Question

The digit at the tens place of a two digit number is three times the digit at the units place. If the sum of this number and the number formed by reversing its digits is $88$, find the number.

Answer 1

Let the digit at tens place be $x$ and the digit unis place be $y$.

Then the original number will be $10x+y$.

The digit at tens place is three times the digit at the units place.

i.e. $x=3y$ ....(1)

The sum of this number and the number formed by reversing its digits is $88$.

$(10x+y)+(10y+x)=88$

$\Rightarrow 11x+11y=88$

$\Rightarrow x+y=8$ ....(2)

Substitute the value of equation (1) in equation (2).

Therefore, $3y+y=8$

$\Rightarrow 4y=8$

$\Rightarrow y=2$

Therefore, $x=3y=3\times 2=6$

Therefore, the original number is $62$.