Playing with Numbers — Exercise 4(C)

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & \text{A} & \text{A} \\ + & \text{B} & 6 \\ \hline \text{C} & 2 & 5 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{A} & \text{A} \\ + & \text{B} & 6 \\ \hline \text{C} & 2 & 5 \\ \hline \end{matrix}$

Firstly, we will find the value of A.

Clearly, A + 6 is a number whose units digit is 5.

⇒ A + 6 = 5, A + 6 = 15; and so on.

⇒ A = 5 - 6, A = 15 - 6; and so on.

⇒ A = -1, A = 9; and so on.

Since A is a digit.

∴ A = 9 and 1 is carried over.

Secondly, we will find the values of B and C.

1 + A + B = 1 + 9 + B = 10 + B, which is a number whose units digit is 2.

⇒ 10 + B = 12

⇒ B = 2 and 1 is carried over.

∴ C = 1

$\begin{matrix} & \overset{1}{9} & 9 \\ + & 2 & 6 \\ \hline 1 & 2 & 5 \\ \hline \end{matrix}$

Hence, A = 9, B = 2 and C = 1.

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & 5 & \text{C} & \text{B} & 8 \\ + & \text{D} & 9 & 7 & \text{A} \\ \hline & 8 & 2 & 4 & 3 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & 5 & \text{C} & \text{B} & 8 \\ + & \text{D} & 9 & 7 & \text{A} \\ \hline & 8 & 2 & 4 & 3 \\ \hline \end{matrix}$

Firstly, we will find the value of A.

Clearly, 8 + A is a number whose units digit is 3.

⇒ 8 + A = 13

⇒ A = 5 and 1 is carried over.

Secondly, we will find the value of B.

1 + B + 7 = 8 + B, which is a number whose units digit is 4.

⇒ 8 + B = 14

⇒ B = 6 and 1 is carried over.

Thirdly, we will find the value of C.

1 + C + 9 = 10 + C, which is a number whose units digit is 2.

⇒ 10 + C = 12

⇒ C = 2 and 1 is carried over.

Lastly, we will find the value of D.

1 + 5 + D = 8

⇒ D = 8 - 6 = 2

$\begin{matrix} & \overset{1}{5} & \overset{1}{2} & \overset{1}{6} & 8 \\ + & 2 & 9 & 7 & 5 \\ \hline & 8 & 2 & 4 & 3 \\ \hline \end{matrix}$

Hence, A = 5, B = 6, C = 2 and D = 2.

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & \text{A} \\ & \text{A} \\ + & \text{A} \\ \hline \text{B} & \text{A} \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{A} \\ & \text{A} \\ + & \text{A} \\ \hline \text{B} & \text{A} \\ \hline \end{matrix}$

Clearly, A + A + A = 3A is a number whose units digit is A.

⇒ 3A = 10B + A

⇒ 2A = 10B

⇒ A = 5B

Since A and B are digits and B is the tens digit of a 2-digit number, so B ≠ 0.

∴ B = 1 and A = 5 × 1 = 5

Checking with A = 5 and B = 1 : 5 + 5 + 5 = 15, which is of the form BA with B = 1 and A = 5.

$\begin{matrix} & 5 \\ & 5 \\ + & 5 \\ \hline 1 & 5 \\ \hline \end{matrix}$

Hence, A = 5 and B = 1.

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & 6 & \text{A} \\ - & \text{A} & \text{B} \\ \hline & 3 & 7 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & 6 & \text{A} \\ - & \text{A} & \text{B} \\ \hline & 3 & 7 \\ \hline \end{matrix}$

Firstly, we will find the value of A.

In the tens column, A is subtracted from 6 and the result is 3.

If nothing is taken away from 6, then 6 - A = 3 ⇒ A = 3, and then in the units column 3 - B = 7 gives B = -4, which is not a digit.

So, 1 is taken away from 6 on the left.

⇒ 6 - 1 - A = 3

⇒ 5 - A = 3

⇒ A = 2

Secondly, we will find the value of B.

Since 1 is taken away from 6, so in the units column,

(10 + A) - B = 7

⇒ (10 + 2) - B = 7

⇒ 12 - B = 7

⇒ B = 5

$\begin{matrix} & 6 & 2 \\ - & 2 & 5 \\ \hline & 3 & 7 \\ \hline \end{matrix}$

Hence, A = 2 and B = 5.

