This cycle repeats forever.

Therefore, the digital roots of consecutive multiples of 3 are 3, 6, 9, 3, 6, 9,...

(ii) Multiples of 4

4 = 4

8 = 8

12 = $1 + 2 = 3$

16 = $1 + 6 = 7$

20 = $2 + 0 = 2$

24 = $2 + 4 = 6$

28 = $2 + 8 = 1 + 0 = 1$

32 = $3 + 2 = 5$

36 = $3 + 6 = 9$

40 = $4 + 0 = 4$

Observations :

The digital roots follow a 9-step cycle: 4, 8, 3, 7, 2, 6, 1, 5, 9. After nine multiples, the pattern resets (notice the 40 returns to a digital root of 4).

Therefore, the digital roots of consecutive multiples of 4 are 4, 8, 3, 7, 2,6 ,1 , 5, 9 .....

(i) Multiples of 6

6 = 6

12 = $1 + 2 = 3$

18 = $1 + 8 = 9$

24 = $2 + 4 = 6$

30 = $3 + 0 = 3$

36 = $3 + 6 = 9$

Observations :

The digital roots follow a 3-step cycle: 6, 3, 9. Since 6 is a multiple of 3, its roots are restricted to the same $\{3, 6, 9\}$

Thus, the digital roots of consecutive multiples of 6 are 3, 9, 6, 3, 9, 6, ....

Question 4

What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice? Try to explain the patterns noticed.

Solution :

To find the digital roots of numbers that are 1 more than a multiple of 6 (which can be written algebraically as $6n + 1$).

The Digital Roots of $6n + 1$

Multiple of 6	no. $6n + 1$	Calculation	Digital Root
0	1	1	1
6	7	7	7
12	13	$1 + 3$	4
18	19	$1 + 9 = 10 \rightarrow 1+0$	1
24	25	$2 + 5$	7
30	31	$3 + 1$	4
36	37	$3 + 7 = 10 \rightarrow 1+0$	1
42	43	$4 + 3$	7
48	49	$4 + 9 = 13 \rightarrow 1+3$	4

The Repeating Cycle: The digital roots follow a specific 3-step repeating pattern: 1, 7, 4.

Relationship to 3: All of these roots (1, 7, 4) leave a remainder of 1 when divided by 3.

Question 5

I’m made of digits, each tiniest and odd, No shared ground with root #1 – how odd! My digits count, their sum, my root – All point to one bold number’s pursuit – The largest odd single-digit I proudly claim. What’s my number? What’s my name?

Solution :

Since the digits are all 1, the sum of the digits is simply the number of digits.

If the sum is 9, there must be nine 1s.

Let's check the digital root: $1+1+1+1+1+1+1+1+1 = 9$.

As we've seen, a sum of 9 results in a digital root of 9.

The number is 111,111,111 (One hundred eleven million, one hundred eleven thousand, one hundred eleven).

All digits odd; 9 digits; sum = 9; digital root = 9. All conditions are met.

Question 1

The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?

Solution :

The digital root of a sum is equal to the digital root of the sum of the individual digital roots.

Digital root of the given number: 5.

Digital root of 10: $1 + 0 = \mathbf{1}$.

To find the digital root of "10 more" than your number, we can use the same property of addition

Sum of the digital roots: $5 + 1 = \mathbf{6}$.

The digital root of 10 more than that number is 6.

Example for Verification :

Let’s pick an easy 8-digit number with a digital root of 5: 10000004

Digital root: $1+0+0+0+0+0+0+4 = 5$.

Now Add 10:

$10,000,004 + 10 = \mathbf{10,000,014}$.

New digital root: $1+0+0+0+0+0+1+4 = \mathbf{6}$.

Question 2

Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

Solution :

The digital root of a sum is equal to the digital root of the sum of the individual digital roots.

