ML Aggarwal: Playing with Numbers - 1

Q.1. Write the following numbers in generalized form:
(i) 89
(ii) 207
(iii) 369

The generalized form is as follows:
(i) 89

= 8 × 10 + 9

(ii) 207

= 2 × 100 + 0 × 10 + 7 × 1

(iii) 369

= 3 × 100 + 6 × 10 + 9 × 1

Q.2. 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

Given:
Sum of two-digit number 34 and the number
obtained by reversing the digit 43 = 34 + 43
= 77

(i) 11

77 ÷ 11 = 7

(ii) Sum of digits

77 ÷ (Sum of digit) = 77 ÷ (4 + 3)
= 77 ÷ 7
= 11 

Q.3. 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) a difference of digits.

Given:
Difference of two digit number 73 and the number obtained by reversing the digits is
= 73 - 37 = 36
(i) 9 

divide by 9
So, 36 ÷ 9 = 4

(ii) a difference of digits.

36 ÷ (7 - 3) = 36 + 4
= 9

Q.4. 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

Given:
Sum of 3-digit number abc and the number obtained by changing the order of digits i.e. bca and cab. 

∴ abc + bca + cab
= 100a + 10b + c + 1006 + 10c + a + 100c + 10a + b
= 111a + 111b + 111c = 111 (a + b + c)

(i) 111

When divided by 111, we get
111 (a + b + c) ÷ 111 = a + b + c

(ii) (a + b + c)

When divided by (a + b + c), we get
111 (a + b + c) ÷ (a + b + c) = 111

(iii) 37

When divided by 37, we get
111 (a + b + c) ÷ 37 = 3(a + b + c)

(iv) 3

When divided by 3, we get
111 (a + b + c) ÷ 3 = 37(a + b + c

Q.5. 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

Given:
Difference of 3-digit number 843 and the number obtained by reversing the digit is 348
= 843 - 348 = 495

(i) 99

divided by 99, we get 495 ÷ 99 = 5

(ii) 5

Divided by 5, we get 495 ÷ 5 = 99

Q.6. 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.

Given:
Sum of two digit number = 11
Let unit's digit be 'x' and tens digit be 'y',
then x + y = 11 ... (i)
and number = x + 10y
By reversing the digits,
Unit digit be 'y' and tens digit be 'x' and number = y + 10x + 9
Now by equating both numbers,
y + 10x + 9 = x + 10y
10x + y - 10y - x = -9
9x - 9y = -9
x - y = -1 ... (ii)
Adding (i) and (ii), we get
2x = 10
x = 10/2
= 5
∴ y = 1 + 5 = 6
By substituting the vales of x and y, we get
Number = x + 10y = 5 + 10 × 6
= 5 + 60
= 65
∴ the number is 65.

Q.7. 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.

Let us consider the unit digit be 'x' and tens digit be 'y'
So, the number is = x + 10y
By reversing the digits
Unit digit be 'y' and tens digit be 'x'
the number is y + 10x = 36
Now by equating both numbers,
x + 10y - y - 10x = 36
-9x + 9y = 36
(y - x) = 36/9
y - x = 4
∴ The difference between digits of the 2-digit number is y - x = 4.

Q.8. 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.

Let us consider unit digit be 'x' and tens digit be 'y'
So the number is x + 10y
By reversing the digits
Unit digit be 'y' and tens digit be 'x'
The number is y + 10x = 55
Now by equating both numbers,
x + 10y + y + 10x = 55
11x + 11y = 55
11(x + y) = 55
x + y = 55/11
x + y = 5
∴ Difference of the digits of the number is 5.

Q.9. 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.

Given:
Ratio in the digits of a three digit number = 1: 2: 3
Let us consider unit digit be 'x'
Tens digit be '2x'
and hundreds digit be '3x'
So the number is x + 10 × 2x + 100 × 3x
= x + 20x + 300x
= 321x
By reversing the digits,
Unit digit be '3x'
Ten's digit be '2x'
Hundreds digit be 'x'
So the number is 3x + 10 × 2x + 100 × x
= 3x + 20x + 100x
= 123x
According to the condition,
321x - 123x = 594
198x = 594
x = 594/198
= 3
∴ The number is = 321x = 321 × 3 = 963

Q.10. 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.

Let us consider the hundreds digit be 'x'
Unit digit be 'x + 1'
and ten's digit be 'x - 1'
So the number = (x + 1) + 10(x - 1) + 100 × x
= x + 1 + 10x - 10 + 100x
= 111x - 9
By reversing the digits,
Unit digit be 'x - 1'
Tens digit be 'x'
Hundred digit be 'x + 1'
So the number = x - 1 + 10x + 100x + 100 = 111x + 99 and sum of original 3-digit number = x + 10(x + 1) + 100(x - 1)
= x + 10x + 10 + 100x - 100
= 111x - 90
Now according to the condition,
111x - 9 + 111x + 99 + 111x - 90 = 2664
333x + 99 - 99 = 2664
333x = 2664
x = 2664/333
= 8
∴ The number = 111x - 9
= 111(8) - 9
= 888 - 9
= 879