rocdocmac 9 Posted November 28, 2018 Report Share Posted November 28, 2018 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? Quote Link to post Share on other sites

1 Solution EventHorizon 15 Posted November 29, 2018 Solution Report Share Posted November 29, 2018 Spoiler 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. Quote Link to post Share on other sites

0 rocdocmac 9 Posted November 29, 2018 Author Report Share Posted November 29, 2018 Excellent Spoiler Excellent! I could only do it by trial and error! Quote Link to post Share on other sites

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## rocdocmac 9

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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