Let's start with the number 7.

adding 11 :

The sequence of numbers by repeatedly adding 11 are

7 , 18 ($7 + 11$), 29 ($18 + 11$), 40 ($29 + 11$), 51 ($40 + 11$), 62 ($51 + 11$), 73 ($62 + 11$), 84 ($73 + 11$), 95 ($84 + 11$),106 ($95 + 11$)

The sequence of Digital Roots by repeatedly adding 11 are :

7, 9, 2, 4, 6, 8, 1, 3, 5, 7

Observations

1. A Full Nine-Step Cycle : The digital roots are repeating after 9 steps.

2. The "Plus 2" Pattern Notice the sequence of digital roots: 7, 9, 2, 4, 6, 8, 1, 3, 5, 7... Every time you add 11 to the number, you add 2 to the digital root.

Question 3

What will be the digital root of the number 9a + 36b + 13?

Solution :

To find the digital root of the expression $9a + 36b + 13$, we can simplify each part of the expression by its own digital root

Analyze the "Multiples of 9"

$9a$: Since this is $9 \times a$, it is always a multiple of 9.

So,

Digital root of 9a = 9.

$36b$: Since 36 is a multiple of 9 ($9 \times 4$), any number multiplied by 36 is also a multiple of 9. Therefore, $36b$ also has a digital root of 9.

So,

Digital root of 36b = 9.

$13$: The digital root of 13 is $1 + 3 = \mathbf{4}$.

Combine the Roots

$$\text{Root of } (9a) + \text{Root of } (36b)$$ $$ + \text{Root of } (13)$$

$$9 + 9 + 4 = 22$$

Finally, we reduce 22 to a single digit:

$$2 + 2 = \mathbf{4}$$

The digital root of $9a + 36b + 13$ is 4.

Question 4

Make conjectures by examining if there are any patterns or relations between

(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.

Solution :

let’s look at a sequence of numbers and compare their properties.

(i) Parity and Digital Root

Parity refers to whether a number is even or odd. Let’s list some numbers and their digital roots (DR).

Number :1 , Parity : Odd, Digital Root : 1

Number :2 , Parity : Even, Digital Root : 2

Number :3 , Parity : Odd, Digital Root : 3

Number :10 , Parity : Even, Digital Root : 1

Number :11 , Parity : Odd, Digital Root : 2

Number :12 , Parity : Even, Digital Root : 3

Observation:

Conjecture: Parity and digital roots are independent in terms of value.

However, there is no fixed rule that all even numbers have certain digital roots and all odd numbers have others.

Both even and odd numbers can have any digital root from 1 to 9.

(ii) Digital Root and Remainders (3 and 9)

Relation with Remainder by 9

If a number has a Digital Root of $r$, then the remainder when divided by 9 is also $r$.

Exception: If the digital root is 9, the remainder is 0 (because the number is exactly divisible by 9).

Conjecture: The digital root of a number is simply its remainder when divided by 9, with 9 replacing 0 for exact multiples.

Relation with Remainder by 3

DR is 3, 6, or 9: The number is divisible by 3 (Remainder = 0).

DR is 1, 4, or 7: The number has a remainder of 1 when divided by 3.

DR is 2, 5, or 8: The number has a remainder of 2 when divided by 3.

Conjecture: The remainder of a number when divided by 3 is the same as the digital root of its digital root when divided by 3.

Question 1

If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.

Solution :

To find the value of the digit $z$ in the number $31z5$, we use the Divisibility Rule for 9.

A number is divisible by 9 if the sum of its digits is a multiple of 9.

$\text{Sum of digits} = 3 + 1 + z + 5 = 9 + z$

Solve for $z$

For the number to be a multiple of 9, the expression $(9 + z)$ must be equal to 9, 18, 27, and so on. Since $z$ is a single digit, it must be between 0 and 9.

If $9 + z = 9$:

$$z = 0$$

If $9 + z = 18$:

$$z = 9$$

If $9 + z = 27$:

$$z = 18$$

This is not possible because $z$ must be a single digit.

The two possible values for $z$ are 0 and 9.

