Playing with Numbers — Exercise 4(D)

If 42*8 is a multiple of 9, then the digit represented by * is

1. 0

2. 1

3. 2

4. 4

Answer

According to the divisibility test of 9, a number is divisible by 9 only when the sum of its digits is divisible by 9.

⇒ 42*8 is a multiple of 9

⇒ 4 + 2 + * + 8 is divisible by 9

⇒ 14 + * is divisible by 9

⇒ 14 + * = 18, 27, ...............

⇒ * = 4, 13, ...............

Since * is a digit,

∴ * = 4.

Hence, option 4 is the correct option.

Consider the following statements:

(i) In a given whole number, if the sum of the odd numbered digits is equal to the sum of even numbered digits, then the number is divisible by 11.

(ii) In a given whole number, if the difference of sums of odd numbered digits and even numbered digits is divisible by 11, then the number is divisible by 11.

Of these statements

1. only (i) is correct

2. both (i) and (ii) are correct

3. only (ii) is correct

4. both (i) and (ii) are wrong

Answer

According to the divisibility test of 11, a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

(i) If the sum of the odd numbered digits is equal to the sum of the even numbered digits, then their difference is 0. So, the number is divisible by 11.

∴ Statement (i) is correct.

(ii) If the difference of the sums of the odd numbered digits and the even numbered digits is divisible by 11, then the number is divisible by 11.

∴ Statement (ii) is correct.

Hence, option 2 is the correct option.

The number 12345674 is divisible by

1. 3

2. 5

3. 9

4. 11

Answer

The given number is 12345674.

Sum of its digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 4 = 32, which is divisible neither by 3 nor by 9.

∴ It is divisible neither by 3 nor by 9.

Its units digit is 4, which is neither 0 nor 5.

∴ It is not divisible by 5.

Sum of its digits at odd places = (4 + 6 + 4 + 2) = 16.

Sum of its digits at even places = (7 + 5 + 3 + 1) = 16.

Difference of the above sums = (16 - 16) = 0.

∴ 12345674 is divisible by 11.

Hence, option 4 is the correct option.

What least value must be assigned to * so that the number 91876*2 is divisible by 8?

1. 1

2. 2

3. 3

4. 4

Answer

According to the divisibility test of 8, a number is divisible by 8 only when the number formed by its last three digits is divisible by 8.

Here, the number formed by the last three digits is 6*2.

When * = 1, 612 is not divisible by 8.

When * = 2, 622 is not divisible by 8.

When * = 3, 632 = 8 × 79, which is divisible by 8.

∴ The least value of * is 3.

Hence, option 3 is the correct option.

The number 111,111,111,111 is divisible by

1. 9 and 11

2. 5 and 9

3. 3 and 11

4. 5 and 11

Answer

The given number is 111111111111, which has twelve digits, each equal to 1.

Sum of its digits = 1 × 12 = 12, which is divisible by 3 but not by 9.

∴ It is divisible by 3 but not by 9.

Its units digit is 1, which is neither 0 nor 5.

∴ It is not divisible by 5.

Sum of its digits at odd places = (1 + 1 + 1 + 1 + 1 + 1) = 6.

Sum of its digits at even places = (1 + 1 + 1 + 1 + 1 + 1) = 6.

Difference of the above sums = (6 - 6) = 0.

∴ It is divisible by 11.

So, the given number is divisible by 3 and 11.

Hence, option 3 is the correct option.

What least value must be given to * so that 97215*6 is divisible by 11?

1. 1

2. 2

3. 3

4. 5

Answer

The given number is 97215*6.

Sum of its digits at odd places = (6 + 5 + 2 + 9) = 22.

Sum of its digits at even places = (* + 1 + 7) = 8 + *.

Difference of the above sums = 22 - (8 + *) = 14 - *, which should be either 0 or divisible by 11.

⇒ 14 - * = 11

⇒ * = 3

Hence, option 3 is the correct option.

In a 3-digit number, if the middle digit is equal to the sum of the other two digits, then it is a multiple of

1. 9

2. 10

3. 11

4. None of these

Answer

Let the hundreds digit be $a$, the tens digit be $b$ and the units digit be $c$.

∴ The 3-digit number = $100a + 10b + c$

Given, the middle digit is equal to the sum of the other two digits.

⇒ $b = a + c$

$\Rightarrow \text{The number} = 100a + 10(a + c) + c \\[1em] = 100a + 10a + 10c + c \\[1em] = 110a + 11c \\[1em] = 11(10a + c)$

So, the number is a multiple of 11.

