The digit at the tens place of a two digit number is three times the digit at ones place . If the sum of this number and the number formed by reversing it's digits is 88 . Find the number .

Digit at the ten's place is three times the digit at ones place.

The sum of the number and the number formed by reversing the digits = 88

the number

Let the digit at ten's place be x and digit at one's place be y.

The number = 10x + y or xy

Digit at the ten's place is three times the digit at ones place.

→ x = 3y -- equation (1)

Also,

The sum of the number and the number formed by reversing the digits = 88

Number formed by reversing the digits = 10y + x

Hence,

→ 10x + y + 10y + x = 88

→ 11x + 11y = 88

→ 11(x + y) = 88

→ x + y = 88/11

Substitute the value of x from equation (1).

→ 3y + y = 8

→ 4y = 8

→ y = 8/4

→ y = 2

Putting the value of y in equation (1) we get,

→ x = 3 * 2

→ x = 6

The number = xy = 62.

Therefore, the required two digit number is 62.

I hope it will help you.

Regards.

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