Why are there two answers?

The Base Sum of the known digits ($3 + 1 + 5$) is 9, which is already perfectly divisible by 9.

Because the "gap" between multiples of 9 is exactly 9, and our available digits range from 0 to 9, we are able to hit two different multiples of 9 using only a single-digit placeholder.

Question 2

“I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.

Solution :

Define the Numbers

First Number : $= 12a + 8$

Second Number : $= 12b - 4$

$\text{Sum} = (12a + 8) + (12b - 4)$

$\text{Sum} = 12a + 12b + (8 - 4)$

$\text{Sum} = 12(a + b) + 4$

Analyze the Result

For the sum to always be a multiple of 8, the expression $12(a + b) + 4$ must be divisible by 8 for any values of $a$ and $b$.

Putting a = 1 and b = 1

$Sum: = 12(1 + 1) + 4$

$Sum: = 24 + 4$

Sum:= $28 $

Is 28 a multiple of 8?

No. $28 \div 8 = 3$ with a remainder of 4.

Now, Putting a = 3 and b = 4

$Sum: = 12(3 + 4) + 4$

$Sum: = 12(7) + 4$

Sum:= $84 + 4 = 88 $

Is 88 a multiple of 8?

Yes. $88 \div 8 = 11$ (multiple of 8).

Conclusion: Snehal's Claim is False

Snehal is incorrect. While the sum of these two numbers will always be a multiple of 4, it is not always a multiple of 8.

Question 3

When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.

Solution :

Define the Pattern

We can generalize this using algebra. Let the two multiples of 3 be $3m$ and $3n$. Their sum is $3(m + n)$.

For $3(m + n)$ to be a multiple of 6, the term $(m + n)$ must be even.

For $3(m + n)$ to be a multiple of 6, it must be divisible by both 2 and 3.

The Different Possible Cases

Case 1: Adding two Even multiples of 3

Example: $6 + 12 = 18$

Since both are already divisible by 6, their sum must be divisible by 6.

Case 2: Adding two Odd multiples of 3

Example: $3 + 9 = 12$

An odd number plus an odd number always equals an even number. Since both are multiples of 3, the sum is a multiple of 3. An even multiple of 3 is always a multiple of 6.

Case 3: Adding one Even and one Odd multiple

Example: $3 + 6 = 9$

An odd number plus an even number always equals an odd number. An odd multiple of 3 can never be a multiple of 6.

The sum of two multiples of 3 is a multiple of 6 if and only if both numbers have the same parity (both are even or both are odd).

If one is even and the other is odd, the sum will only be a multiple of 3, not 6.

Question 4

Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”.

(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?

Solution :

(i) Examining the Conjecture

The rule for divisibility by 9 states that a number is divisible by 9 if and only if the sum of its digits is a multiple of 9.

Take the number 72 .Sum of digits: $7 + 2 = 9$

Reverse the number 27 .Sum of digits: $7 + 7 = 9$

When you reverse a number, you are simply changing the position of the digits, not the digits themselves.

Since the set of digits remains identical, their sum remains exactly the same.

This conjecture is true for any multiple of 9, regardless of how many digits the number has.

(ii) Are other digit shuffles possible?

Yes, other digit shuffles are also possible.

Because the divisibility rule only cares about the sum of the digits, and addition is commutative (meaning $a + b = b + a$), the order of the digits has no impact on divisibility.

Swapping adjacent digits: $189 \rightarrow 198$

Any rearrangement of the digits of a number divisible by 9 will again be divisible by 9, as the sum of the digits does not change.

Question 5

If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

Solution :

For a number to be a multiple of 18, it must satisfy two conditions:

1 : It must be even (divisible by 2).

2 : The sum of its digits must be a multiple of 9.

