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

let the digit at one 's place =x

then the digit at ten 's place = 3*x

then original no. will be= 10*(3x)+x


by reversing the no.

one 's digit will be =3x

ten 's digit will be= x

then no. will be =10*x+3x


then according to the problem



then original no will be 31*2=62

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