Playing with Numbers — Exercise 5.1

Write the following numbers in generalized form:

(i) 89

(ii) 207

(iii) 369

Answer

(i) 89 = 10 × 8 + 9

(ii) 207 = 100 × 2 + 10 × 0 + 7

(iii) 369 = 100 × 3 + 10 × 6 + 9

Write the quotient, when the sum of a 2-digit number 34 and number obtained by reversing the digits is divided by

(i) 11

(ii) sum of digits

Answer

Given number is 34.

Number obtained by reversing the digits is 43.

Sum = 34 + 43 = 77 = 11 × 7

(i) According to the property, when ab + ba is divided by 11, the quotient = a + b.

Hence, when 34 + 43 is divided by 11, the quotient = 3 + 4 = 7.

(ii) According to the property, when ab + ba is divided by a + b, the quotient = 11.

Hence, when 34 + 43 is divided by 7 (3 + 4 = 7), the quotient = 11.

Write the quotient when the difference of a 2-digit number 73 and number obtained by reversing the digits is divided by

(i) 9

(ii) difference of digits.

Answer

Given number is 73.

Number obtained by reversing the digits is 37.

Difference = 73 - 37 = 36 = 9 × 4

(i) According to the property, when ab - ba is divided by 9, the quotient = a - b.

Hence, when 73 - 37 is divided by 9, the quotient = 7 - 3 = 4.

(ii) According to the property, when ab - ba is divided by a - b, the quotient = 9.

Hence, when 73 - 37 is divided by 4 (7 - 3 = 4), the quotient = 9.

Without actual calculation, write the quotient when the sum of a 3-digit number abc and the number obtained by changing the order of digits cyclically i.e. bca and cab is divided by

(i) 111

(ii) (a + b + c)

(iii) 37

(iv) 3

Answer

⇒ abc + bca + cab

⇒ (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

⇒ (100a + 10a + a) + (10b + 100b + b) + (c + 10c + 100c)

⇒ 111a + 111b + 111c

⇒ 111(a + b + c)

⇒ 3 × 37 × (a + b + c)

(i) When the sum is divided by 111, the quotient = (a + b + c).

(ii) When the sum is divided by (a + b + c), the quotient = 111.

(iii) Since 111(a + b + c) = 37 × 3(a + b + c), when the sum is divided by 37, the quotient = 3(a + b + c).

(iv) Since 111(a + b + c) = 3 × 37(a + b + c), when the sum is divided by 3, the quotient = 37(a + b + c).

Write the quotient when the difference of a 3-digit number 843 and number obtained by reversing the digits is divided by

(i) 99

(ii) 5

Answer

Given number is 843.

Number obtained by reversing the digits is 348.

∴ Difference = 843 - 348 = 495 = 99 × 5

(i) When the difference is divided by 99, the quotient = 5.

(ii) When the difference is divided by 5, the quotient = 99.

The sum of digits of a 2-digit number is 11. If the number obtained by reversing the digits is 9 less than the original number, find the number.

Answer

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Given, sum of digits = 11

⇒ a + b = 11 .....(1)

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the number obtained by reversing the digits is 9 less than the original number.

⇒ (10a + b) - (10b + a) = 9

⇒ 9a - 9b = 9

⇒ 9(a - b) = 9

⇒ a - b = 1 .....(2)

Adding (1) and (2), we get

2a = 12

⇒ a = 6

Putting a = 6 in (1), we get b = 11 - 6 = 5

∴ 2-digit number = 10 × 6 + 5 = 65

Hence, the required 2-digit number is 65.

If the difference of two digit number and the number obtained by reversing the digits is 36, find the difference between the digits of the 2-digit number.

Answer

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the difference of the 2-digit number and the number obtained by reversing the digits is 36.

⇒ (10a + b) - (10b + a) = 36

⇒ 9a - 9b = 36

⇒ 9(a - b) = 36

⇒ a - b = 4

Hence, the difference between the digits of the 2-digit number is 4.

If the sum of two digit number and number obtained by reversing the digits is 55, find the sum of the digits of the 2-digit number.

Answer

Let the 2-digit number be ab i.e. 10a + b, where unit's digit is b and ten's digit is a.

Number obtained by reversing the digits = ba i.e. 10b + a

Given, the sum of the 2-digit number and the number obtained by reversing the digits is 55.

⇒ (10a + b) + (10b + a) = 55

⇒ 11a + 11b = 55

⇒ 11(a + b) = 55

⇒ a + b = 5

Hence, the sum of the digits of the 2-digit number is 5.

The middle digit of a three digit number is 0 and the sum of the other two digits is 11. If the number obtained by reversing the digits exceeds the original number by 495, find the number.

Answer

Let the 3-digit number be a0c i.e. 100a + 10 × 0 + c = 100a + c, where the middle digit is 0.

Given, sum of the other two digits = 11

⇒ a + c = 11 .....(1)

Number obtained by reversing the digits = c0a i.e. 100c + a

Given, the number obtained by reversing the digits exceeds the original number by 495.

⇒ (100c + a) - (100a + c) = 495

⇒ 99c - 99a = 495

⇒ 99(c - a) = 495

⇒ c - a = 5 .....(2)

Adding equations (1) and (2), we get:

⇒ (a + c) + (c - a) = 11 + 5

⇒ 2c = 16

⇒ c = 8

Putting c = 8 in (1), we get a = 11 - 8 = 3

∴ 3-digit number = 100 × 3 + 10 × 0 + 8 = 308

Hence, the required number is 308.

In a 3-digit number, unit's digit, ten's digit and hundred's digit are in the ratio 1 : 2 : 3. If the difference of original number and the number obtained by reversing the digits is 594, find the number.

Answer

Given, unit's digit, ten's digit and hundred's digit are in the ratio 1 : 2 : 3.

Let unit's digit = x, ten's digit = 2x and hundred's digit = 3x.

∴ 3-digit number = 100 × 3x + 10 × 2x + x = 300x + 20x + x = 321x

Number obtained by reversing the digits = 100 × x + 10 × 2x + 3x = 100x + 20x + 3x = 123x

Given, the difference of the original number and the number obtained by reversing the digits is 594.

⇒ 321x - 123x = 594

⇒ 198x = 594

⇒ x = $\dfrac{594}{198}$

⇒ x = 3

∴ Unit's digit = 3, ten's digit = 2 × 3 = 6 and hundred's digit = 3 × 3 = 9

∴ 3-digit number = 100 × 9 + 10 × 6 + 3 = 963

Hence, the required number is 963.

In a 3-digit number, unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit. If the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664, find the number.

Answer

Let hundred's digit = a.

Given, unit's digit is one more than the hundred's digit and ten's digit is one less than the hundred's digit.

∴ Ten's digit = a - 1 and unit's digit = a + 1

According to the property, the sum of a 3-digit number and the numbers obtained by changing the order of digits cyclically = 111 × (sum of digits).

Given, this sum is 2664.

⇒ 111 × (sum of digits) = 2664

⇒ Sum of digits = $\dfrac{2664}{111}$

⇒ Sum of digits = 24

⇒ a + (a - 1) + (a + 1) = 24

⇒ 3a = 24

⇒ a = 8

∴ Hundred's digit = 8, ten's digit = 8 - 1 = 7 and unit's digit = 8 + 1 = 9

∴ 3-digit number = 100 × 8 + 10 × 7 + 9 = 879

Hence, the required number is 879.