Find possible values for $b$

Since the number $48a23b$ must be even, the last digit $b$ must be an even digit.

$$b \in \{0, 2, 4, 6, 8\}$$

Apply the Divisibility Rule for 9

The sum of the digits must be a multiple of 9:

$$\text{Sum} = 4 + 8 + a + 2 + 3 + b$$

$$\text{Sum} = 17 + a + b$$

For the number to be a multiple of 9, the digital root of $(17 + a + b)$ must be 9.

List of Possible Pairs $(a, b)$

If $b = 0$, then $a = 1$. (Pair: 1, 0), Resulting Number : 481230

If $b = 2$, then $a = 8$. (Pair: 8, 2), Resulting Number : 488232

If $b = 4$, then $a = 6$. (Pair: 6, 4), Resulting Number : 486234

If $b = 6$, then $a = 4$. (Pair: 4, 6), Resulting Number : 484236

If $b = 8$, then $a = 2$. (Pair: 2, 8), Resulting Number : 482238

Question 6

If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

Solution :

For a number to be a multiple of 44, it must satisfy one condition

1 : it must be divisible by both 4 and 11 (since $4 \times 11 = 44$ and they are co-prime).

Given by question, 3p7q8 is divisible by 44.

Apply the Divisibility Rule for 4

A number is divisible by 4 if its last two digits form a multiple of 4.The last two digits are $q8$

We need to find values of the digit $q$ such that $q8$ is divisible by 4:

If $q=0 \rightarrow 08$ (Yes)

If $q=1 \rightarrow 18$ (No)

If $q=2 \rightarrow 28$ Yes) and ..........

as per define concept :

Possible values for $q$: $\{0, 2, 4, 6, 8\}$

Apply the Divisibility Rule for 11

A number is divisible by 11 if the difference between the sum of digits at odd positions and the sum of digits at even positions is 0 or a multiple of 11.

For the number $3 \ p \ 7 \ q \ 8$:

Sum of odd positions (1st, 3rd, 5th): $3 + 7 + 8 = 18$

Sum of even positions (2nd, 4th): $p + q$

Difference: $18 - (p + q)$

For this to be a multiple of 11:

$18 - (p + q) = 11 \implies p + q = 7$

Find possible values for $p, q$

The solution’s trials yield the following pairs:

If $q = 0 \implies p = 7$ (7, 0)

If $q = 2 \implies p = 5$ (5, 2)

If $q = 4 \implies p = 3$ (3, 4)

If $q = 6 \implies p = 1$ (1, 6)

Question 7

Find three consecutive numbers such that the first number is a multiple of 2,the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?

Solution :

To find these numbers, let’s represent the three consecutive numbers as $n$, $n+1$, and $n+2$.

Find the first occurrence

As per given conditions in the question

Let's test small even numbers for $n$:

If $n = 2$: Sequence is 2, 3, 4.

Find the next occurrence

Let's continue testing even numbers for $n$:

If $n = 14$: Sequence is 14, 15, 16.

The second set of numbers is 14, 15, 16.

Find the pattern

To find how often they occur, we look at the difference between the starting numbers: $14 - 2 = 12$.

For the pattern to repeat, the new number $n$ must satisfy the same divisibility conditions relative to the original $n$.

This involves the Least Common Multiple (LCM).

$LCM(2, 3, 4) = 12$.

The general form: The sequences start at $12k + 2$ (where $k = 0, 1, 2, \dots$).

Thus, the three consecutive numbers are (14, 15, 16)

(26, 27, 28) and(38, 39, 40)

Question 8

Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.

Solution :

For a number to be a multiple of 36, it must satisfy one condition

1 : it must be divisible by both 4 and 9 (since $4 \times 9 = 36$ ).

To find these numbers efficiently, we will use a variation of the "Gap Method", we will find a "starting point" near 45,000 and then jump forward in steps of 36.

Find the first multiple near 45,000

First,

$$45,000 \div 36 = 1,250$$

Since this division results in a whole number (1,250), it means 45,000 is itself a multiple of 36.

Verification

Divisibility Rules for for 4: The last two digits must be a multiple of 4

Last two digits of 45000 = 00, it is divisible by 4.

