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rocdocmac

November Quickie II

Question

Can one solve this mathematically i.e. not by trial and error?

In 1988 my granny was quite a bit older than my grandpa.  The difference between the square of their ages was 1988. How old were they?

 

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Let granny's age be 'a' and grandpa's be 'b'.

If only one of a and b were odd, the difference of their squares would be odd since squaring doesn't change even/odd.  So either both a and b are even, or they are both odd.  This means a-b is even.  This also means the a+b is even.

The prime factorization of 1988 is 2,2,7,71.

1988= a^2 - b^2 = (a-b)(a+b)

So the prime factors of 1988 must be partitioned into the factors (a-b) and (a+b)

Since both are even, a 2 must be put into both factors.

7 and 71 cannot both be put into (a+b), the obviously bigger factor, or the sum of their ages would be 1988/2 = 994.  Aside from being too large for human ages, that would leave (a-b)=2... and granny wouldn't be "quite a bit older than" grandpa.

(a-b)=2*7=14

(a+b)=2*71=142

Solving the system of equations gives...

a+b=142
a-b=14 => a=14+b
(14+b)+b=142
14+2b=142
2b=128
b=64
a=14+(64)
a=78

So granny is 78 years old and grandpa is 64.

 

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