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 Mother of nine (Posted on 2010-12-03)
A woman, under sixty, told me, that the square of her age equals the sum of the squares of her children's ages.

How old is she , given that the children's ages form an arithmetic series of nine integers?

 No Solution Yet Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

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 just a bit of brute force | Comment 2 of 5 |
Call the middle child age m and the difference between them d.
Thus the ages are m-4d, m-3d, m-2d, m-d, m, m+d, m+2d, m+3d, m+4d
and the sums of the squared ages is
9m^2 + 60d^2
which we require to be a square of a number less than 60.
also since ages cannot be negative m ≥ 4d

letting m=4d
9(4d)^2 + 60d^2 < 60^2
204d^2 < 3600
d^2 < 17.6
d < 4.2
also d = m/4
9m^2 + 60(m/4)^2 < 60^2
12.75m^2 < 3600
m^2 < 282.35
m < 16.8

So there really arent that many possibilities to check
sqrt(9m^2 + 60*1^2) gives no integers < 60
sqrt(9m^2 + 60*2^2) gives no integers < 60
sqrt(9m^2 + 60*3^2) gives integers <60 for m=2 [24] and m=14 [48]
sqrt(9m^2 + 60*4^2) gives no integers < 60

m=2, d=3 doesnt fit m<4d

so the solution is m=14, d=3 and the ages are
2, 5, 8, 11, 14, 17, 20, 23, 26
the sum squared is 2304 which is 48^2

 Posted by Jer on 2010-12-03 17:29:16

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