Divisibility for 9: The digital root must be 9. Let's check our first and last numbers:

45,036: $4 + 5 + 0 + 0 + 0 = 9 \rightarrow \mathbf{9}$ (Correct)

Thus, 45000 is completely divisible by 36.

Generate the sequence

Since we need multiples between 45,000 and 47,000, we simply start from 45,000 and add 36 repeatedly to find the next five values.

$45,000 + 36 = \mathbf{45,036}$

$45,036 + 36 = \mathbf{45,072}$

$45,072 + 36 = \mathbf{45,108}$

$45,108 + 36 = \mathbf{45,144}$

$45,144 + 36 = \mathbf{45,180}$

Question 9

The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.

Solution :

In a sequence of consecutive even numbers, each number is exactly 2 units away from the one next to it.

If the middle number is given, we can find the numbers before it by subtracting 2, and the numbers after it by adding 2.

Step-by-Step Expression

Middle Number (3rd)= $\rightarrow 5p $

Second Number: Subtract 2 from the middle number = $\rightarrow 5p - 2$

First Number: Subtract 2 from the second number = $\rightarrow (5p - 2) - 2 = 5p - 4$

Fourth Number: Add 2 to the middle number = $\rightarrow 5p + 2$

Fifth Number: Add 2 to the fourth number = $\rightarrow (5p + 2) + 2 = 5p + 4$

The Full Sequence

$5p - 4, \quad 5p - 2, \quad 5p, \quad 5p + 2, \quad 5p + 4$

Question 10

Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6

Solution :

To solve this, we need to find a 6-digit number that satisfies the divisibility rules for both 15 and 6 (when reversed).

Conditions for Divisibility by 15

A number is divisible by 15 if it is divisible by 3 and 5.

Divisible by 5: The last digit must be 0 or 5.

Divisible by 3: The sum of the digits must be a multiple of 3.

Conditions for the Reversed Number (Divisible by 6)

When we reverse the digits, the new number must be divisible by 6, meaning it must be divisible by 2 and 3.

Divisible by 2: The new last digit must be even ($0, 2, 4, 6, 8$). This means the first digit of our original number must be even.

Divisible by 3: The sum of the digits must be a multiple of 3. (Note: Since reversing digits doesn't change the sum, if the original is divisible by 3, the reversal is too!)

Finding the Number

Let our original number be $ABCDEF$.

Requirement for $F$: $F$ must be 0 or 5.

Requirement for $A$: $A$ must be even. However, a 6-digit number cannot start with 0, so $A$ must be 2, 4, 6, or 8.

The Sum: $A+B+C+D+E+F$ must be a multiple of 3.

Let's try

if $F = 0$ and $A = 2$

then Number = 210000

Sum = $2 + 1 + 0 + 0 + 0 + 0 = 3$,

210000 fulfills all the above condition, So, 210000 is divisible by 15.

Reversed the digits:

000,012 (or just 12)

One’s place = 2, so number 000,012 is divisible by 2.

Sum of the digits = $0 + 0 + 0 + 0 + 1 + 2 = 3$

It is also divisible by 3

Reversed Number: 000,012 is divisible by 6.

Therefore, One such 6-digit number is 210,000.

Question 11

Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.

Solution :

When we say a number is a "multiple of 11," it means the number can be written in the form:

$$11 \times n$$

The multiples of 11 are: 11, 22, 33, 44, 55.. . ....

If we double that number, we multiply it by 2:

$$2 \times (11 \times n)$$

When doubled, 22, 44, 66, 88, 110,....

Using the associative property of multiplication, we can rewrite this as:

$$11 \times (2 \times n)$$

i.e. 11 × 2, 11 × 4, 11 × 6, 11 × 8, 11 × 10,.... are also multiples of 11.

Since $(2 \times n)$ is still a whole number, the result is guaranteed to be a multiple of 11.

Deepak's conjecture is false. All multiples of 11 remain multiples of 11 when doubled (or tripled, or multiplied by any integer).

