Tammy's Age
Math brain teasers require computations to solve.
I was talking to a friend Tammy, the middle child of five. She has a younger sister Tracey, and three brothers Tommy (oldest child), Timmy and Tony (youngest child). I asked her how old she was, but she hardly ever gives a straight answer. The following is the conversation we had.
"How old are you Tammy?"
"I am three times as old as Timmy was when Tony was born. Actually, it's funny, but Timmy's, Tracey's, Tommy's and my ages were all factors of Mum's age when Tony was born. The only other year where the ages of more than two of us were factors of Mum's age was when Tracey was half as old as I am now."
"How old is your Mum now?"
"There's another funny thing. When Mum turns 50, Tony will be the same age Mum was when she gave birth to Tommy."
Assuming Tammy was talking in terms of whole years, and with my mathematical and reasoning abilities I was able to quickly deduce not just Tammy's age, but all of their ages. So how old are Tommy, Timmy, Tammy, Tracey, Tony and Mum now?
Answer
Tommy is 23, Timmy is 19, Tammy is 18, Tracey is 16, Tony is 13 and Mum is 43.
Solution:
x = Mum's age when she gave birth to Tommy = Tony's age when Mum turns 50;
y = Tommy's age when Tony was born
z = Mum's age when Tony was born
From Tammy's last statement, 2x + y = 50 and it can be deduced that x = z  y, where y is a factor of z. By combining the two equations you get 2z = 50 + y The only possible solutions for that equation so that y is also a factor of z are (z = 26, y = 2) and (z = 30, y = 10). Only 30 has four or more factors, so Mum was 30 and Tommy was 10 when Tony was born.
This means that of the remaining factors of 30, Timmy was 6, 5 or 3, Tammy was 5, 3 or 2 and Tracey was 3, 2 or 1. This makes Tammy 18, 15 or 9 now. The only other time where the ages of more than two of the children were factors of Mum's age was when Tracey was half what Tammy is now, so Tammy must be 18 now.
In the second multiplefactorsofMum'sage year, Tracey was therefore 9. When Tony was born, Tracey was either 3, 2 or 1, giving possible time differences of 6, 7 or 8 years, which correspond to Tony's age in the second multiplefactorsofMum'sage year. It also means Mum was 36, 37 or 38 that year. 37 is prime and 38 has factors of 1, 2 and 19, and as Tony was at least 6, it means that Mum must have been 36 that year.
Going back to when Tony was born we have that Mum was 30, Tommy was 10, Timmy was 6, Tracey was 3, and as the middle child Tammy must have been 5. If Tammy is 18 now then Tony is 13, Tracey is 16, Timmy is 19, Tommy is 23 and Mum is 43.
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Comments
badolina
Oct 19, 2002
 Wwwwoooowww, GR8, took me hours just to understand your answer. 
snaps
Oct 19, 2002
 Took me ages to understand the answer as well, and I wrote the thing! 
bluelightnign
Apr 11, 2003
 Great Question, It took me about 40 minutes of guessing and checking to get the correct answer. I suppose it would be easier if I had learned the right algebra, but I'll do that in year 11 or 12 probably. 
krishnan
Mar 06, 2005
 Fun, enjoyable teaser. 
Rebeca
Apr 21, 2005
 Loved it! I had a lot of fun with it for a couple hours, and if felt great when the answer suddenly appeared! Phew! Thanks!!! 
javaguru
Jan 11, 2009
 Wow. One of the hardest yet. Awesome teaser! 
Jimbo
Jun 04, 2012
 Great teaser. Quite hard. I first deduced that I had to find 4 discrete factors of a number < 50. Turns out all combinations begin with either 1 x 2 x 3 or 1 x 2 x 4.
The first set leads to Tammy being 9 but Tammy must be an even age (reference to half her age). Therefore It must be 1 x 2 x 4 and Tammy is 12. Thus the Youngest children are 10, 11, 12 and 14. five years ago they were 5, 6, 7,9. Any combination of 3 factors is too large unless they overlap so 6 and 9 must be two of the factors. So eldest was a multiple of 6 and 9 i.e. 18 five years ago. Now we have the answers.
Excellent teaser. 
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