Question

sum of two digit numbers and the number obtained by reversing the digits is 66 if the digits of the number differed by 2. find the number how many such numbers are there....

Answer 1

Please see the attachment

Answer 2

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Let the digit in the unit’s place be x and the digit in the tens place be y.

Then, the number = 10y + x

The number obtained by reversing the order of the digits = 10x + y

According to given conditions,

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

⇒ 11(x + y) = 66

⇒ (x + y) = 6

According to second situation, digits differ by 2

So, either x – y = 2 or y – x = 2

Thus , we have the following sets of simuntaneous equations

x + y = 6 …I

x – y = 2 …II

or,

x + y = 6 …III

x – y = 2 …IV

solving equation I and II, we get x = 2 and y = 4

solving equation III and IV , we get x = 4 and y = 2

When x = 4 and y = 2,

Two digit number = (10y + x) = 10(4) + 2 = 42

When x = 2 and y = 4,

Two digit number = (10y + x) = 10(2) + 4 = 24

Hence, the required number is either 24 or 42.