Question 12

Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning.

(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b – 3) – 4(11b + 1) is a multiple of 12.

Solution :

(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9

Always True,

Reasoning:

A multiple of 6 can be written as $6n$

A multiple of 3 can be written as $3m$.

Their product is $6n \times 3m = 18nm$.

Since 18 is a multiple of 9 ($9 \times 2$), the entire product $18nm$ must be divisible by 9.

(ii) The sum of three consecutive even numbers will be divisible by 6

Always True,

Reasoning:

Let the three consecutive even numbers be $2n$, $2n+2$, and $2n+4$.

Sum: $2n + (2n + 2) + (2n + 4) = 6n + 6$.

We can factor out the 6: = $6(n + 1)$.

Because the total sum can always be expressed as $6 \times (\text{an integer})$, the sum is always a multiple of 6.

(iii) If $abcdef$ is a multiple of 6, then $badcef$ is a multiple of 6.

Always True,

Reasoning:

A number is divisible by 6 if it is divisible by both 2 and 3.

Divisibility by 3: This depends on the sum of digits. Since both numbers have the same digits, their sums are equal.

Divisibility by 2: This depends only on the last digit. For $abcdef$ to be divisible by 2, $f$ must be even. For $badcef$ to be divisible by 2, $f$ must also be even.

Reordering preserves digit-sum (mod 3) and the last digit (even/odd), so divisibility by 6 remains.

(iv) $8(7b - 3) - 4(11b + 1)$ is a multiple of 12.

Never True

Reasoning:

Let's simplify the expression:

$(56b - 24) - (44b + 4)$

$12b - 28$

Now, let's check for divisibility by 12:

$12b$ is always a multiple of 12.

28 is not a multiple of 12 ($28 = 12 \times 2 + 4$).

If $b$ must be an integer .

When you subtract a non-multiple (28) from a multiple ($12b$), the result can never be a multiple of 12.

Question 13

Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.

Solution :

To determine when the sum of any three numbers is divisible by 3, we should look at the remainder each number leaves when divided by 3.

In the world of divisibility by 3, every integer falls into one of three categories:

Type 0: Remainder 0 (Multiples of 3, e.g., 3, 6, 9)

Type 1: Remainder 1 (e.g., 1, 4, 7)

Type 2: Remainder 2 (e.g., 2, 5, 8)

Exploring the Cases

For the sum of three numbers to be divisible by 3, the sum of their remainders must also be divisible by 3

1: All three numbers are of the same type

Three Type 0s: $0 + 0 + 0 = 0$ (Sum is divisible by 3).

Example: $3 + 6 + 9 = 18$.

Three Type 1s: $1 + 1 + 1 = 3$ (Sum is divisible by 3).

Example: $1 + 4 + 7 = 12$.

Three Type 2s: $2 + 2 + 2 = 6$ (Sum is divisible by 3).

Example: $2 + 5 + 8 = 15$.

2: All three numbers are of different types

One Type 0, One Type 1, and One Type 2:

The sum of remainders is $0 + 1 + 2 = 3$. Since 3 is divisible by 3, the total sum will be as well.

Example: $3 + 4 + 5 = 12$.

3: Two are the same, one is different

Two Type 1s and one Type 0: $1 + 1 + 0 = 2$ (Not divisible).

Example: $1 + 4 + 3 = 8$.

Two Type 2s and one Type 1: $2 + 2 + 1 = 5$ (Not divisible).

Example: $2 + 5 + 4 = 11$.

Any "mixed" pair with a third "odd one out":

These will never result in a remainder sum of 0, 3, or 6.

Question 14

Is the product of two consecutive integers always a multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?

Solution :

1. Product of 2 Consecutive Integers Is it always a multiple of 2?

Yes.

The product of two consecutive integers is always a multiple of 2.

In any two consecutive integers, one must be even and one must be odd.

1 × 2 = 2, 2 × 3 = 6, 5 × 6 = 30, 10 × 11 = 110, ....

