SOLUTION: A guy is her mom's age with digits reversed (e.g. 34 and 43). A year ago he was half her mom's age. How old are they now?

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Question 696107: A guy is her mom's age with digits reversed (e.g. 34 and 43). A year ago he was half her mom's age. How old are they now?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE ALGEBRA WAY TO SOLVE IT:
Setting up:
a = first (tens) digit of that guy's age.
b = second (ones) digit of that guy's age.
The guy's age is 10a%2Bb.
His mother's age is 10b%2Ba.
Last year, the guy was 10a%2Bb-1
and his mother was 10b%2Ba-1,
which was twice the guy's age, so
10b%2Ba-1=2%2810a%2Bb-1%29

Solving:
10b%2Ba-1=2%2810a%2Bb-1%29 --> 10b%2Ba-1=20a%2B2b-2 --> 10b%2Ba-1%2B2-a-2b=20a%2B2b-2%2B2-a-2b --> highlight%288b%2B1=19a%29
At this point we could start trying values for a,
knowing that it is a small odd number.
It must be odd to make 19a equal to the obviously odd 8b%2B1.
It is small because it is the first digit of the son's age,
and the mother's is more than twice that, but no more than 99.

Otherwise, 8b%2B1=19a <--> 8b%2B1=16a%2B3a
We see that 3a must be a multiple of 8 plus 1.
3a=8n%2B1
At this point, we should see n=1 with highlight%28a=3%29 as an obvious answer.

(Otherwise, we could try n values starting with n=1 to find that only n=1 works,
because n=2 and n=3 would yield fractional values for a,
and n%3E=4 would yield a%3E=11).

Substituting into highlight%288b%2B1=19a%29, we get
8b%2B1=19%2A3 --> 8b%2B1=57 --> 8b%2B1-1=57-1 --> 8b=56 --> 8b%2Fb=56%2Fb --> highlight%28b=7%29
The guy is highlight%2837%29 and his mom is highlight%2873%29.