Determine all possible pairs (p, q) of base ten positive integers that satisfy this system of equations:
- arithmetic mean (p, q) = 10x+y
- geometric mean (p, q) = 10y+x
where, each of p and q is a nonzero base ten integer, with p≠q, and each of x and y is a base ten digit.
(p+q)/2 = 10x + y
sqrt(pq) = 10y +x
(p+q)^2 = 4*(10x + y)^2
4pq = 4*(10y + x)^2
Subtract:
(p-q)^2 = 4*9*11*(x+y)(x-y)
RHS is square so 11*(x+y)(x-y) must be square making (x+y) = 11 and (x-y) = 1 and (x,y)=(6,5).
Then p-q = 66
p+q = 130
(p,q) = (98,32) and (32,98)
so that AP=65 and GP=56.
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Posted by xdog
on 2023-03-20 15:16:43 |
solution