Since an even number is always a multiple of 2 ($2n$), the product of an even and an odd number will always contain a factor of 2.

2. Product of 3 Consecutive Integers Is it always a multiple of 6?

Yes

For a number to be a multiple of 6, it must be divisible by both 2 and 3.

In any three consecutive numbers, exactly one must be a multiple of 3.

Since the product contains at least one factor of 2 and exactly one factor of 3, it must be divisible by $2 \times 3 = 6$.

Example: $4 \times 5 \times 6 = 120$ (Multiple of 6)

Example: $2 \times 3 \times 4 = 24$ (Multiple of 6)

3: Product of 4 Consecutive Integers

The product of 4 consecutive integers is always a multiple of 24.

Using the same logic, in any four consecutive numbers:

One must be a multiple of 4.

Another must be at least a multiple of 2.

One must be a multiple of 3.

The total product must be divisible by $4 \times 2 \times 3 = 24$.

Example: $3 \times 4 \times 5 \times 6 = 360$ ($24 \times 15$)

4: Product of 5 Consecutive Integers

The product of 5 consecutive integers is always a multiple of 120.

Using the same logic, in any five consecutive numbers:

At least one number is a multiple of 5.

One must be a multiple of 4.

Another must be at least a multiple of 2.

One must be a multiple of 3.

This is equivalent to $5!$ (5 factorial), which is $5 \times 4 \times 3 \times 2 \times 1 = 120$.

Example: 2 × 3 × 4 × 5 × 6 = 720

Question 15

Solve the cryptarithms

(i) EF × E = GGG
(ii) WOW × 5 = MEOW

Solution :

Cryptarithms are mathematical puzzles where each letter represents a unique digit (0–9). Here are the step-by-step solutions for both:

(i) EF × E = GGG

The Logic

GGG :: A three-digit number with identical digits can be written as $G \times 111$

Factorize 111: $111 = 3 \times 37$.

So, $GGG = G \times 3 \times 37$.

Analyze the Equation:

$EF \times E = G \times 3 \times 37$

$E$ is a single digit, so $EF$ must be a multiple of 37.

The Solution:

E = 3 , F = 7, G = 1

Equation: $37 \times 3 = 111$

(ii) WOW × 5 = MEOW

The Logic

Analyze the Last Digit: $W \times 5$

$W \times 5$ ends in $W$

Multiples of 5 end in either 0 or 5.

If $W = 0$, the number $WOW$ would start with 0, which isn't allowed for a 3-digit number.

Therefore, W = 5.

Substitute W: $5O5 \times 5 = MEO5$.

Determine the Range of M:

The smallest $5O5$ is $505 \times 5 = 2525$.

The largest $5O5$ is $595 \times 5 = 2975$.

In both cases, the first two digits start with 2. Thus, M = 2.

Substitute M: $5O5 \times 5 = 2EO5$.

On this logic we Find O and E:

Let's test digits for $O$ in $5O5 \times 5$:

check if O = 1,2,3,4,56,7,8

So, The Solution:

W = 5

O = 7

M = 2

E = 8

Equation: $575 \times 5 = 2875$

Question 16

Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

Solution :

Based on the mathematical relationship between the numbers 4, 8, and 32, the correct Venn diagram is (iv).

Every multiple of 32 is also a multiple of 8 (since $32 = 8 \times 4$), and every multiple of 8 is also a multiple of 4 (since $8 = 4 \times 2$).

In a Venn diagram, if every member of one set belongs to another, the first set must be drawn entirely inside the second.

Diagram (iv) Structure:

Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44...

Multiples of 8 are: 8, 16, 24, 32, 40, 48....

Multiples of 32 are: 32, 64, 96, 128,....

The innermost circle represents Multiples of 32, as it is the smallest, most specific set.

The middle circle represents Multiples of 8, which contains all multiples of 32.

The outermost circle represents Multiples of 4, which contains all multiples of both 8 and 32.