# Math Help - Reverse Digit problem help

1. ## Reverse Digit problem help

I need help on understanding and do a reverse digit problem.

problem is:

If you take two-digit number to make a second two-digit number, and add these numbers together, their sun will be 121. What is the original number??

2. First number is $10a+b$
Reverse the digits, you get $10b+a$
add the two to get $10(a+b)+(a+b)=11(a+b)=121$

so you must have $a+b=11$. Possible choices (up to reversing the digits) are $29,38,47,56$.

3. Thanks for the answer and i understand how to do the problem now.

4. I'm assuming that when you say you "take two-digit number to make a second two-digit number", you make the second number by reversing the digits.

Let x be the first digit, and y be the second digit.

Your first number is x + 10y, so your second number is 10x + y

When you add the two numbers together you have

(x + 10y) + (10x + y) = 10x + x + 10y + y = 11x + 11y = 11(x + y) = 121

So x + y = 11

Remember, x and y represent the digits, so they are single digit numbers.

So, we have:
x y
2 9
3 8
4 7
5 6
6 5
7 4
8 3
9 2

i.e. 29 + 92 = 38 + 83 = 47 + 74 = 56 + 65 = 65 + 56 = 74 + 47 = 83 + 38 = 92 + 29 = 121