Hence, option 3 is the correct option.

Find the multiple of 11 in the following numbers.

1. 112144

2. 447355

3. 869756

4. 978626

Answer

According to the divisibility test of 11, a number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

1. For 112144, sum of digits at odd places = (4 + 1 + 1) = 6 and sum of digits at even places = (4 + 2 + 1) = 7. Their difference = (7 - 6) = 1, which is not divisible by 11.

2. For 447355, sum of digits at odd places = (5 + 3 + 4) = 12 and sum of digits at even places = (5 + 7 + 4) = 16. Their difference = (16 - 12) = 4, which is not divisible by 11.

3. For 869756, sum of digits at odd places = (6 + 7 + 6) = 19 and sum of digits at even places = (5 + 9 + 8) = 22. Their difference = (22 - 19) = 3, which is not divisible by 11.

4. For 978626, sum of digits at odd places = (6 + 6 + 7) = 19 and sum of digits at even places = (2 + 8 + 9) = 19. Their difference = (19 - 19) = 0.

∴ 978626 is a multiple of 11.

Hence, option 4 is the correct option.

What least value must be given to $n$ so that the number $6135n2$ becomes divisible by 9?

1. 1

2. 3

3. 3

4. 4

Answer

According to the divisibility test of 9, a number is divisible by 9 only when the sum of its digits is divisible by 9.

∵ $6135n2$ is divisible by 9

⇒ $6 + 1 + 3 + 5 + n + 2$ is divisible by 9

⇒ $17 + n$ is divisible by 9

⇒ $17 + n$ = 18, 27, ...............

⇒ $n$ = 18 - 17, 27 - 17, ........

⇒ $n$ = 1, 10, ...............

Since $n$ is a digit, ∴ $n = 1$.

Hence, option 1 is the correct option.

The number 715938* is divisible by 4. The unknown non-zero digit marked as * will be

1. 2

2. 3

3. 4

4. 6

Answer

According to the divisibility test of 4, a number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

Here, the number formed by the last two digits is 8*.

When * = 2, 82 is not divisible by 4.

When * = 3, 83 is not divisible by 4.

When * = 4, 84 = 4 × 21, which is divisible by 4.

When * = 6, 86 is not divisible by 4.

∴ * = 4

Hence, option 3 is the correct option.

Consider the following statements: The numbers 24984, 26784 and 28584 are

(i) divisible by 3

(ii) divisible by 4

(iii) divisible by 9

Which of these are correct?

1. (i) and (ii)

2. (ii) and (iii)

3. (i) and (iii)

4. (i), (ii) and (iii)

Answer

Sum of digits of 24984 = 2 + 4 + 9 + 8 + 4 = 27, which is divisible by both 3 and 9.

Sum of digits of 26784 = 2 + 6 + 7 + 8 + 4 = 27, which is divisible by both 3 and 9.

Sum of digits of 28584 = 2 + 8 + 5 + 8 + 4 = 27, which is divisible by both 3 and 9.

∴ All the three numbers are divisible by 3 and by 9.

Also, the number formed by the last two digits of each of them is 84 = 4 × 21, which is divisible by 4.

∴ All the three numbers are divisible by 4.

So, the statements (i), (ii) and (iii) are all correct.

Hence, option 4 is the correct option.

How many of the following numbers are divisible by 3 but not by 9?

2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276

1. 5

2. 6

3. 7

4. None of these

Answer

A number is divisible by 3 only when the sum of its digits is divisible by 3, and it is divisible by 9 only when the sum of its digits is divisible by 9.

2133 — Sum of digits = 2 + 1 + 3 + 3 = 9, which is divisible by both 3 and 9.

2343 — Sum of digits = 2 + 3 + 4 + 3 = 12, which is divisible by 3 but not by 9.

3474 — Sum of digits = 3 + 4 + 7 + 4 = 18, which is divisible by both 3 and 9.

4131 — Sum of digits = 4 + 1 + 3 + 1 = 9, which is divisible by both 3 and 9.

5286 — Sum of digits = 5 + 2 + 8 + 6 = 21, which is divisible by 3 but not by 9.

5340 — Sum of digits = 5 + 3 + 4 + 0 = 12, which is divisible by 3 but not by 9.

6336 — Sum of digits = 6 + 3 + 3 + 6 = 18, which is divisible by both 3 and 9.

