SOLUTION: Joan is 14 years old and her mom is 41 years old. Both their ages have the same two digits. How many years until their ages have two digits again? I have tried to work this one

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Question 680872: Joan is 14 years old and her mom is 41 years old. Both their ages have the same two digits. How many years until their ages have two digits again?
I have tried to work this one out on my own, but it is a very hard question that just takes a lot of time. I think there is a trick to it, but I do not know for sure. Thanks for your time.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Note that the difference in their ages, which is
+41+-+14+=+27+, will always stay the same
in the future.
I can express mom's future age as
+10a+%2B+b+
and, Joan's age as
+10b+%2B+a+
So, I have
+10a+%2B+b+-+%28+10b+%2B+a+%29+=+27+
+10a+%2B+b+-+10b+-+a+=+27+
+9a+-+9b+=+27+
+a+-+b+=+3+
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This is true now for Mom, since
+4+-+1+=+3+
note tyhat
+5+-+2+=+3+
--------------
Mom will be 52 and Joan is 25 when digits are swapped
Note that +52+-+25+=+27+ as it should be
--------------
The next time would be when
Mom is 63 and Joan is 36
+63+-+36+=+27+