Question

26. The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.

Answer 1

Let x be the tenths digit and x-3 be the ones digit. We have, The digit=(10*x)+x-3=10x+x-3=11x-3. The reversed digit=10(x-3)+x=10x-30+x=11x-30 Now, Sum of the digit+Sum of the reversed digit=121 11x-3+11x-30=121. 22x-33=121. 22x=121+33. x=154/22=7. The digit=11x-3=11*7-3=77-3=74