7347 — Sum of digits = 7 + 3 + 4 + 7 = 21, which is divisible by 3 but not by 9.

8115 — Sum of digits = 8 + 1 + 1 + 5 = 15, which is divisible by 3 but not by 9.

9276 — Sum of digits = 9 + 2 + 7 + 6 = 24, which is divisible by 3 but not by 9.

∴ The numbers divisible by 3 but not by 9 are 2343, 5286, 5340, 7347, 8115 and 9276, which are 6 in number.

Hence, option 2 is the correct option.

If the number $357\star25\star$ is divisible by both 3 and 5, then the missing digits in the units place and the thousands place respectively are

1. 0, 2

2. 0, 6

3. 5, 1

4. 5, 4

Answer

In the number $357\star25\star$, the first $\star$ is at the thousands place and the second $\star$ is at the units place.

Let the digit at the thousands place be $t$ and the digit at the units place be $u$.

According to the divisibility test of 5, the number is divisible by 5 only when $u = 0 \text{ or } u = 5$.

According to the divisibility test of 3, the sum of the digits must be divisible by 3.

Sum of the digits = $3 + 5 + 7 + t + 2 + 5 + u = 22 + t + u$

Now, we check the given options :

1. $u = 0, t = 2$ : Sum = 22 + 2 + 0 = 24, which is divisible by 3.

2. $u = 0, t = 6$ : Sum = 22 + 6 + 0 = 28, which is not divisible by 3.

3. $u = 5, t = 1$ : Sum = 22 + 1 + 5 = 28, which is not divisible by 3.

4. $u = 5, t = 4$ : Sum = 22 + 4 + 5 = 31, which is not divisible by 3.

∴ The missing digits in the units place and the thousands place are 0 and 2 respectively.

Hence, option 1 is the correct option.

3 - 3 + 3 - 3 + ....... 101 terms = ?

1. -3

2. 0

3. 3

4. None of these

Answer

In the given expression, the terms are alternately 3 and -3, and the first term is 3.

Since 101 is an odd number, the number of terms equal to 3 is 51 and the number of terms equal to -3 is 50.

∴ Sum = 51 × 3 + 50 × (-3)

= 153 - 150

= 3

Hence, option 3 is the correct option.

If $m, n, p$ be the digits of a number beginning from the left, the number is

1. $mnp$

2. $m + 10n + 100p$

3. $10m + n + 100p$

4. $100m + 10n + p$

Answer

Since $m, n, p$ are the digits of the number beginning from the left, so $m$ is the hundreds digit, $n$ is the tens digit and $p$ is the units digit.

∴ The number = $100 \times m + 10 \times n + p = 100m + 10n + p$

Hence, option 4 is the correct option.

If $x$ and $y$ are both odd numbers, which of the following is an even number?

1. $x + y$

2. $x + y + 1$

3. $xy + 2$

4. $xy$

Answer

The sum of two odd numbers is always an even number.

Since $x$ and $y$ are both odd, so $(x + y)$ is an even number.

For example, if $x = 3 \text{ and } y = 5$, then $x + y = 8$, which is even.

(Also, $x + y + 1 = 9, xy + 2 = 17 \text{ and } xy = 15$ are all odd numbers.)

Hence, option 1 is the correct option.

If $(p - 1)$ is an odd number, what are the two other odd numbers nearest to it?

1. $p + 1, p + 3$

2. $p, p - 2$

3. $p - 3, p + 1$

4. $p - 3, p - 5$

Answer

Given, $(p - 1)$ is an odd number.

Consecutive odd numbers differ by 2.

∴ The odd number just before $(p - 1)$ is $(p - 1) - 2 = p - 3$.

And, the odd number just after $(p - 1)$ is $(p - 1) + 2 = p + 1$.

∴ The two other odd numbers nearest to $(p - 1)$ are $(p - 3) \text{ and } (p + 1)$.

Hence, option 3 is the correct option.

Find the missing number in the following addition problem:

$\begin{matrix} & 8 & 3 & 5 \\ & 4 & * & 8 \\ + & 9 & * & 4 \\ \hline 2 & 2 & * & 7 \\ \hline \end{matrix}$

1. 0

2. 4

3. 6

4. 9

Answer

Given,

$\begin{matrix} & 8 & 3 & 5 \\ & 4 & * & 8 \\ + & 9 & * & 4 \\ \hline 2 & 2 & * & 7 \\ \hline \end{matrix}$

Let the missing number be *.

