Playing with Numbers — Exercise 4(A)

In a two-digit number the units digit is four times the tens digit and the sum of the digits is 10. Find the number.

Answer

Let the tens digit be $x$.

Then, units digit = $4x$.

Given, sum of the digits is 10.

$\Rightarrow x + 4x = 10 \\[1em] \Rightarrow 5x = 10 \\[1em] \Rightarrow x = 2$

So, tens digit = 2 and units digit = (4 × 2) = 8.

∴ Required number = 10 × 2 + 8 = 20 + 8 = 28

Hence, the required number is 28.

The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number.

Answer

Let the tens digit be $x$.

Given, units digit = 3.

∴ The number = $10x + 3$ and sum of its digits = $x + 3$.

Given, seven times the sum of the digits is the number itself.

$\Rightarrow 7(x + 3) = 10x + 3 \\[1em] \Rightarrow 7x + 21 = 10x + 3 \\[1em] \Rightarrow 10x - 7x = 21 - 3 \\[1em] \Rightarrow 3x = 18 \\[1em] \Rightarrow x = 6$

So, tens digit = 6 and units digit = 3.

∴ Required number = 10 × 6 + 3 = 60 + 3 = 63

Hence, the required number is 63.

In a two-digit number, the digit at the units place is thrice the digit in the tens place. The number exceeds the sum of its digits by 27. Find the number.

Answer

Let the tens digit be $x$.

Then, units digit = $3x$.

∴ The number = $10x + 3x = 13x$ and sum of its digits = $x + 3x = 4x$.

Given, the number exceeds the sum of its digits by 27.

$\Rightarrow 13x - 4x = 27 \\[1em] \Rightarrow 9x = 27 \\[1em] \Rightarrow x = 3$

So, tens digit = 3 and units digit = (3 × 3) = 9.

∴ Required number = 10 × 3 + 9 = 30 + 9 = 39

Hence, the required number is 39.

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

Answer

Let the tens digit be $x$ and the units digit be $y$.

∴ The number = $10x + y$ and sum of its digits = $x + y$.

Given, the number is 3 more than 4 times the sum of its digits.

$\Rightarrow 10x + y = 4(x + y) + 3 \\[1em] \Rightarrow 10x + y = 4x + 4y + 3 \\[1em] \Rightarrow 10x - 4x + y - 4y = 3 \\[1em] \Rightarrow 6x - 3y = 3 \\[1em] \Rightarrow 2x - y = 1 \quad \text{.....(1)}$

Number obtained by reversing the digits = $10y + x$

Given, if 18 is added to the number, its digits are reversed.

$\Rightarrow (10x + y) + 18 = 10y + x \\[1em] \Rightarrow 10x - x + y - 10y = -18 \\[1em] \Rightarrow 9x - 9y = -18 \\[1em] \Rightarrow x - y = -2 \quad \text{.....(2)}$

Subtracting (2) from (1), we get

$\Rightarrow (2x - y) - (x - y) = 1 - (-2) \\[1em] \Rightarrow x = 3$

Putting $x = 3$ in (2), we get $3 - y = -2 \Rightarrow y = 5$

So, tens digit = 3 and units digit = 5.

∴ Required number = 10 × 3 + 5 = 30 + 5 = 35

Hence, the required number is 35.

The sum of the digits of a two-digit number is 13. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

Answer

Let the tens digit be $x$ and the units digit be $y$.

Given, sum of the digits is 13.

⇒ $x + y = 13 \quad \text{.....(1)}$

Original number = $10x + y$.

Number obtained by interchanging its digits = $10y + x$

Given, the number obtained by interchanging the digits exceeds the given number by 9.

$\Rightarrow (10y + x) - (10x + y) = 9 \\[1em] \Rightarrow 9y - 9x = 9 \\[1em] \Rightarrow y - x = 1 \quad \text{.....(2)}$

Adding (1) and (2), we get

$\Rightarrow 2y = 14 \\[1em] \Rightarrow y = 7$

Putting $y = 7$ in (1), we get $x = 13 - 7 = 6$

So, tens digit = 6 and units digit = 7.

∴ Original number = 10 × 6 + 7 = 60 + 7 = 67

Hence, the original number is 67.

The difference between a 2-digit number and the number obtained by interchanging its digits is 63. What is the difference between the digits of the number?

Answer

Let the tens digit be $x$ and the units digit be $y$.

∴ Original number = $10x + y$.

Number obtained by interchanging its digits = $10y + x$

Given, the difference between the number and the number obtained by interchanging its digits is 63.

$\Rightarrow (10x + y) - (10y + x) = 63 \\[1em] \Rightarrow 9x - 9y = 63 \\[1em] \Rightarrow 9(x - y) = 63 \\[1em] \Rightarrow x - y = 7$

Hence, the difference between the digits of the number is 7.

In a 3-digit number, the hundreds digit is twice the tens digit while the units digit is thrice the tens digit. Also, the sum of its digits is 18. Find the number.

Answer

Let the tens digit be $x$.

Then, hundreds digit = $2x$ and units digit = $3x$.

Given, sum of its digits is 18.

$\Rightarrow 2x + x + 3x = 18 \\[1em] \Rightarrow 6x = 18 \\[1em] \Rightarrow x = 3$

So, hundreds digit = (2 × 3) = 6, tens digit = 3 and units digit = (3 × 3) = 9.

∴ Required number = 100 × 6 + 10 × 3 + 9

= 600 + 30 + 9

= 639

Hence, the required number is 639.