The sum of the digits of a two-digit number is 11. When the digits are reversed, the number increases by 27. Find the original number.

Question

this is difficult for me

Answer 1

If the digits are a and b, then the number is 10a+b (a is in the tens place and b is in the ones place).

So a+b=11. If we reverse the digits we get the number 10b+a. So 10a+b+27=10b+a.

Subtract a from each side: 9a+b+27=10b, now subtract 10b from each side: 9a-9b+27=0.

We can divide through by 9: a-b+3=0 and if we add b-a to each side we get 3=b-a. So the difference between a and b is 3: b-a=3. But a+b=11, and if we add the two equations together, the a’s cancel out and we get 2b=14, so b=7. If b=7 then a must be 4. So the number is 47.