In the units column, 5 + 8 + 4 = 17. So, 7 is written and 1 is carried over.

In the tens column, 1 (carry over) + 3 + * + * = 4 + 2*, which is a number whose units digit is *.

⇒ 4 + 2* = 10 + *

⇒ * = 6

So, 4 + 2 × 6 = 16, i.e. 6 is written and 1 is carried over.

In the hundreds column, 1 (carry over) + 8 + 4 + 9 = 22, which gives the digits 2 and 2 of the sum, as required.

Checking with * = 6 : 835 + 468 + 964 = 2267.

Hence, option 3 is the correct option.

What number should replace M in the following multiplication problem?

$\begin{matrix} & 3 & \text{M} & 4 \\ \times & & & 4 \\ \hline 1 & 2 & 1 & 6 \\ \hline \end{matrix}$

1. 0

2. 2

3. 4

4. 8

Answer

Given,

$\begin{matrix} & 3 & \text{M} & 4 \\ \times & & & 4 \\ \hline 1 & 2 & 1 & 6 \\ \hline \end{matrix}$

Here, 3M4 × 4 = 1216

⇒ 3M4 = $\dfrac{1216}{4}$ = 304

∴ M = 0

Checking with M = 0 : 304 × 4 = 1216, as required.

Hence, option 1 is the correct option.

Fill in the blanks:

(i) If $84y6$ is exactly divisible by 3, then the least value of $y$ is ......

(ii) If $73m4$ is exactly divisible by 9, then the least value of $m$ is ......

(iii) If $m5n7$ is exactly divisible by 3, then the least value of $(m + n)$ is ......

(iv) If $x4y5z$ is exactly divisible by 9, then the least value of $(x + y + z)$ is ......

(v) If $7912p$ is exactly divisible by 8, then the least value of $p$ is ......

Answer

(i) Sum of digits of $84y6$ = $8 + 4 + y + 6 = 18 + y$, which is divisible by 3 when $y$ = 0, 3, 6 or 9.

If $84y6$ is exactly divisible by 3, then the least value of $y$ is 0.

(ii) Sum of digits of $73m4$ = $7 + 3 + m + 4 = 14 + m$, which is divisible by 9 when $14 + m = 18$, i.e. $m = 4$.

If $73m4$ is exactly divisible by 9, then the least value of $m$ is 4.

(iii) Sum of digits of $m5n7$ = $m + 5 + n + 7 = 12 + (m + n)$, which is divisible by 3 when $(m + n)$ = 0, 3, 6, ...............

Since $m$ is the thousands digit of a 4-digit number, so $m \neq 0$ and hence $(m + n) \neq 0$.

If $m5n7$ is exactly divisible by 3, then the least value of $(m + n)$ is 3.

(iv) Sum of digits of $x4y5z$ = $x + 4 + y + 5 + z = 9 + (x + y + z)$, which is divisible by 9 when $(x + y + z)$ = 0, 9, 18, ...............

Since $x$ is the ten thousands digit of a 5-digit number, so $x \neq 0$ and hence $(x + y + z) \neq 0$.

If $x4y5z$ is exactly divisible by 9, then the least value of $(x + y + z)$ is 9.

(v) The number formed by the last three digits of $7912p$ is $12p$, which is divisible by 8 when $12p = 120 \text{ or } 128$, i.e. $p = 0 \text{ or } 8$.

If $7912p$ is exactly divisible by 8, then the least value of $p$ is 0.

Assertion (A): The 5-digit number 82592 is divisible by 4 since 92 is divisible by 4.

Reason (R): A 5-digit number $abcde$ is divisible by 4 if $de$ is divisible by 4.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

According to the divisibility test of 4, a number is divisible by 4 only when the number formed by its last two digits is divisible by 4.

So, a 5-digit number $abcde$ is divisible by 4 if $de$ is divisible by 4.

∴ Reason (R) is true.

For the number 82592, the number formed by its last two digits = 92 = 4 × 23, which is divisible by 4.

∴ 82592 is divisible by 4, i.e. Assertion (A) is true.

Also, Reason (R) is exactly the rule used in Assertion (A).

∴ Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Assertion (A): The number $7N + 1$ will leave remainder 1, when divided by 7.

Reason (R): A number divisible by both 2 and 7 is also divisible by 14.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

On dividing $7N + 1$ by 7, we get quotient = $N$ and remainder = 1, since $7N$ is a multiple of 7.

