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  • The sum of a two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number.

The sum of a two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number.

Answers (1)

Let’s assume the digit at unit’s place as x and ten’s place as y. Thus from the question, the number needed  is 10y + x.

according to the question,  the two digits of the number are differing by 2. Thus, we can write

x – y =2………….. (i)

or x-y=-2--------(ii)

Now after reversing the order of the digits, the number becomes 10x + y.

Again from the question 

(10x+ y) + (10y+x) = 66

⇒ 10x + y + 10y + x = 66

⇒ 11x +11y = 66

⇒ 11(x + y) = 66

⇒ x + y = 66/11

⇒ x + y = 6………….. (iii)

On adding the equations (i) and (iii), we get;

(x – y) + (x + y) = 2+6

⇒ x – y + x + y = 8

⇒ 2x =8

⇒ x = 8/2

⇒ x = 4

Putting the value of x in equation (iii), we get

4 – y = 2

⇒ y = 4 – 2

⇒ y = 2

Hence, the required number is 10 × 2 +4 = 24

Now,

On adding the equations (ii) and (iii), we get

(x – y)+(x + y )= -2 + 6

⇒ x – y + x + y = 4

⇒ 2x = 4

⇒ x = 4/2

⇒ x = 2

Putting the value of x in equation (ii), we get;

2 – y = -2

⇒ y = 2+2

⇒ y = 4

Hence, the required number is 10×4+ 2 = 42

Therefore, there are two such possible numbers i.e, 24 and 42.

Posted by

shubham.krishnan

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