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & \text{C} & \text{B} & 6 \\ - & 2 & 9 & \text{A} \\ \hline & 5 & 3 & 9 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{C} & \text{B} & 6 \\ - & 2 & 9 & \text{A} \\ \hline & 5 & 3 & 9 \\ \hline \end{matrix}$

Firstly, we will find the value of A.

Since 6 - A cannot be 9, so 1 is taken away from B on the left.

⇒ (10 + 6) - A = 9

⇒ 16 - A = 9

⇒ A = 7

Secondly, we will find the value of B.

Since 1 is taken away from B and 9 cannot be subtracted from (B - 1), so 1 is taken away from C on the left.

⇒ (10 + B - 1) - 9 = 3

⇒ B = 3

Lastly, we will find the value of C.

Since 1 is taken away from C,

C - 1 - 2 = 5

⇒ C - 3 = 5

⇒ C = 8

$\begin{matrix} & 8 & 3 & 6 \\ - & 2 & 9 & 7 \\ \hline & 5 & 3 & 9 \\ \hline \end{matrix}$

Hence, A = 7, B = 3 and C = 8.

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & \text{A} & \text{B} \\ \times & & 3 \\ \hline \text{C} & \text{A} & \text{B} \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{A} & \text{B} \\ \times & & 3 \\ \hline \text{C} & \text{A} & \text{B} \\ \hline \end{matrix}$

Here, AB = 10A + B and CAB = 100C + 10A + B.

∴ (10A + B) × 3 = 100C + 10A + B

⇒ 30A + 3B = 100C + 10A + B

⇒ 20A + 2B = 100C

⇒ 10A + B = 50C

⇒ AB = 50C

Since AB is a 2-digit number, so 50C must be a 2-digit number.

∴ C = 1 and AB = 50 × 1 = 50

⇒ A = 5 and B = 0

Checking with A = 5, B = 0 and C = 1 : 50 × 3 = 150, which is of the form CAB.

$\begin{matrix} & 5 & 0 \\ \times & & 3 \\ \hline 1 & 5 & 0 \\ \hline \end{matrix}$

Hence, A = 5, B = 0 and C = 1.

Replace A, B, C, D by suitable numerals:

$\begin{matrix} & \text{A} & \text{B} \\ \times & \text{B} & \text{A} \\ \hline (\text{B}+1) & \text{C} & \text{B} \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & \text{A} & \text{B} \\ \times & \text{B} & \text{A} \\ \hline (\text{B}+1) & \text{C} & \text{B} \\ \hline \end{matrix}$

In the units column, A × B is a number whose units digit is B.

This happens when A = 1, or when A = 6 and B is even, or when B = 5 and A is odd. (B cannot be 0, as BA would then not be a 2-digit number.)

If A = 6 and B is even, AB × BA is a 4-digit number (the smallest such product is 62 × 26 = 1612).

If B = 5 and A is odd (A ≠ 1), AB × BA is also a 4-digit number (the smallest such product is 35 × 53 = 1855).

But the given product is a 3-digit number.

∴ A = 1

Thus, we have,

$\begin{matrix} & 1 & \text{B} \\ \times & \text{B} & 1 \\ \hline & 1 & \text{B} \\ \text{B} & \text{B}^2 & \times \\ \hline \text{B} & (1+\text{B}^2) & \text{B} \\ \hline \end{matrix}$

Now, the hundreds digit of the product is (B + 1), i.e. it is 1 more than B.

So, the tens column must give a carry over of 1.

⇒ (1 + B²) is a 2-digit number whose tens digit is 1.

⇒ 10 ≤ 1 + B² ≤ 19

⇒ 9 ≤ B² ≤ 18

⇒ B = 3 or B = 4

When B = 3 : 1 + B² = 1 + 9 = 10. So, C = 0 and 1 is carried over, giving hundreds digit = 3 + 1 = 4 = B + 1.

Here, 13 × 31 = 403, which is of the form (B + 1)CB.

When B = 4 : 1 + B² = 1 + 16 = 17. So, C = 7 and 1 is carried over, giving hundreds digit = 4 + 1 = 5 = B + 1.

Here, 14 × 41 = 574, which is of the form (B + 1)CB.

Hence, A = 1, B = 3, C = 0 or A = 1, B = 4, C = 7.

Replace A, B, C, D by suitable numerals:

$\begin{array}{l} 6\overline{\smash{\big)} 5\text{AB}\smash{\big(}}\ 9\text{C} \\ \phantom{}\underline{\ -54} \\ \phantom{+()}3\text{B} \\ \phantom{+}\underline{-36} \\ \phantom{++}\times \\ \end{array}$

Answer

Given,

$\begin{array}{l} 6\overline{\smash{\big)} 5\text{AB}\smash{\big(}}\ 9\text{C} \\ \phantom{}\underline{\ -54} \\ \phantom{+()}3\text{B} \\ \phantom{+}\underline{-36} \\ \phantom{++}\times \\ \end{array}$

Here, the divisor is 6, the dividend is 5AB and the quotient is 9C.