∴ Assertion (A) is true.

Since 2 and 7 are co-prime numbers and 2 × 7 = 14, a number divisible by both 2 and 7 is also divisible by 14.

∴ Reason (R) is true.

But Reason (R) talks about divisibility by 14, while Assertion (A) is about the remainder left on dividing by 7.

∴ Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

On dividing the sum of 472, 724 and 247, by 111, the quotient is:

1. an even number

2. an odd number

3. a prime number

4. both (b) and (c)

Answer

The numbers 472, 724 and 247 are obtained from one another by changing the order of the digits 4, 7 and 2 cyclically.

∴ Sum = 472 + 724 + 247 = 1443

Now, 1443 ÷ 111 = 13

∴ Quotient = 13

Here, 13 is an odd number and it is also a prime number, since its only factors are 1 and 13.

Hence, option 4 is the correct option.

Which of the following is the greatest?

1. 333

2. 3³³

3. 33³

4. 3³³

Answer

We have,

333 is a 3-digit number.

33³ = 33 × 33 × 33 = 35937

3³³ = 3²⁷

Now, 3²⁷ > 3¹⁰ = 59049 > 35937 > 333

∴ 3²⁷ is greater than 333 and 33³.

Also, 3³³ and 3²⁷ have the same base 3 and 33 > 27.

∴ 3³³ > 3²⁷

So, 3³³ is the greatest.

Hence, option 2 is the correct option.

When a number is divided by 91, the remainder is 43. On dividing the same number by 13, the remainder will be:

1. 1

2. 2

3. 4

4. 11

Answer

Let the number be $N$ and the quotient be $k$.

Given, on dividing $N$ by 91 the remainder is 43.

∴ $N = 91k + 43$

Now, 91 = 13 × 7 and 43 = 13 × 3 + 4

$\Rightarrow N = 13 \times 7 \times k + 13 \times 3 + 4 \\[1em] = 13(7k + 3) + 4$

So, on dividing $N$ by 13, the remainder is 4.

Hence, option 3 is the correct option.

$\left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) \dots \left(1 - \dfrac{1}{25}\right)$ is equal to:

1. $\dfrac{1}{25}$

2. $\dfrac{2}{25}$

3. $\dfrac{3}{25}$

4. $\dfrac{25}{2}$

Answer

Solving,

$\Rightarrow \left(1 - \dfrac{1}{3}\right)\left(1 - \dfrac{1}{4}\right)\left(1 - \dfrac{1}{5}\right) \dots \left(1 - \dfrac{1}{25}\right) \\[1em] \Rightarrow \dfrac{2}{\cancel{3}} \times \dfrac{\cancel{3}}{\cancel{4}} \times \dfrac{\cancel{4}}{\cancel{5}} \times \dots \times \dfrac{\cancel{24}}{25} \\[1em] \Rightarrow \dfrac{2}{25}$

Hence, option 2 is the correct option.

The units digit in the sum (124)¹²⁴ + (125)¹²⁵ is:

1. 1

2. 2

3. 3

4. 9

Answer

The units digit of (124)¹²⁴ is the same as the units digit of 4¹²⁴.

The units digits of the powers of 4 are 4, 6, 4, 6, ............... i.e. the units digit is 4 when the power is odd and 6 when the power is even.

Since 124 is an even number, the units digit of (124)¹²⁴ is 6.

The units digit of (125)¹²⁵ is the same as the units digit of 5¹²⁵, and every power of 5 ends in 5.

∴ The units digit of (125)¹²⁵ is 5.

∴ Units digit of the sum = units digit of (6 + 5) = units digit of 11 = 1

Hence, option 1 is the correct option.

Numbers that read the same backwards and forwards are called palindromes; e.g., 12321 is a palindrome. How many palindromes are there from 100 to 500?

1. 40

2. 50

3. 70

4. 100

Answer

Every number from 100 to 500 is a 3-digit number.

Let a 3-digit palindrome be $aba$, where $a$ is the hundreds digit as well as the units digit and $b$ is the tens digit.

Since the number lies between 100 and 500, so $a$ = 1, 2, 3 or 4, i.e. $a$ can take 4 values.

(If $a = 5$, then the number $5b5$ is at least 505, which is more than 500.)

Also, $b$ can be any one of the digits 0, 1, 2, ..., 9, i.e. $b$ can take 10 values.

∴ Total number of palindromes = 4 × 10 = 40

Hence, option 1 is the correct option.