Firstly, we will find the value of A.

We have, 6 × 9 = 54 and this 54 is subtracted from 5A leaving 3.

⇒ 5A - 54 = 3

⇒ 5A = 57

⇒ A = 7

Secondly, we will find the values of B and C.

B is brought down and 3B is divided by 6, and 36 is subtracted from 3B leaving no remainder.

⇒ 3B = 36

⇒ B = 6

Also, 6 × C = 36

⇒ C = 6

Checking with A = 7, B = 6 and C = 6 : 576 ÷ 6 = 96, which is of the form 9C.

$\begin{array}{l} 6\overline{\smash{\big)} 576\smash{\big(}}\ 96 \\ \phantom{}\underline{\ -54} \\ \phantom{+()}36 \\ \phantom{()}\underline{-36} \\ \phantom{++}\times \\ \end{array}$

Hence, A = 7, B = 6 and C = 6.

Find the quotient when you divide the number of four-digit numbers by the number of three-digit numbers.

Answer

The smallest 4-digit number is 1000 and the largest 4-digit number is 9999.

∴ Number of 4-digit numbers = (9999 - 1000) + 1 = 8999 + 1 = 9000

The smallest 3-digit number is 100 and the largest 3-digit number is 999.

∴ Number of 3-digit numbers = (999 - 100) + 1 = 899 + 1 = 900

∴ Required quotient = $\dfrac{9000}{900}$ = 10

Hence, the required quotient is 10.

Find the largest natural number by which the difference between a three-digit number and the number obtained by reversing its digits is always divisible.

Answer

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

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

Number obtained by reversing its digits = $100c + 10b + a$

$\Rightarrow \text{Difference} = (100a + 10b + c) - (100c + 10b + a) \\[1em] = 100a + 10b + c - 100c - 10b - a \\[1em] = 99a - 99c \\[1em] = 99(a - c)$

Since $(a - c)$ is an integer, so the difference is always a multiple of 99.

Also, $(a - c)$ can be equal to 1 (for example, 211 - 112 = 99), so no natural number greater than 99 can always divide this difference.

Hence, the required largest natural number is 99.

On multiplying a number by 7, the product is a number each of whose digits is 3. Find the smallest such number.

Answer

We go on dividing a string of 3's by 7 till 0 is obtained as the remainder.

3 ÷ 7 → remainder = 3

33 ÷ 7 → quotient = 4, remainder = 5

333 ÷ 7 → quotient = 47, remainder = 4

3333 ÷ 7 → quotient = 476, remainder = 1

33333 ÷ 7 → quotient = 4761, remainder = 6

333333 ÷ 7 → quotient = 47619, remainder = 0

∴ The required number = 47619

Checking : 47619 × 7 = 333333, each of whose digits is 3.

Hence, the smallest such number is 47619.

The letters A, B, C, D, E, F, G, H and I are assigned values from 1 to 9 but not in that order. 4 is assigned to E. The difference between E and I is 5. The difference between C and I is 3. What is the value assigned to C?

Answer

Given, E = 4.

Given, the difference between E and I is 5.

⇒ I = 4 + 5 = 9 or I = 4 - 5 = -1

Since the values assigned are from 1 to 9, so I ≠ -1.

∴ I = 9

Given, the difference between C and I is 3.

⇒ C = 9 + 3 = 12 or C = 9 - 3 = 6

Since the values assigned are from 1 to 9, so C ≠ 12.

∴ C = 6

Hence, the value assigned to C is 6.

Find the values of A and B in the division sum shown below:

$\begin{array}{l} \text{AB}\overline{\smash{\big)} 252\smash{\big(}}\ \text{BA} \\ \phantom{+}\underline{ \\ -24} \\ \phantom{++(}12 \\ \phantom{+()}\underline{ -12} \\ \phantom{++()}\times \\ \end{array}$

Answer

Given,

$\begin{array}{l} \text{AB}\overline{\smash{\big)}252\smash{\big(}}\ \text{BA} \\ \phantom{+}\underline{ \\ -24} \\ \phantom{++(}12 \\ \phantom{+()}\underline{ -12} \\ \phantom{++()}\times \\ \end{array}$

Here, the divisor is AB, the dividend is 252 and the quotient is BA.

From the division sum, AB × B = 24 and AB × A = 12.

$\Rightarrow \dfrac{\text{AB} \times \text{B}}{\text{AB} \times \text{A}} = \dfrac{24}{12} \\[1em] \Rightarrow \dfrac{\text{B}}{\text{A}} = \dfrac{2}{1}$

⇒ B = 2A

Now, AB = 10A + B = 10A + 2A = 12A

Also, AB × A = 12

⇒ 12A × A = 12

⇒ 12A² = 12

⇒ A² = 1

⇒ A = 1

∴ B = 2 × 1 = 2

Checking with A = 1 and B = 2 : AB = 12, BA = 21 and 12 × 21 = 252.

$\begin{array}{l} 12\overline{\smash{\big)}252\smash{\big(}}\ 21 \\ \phantom{(}\underline{\ -24} \\ \phantom{++}12 \\ \phantom{+(}\underline{ -12} \\ \phantom{++}\times \\ \end{array}$

Hence, A = 1 and B = 2.

If X and Y represent digits, what is the maximum possible value of Y in the following?

$\begin{matrix} & 5 & \text{X} & 9 \\ & 3 & 2 & 7 \\ + & 2 & \text{Y} & 8 \\ \hline 1 & 1 & 1 & 4 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & 5 & \text{X} & 9 \\ & 3 & 2 & 7 \\ + & 2 & \text{Y} & 8 \\ \hline 1 & 1 & 1 & 4 \\ \hline \end{matrix}$

In the units column, 9 + 7 + 8 = 24. So, 4 is written and 2 is carried over.

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

⇒ X + Y + 4 = 11

⇒ X + Y = 7

So, 1 is carried over to the hundreds column.

(In the hundreds column, 1 + 5 + 3 + 2 = 11, which gives the digits 1 and 1 of the sum, as required.)

Now, X + Y = 7, where X and Y are digits.

∴ Y is maximum when X is minimum, i.e. when X = 0.

⇒ Y = 7 - 0 = 7

Hence, the maximum possible value of Y is 7.

What would be the maximum value of Q in the following addition sum?

$\begin{matrix} & 5 & \text{P} & 7 \\ & 8 & \text{Q} & 9 \\ + & \text{R} & 3 & 2 \\ \hline 1 & 9 & 2 & 8 \\ \hline \end{matrix}$

Answer

Given,

$\begin{matrix} & 5 & \text{P} & 7 \\ & 8 & \text{Q} & 9 \\ + & \text{R} & 3 & 2 \\ \hline 1 & 9 & 2 & 8 \\ \hline \end{matrix}$

In the units column, 7 + 9 + 2 = 18. So, 8 is written and 1 is carried over.

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

⇒ P + Q + 4 = 12 or P + Q + 4 = 22

⇒ P + Q = 8 or P + Q = 18

Since P and Q are digits, P + Q = 18 is possible only when P = Q = 9. But P and Q are different letters, so they stand for different digits.

∴ P + Q = 8 and 1 is carried over.

Q is maximum when P is minimum, i.e. when P = 0.

⇒ Q = 8 - 0 = 8

In the hundreds column, 1 (carry over) + 5 + 8 + R = 14 + R, which is a number whose units digit is 9.

⇒ 14 + R = 19

⇒ R = 5 and 1 is carried over, which gives the thousands digit 1 of the sum, as required.

Checking with P = 0, Q = 8 and R = 5 : 507 + 889 + 532 = 1928, as required.

$\begin{matrix} & \overset{1}{5} & \overset{1}{0} & 7 \\ & 8 & 8 & 9 \\ + & 5 & 3 & 2 \\ \hline 1 & 9 & 2 & 8 \\ \hline \end{matrix}$

Hence, the maximum value of Q is 8.

Find two numbers whose product is a 1-digit number and the sum is a 2-digit number.

Answer

Let us take the two numbers as 1 and 9.

Product of the numbers = 1 × 9 = 9, which is a 1-digit number.

Sum of the numbers = 1 + 9 = 10, which is a 2-digit number.

Hence, the required two numbers are 1 and 9.

Find three natural numbers whose sum and product are equal.

Answer

Let us take the three natural numbers as 1, 2 and 3.

Sum of the numbers = 1 + 2 + 3 = 6

Product of the numbers = 1 × 2 × 3 = 6

∴ Sum of the numbers = Product of the numbers

Hence, the required three natural numbers are 1, 2 and 3.