Let p, q, r, s, t be 5 consecutive positive integers such that p+q+r+s+t is a perfect square and q+r+s is a perfect cube. What is sqrt(p+q+r+s+t)?
Let r = n. Then p+q+r+s+t = 5n = u^2 and q+r+s = 3n = v^3 for some positive integers u and v. We can see that 5 | u and 3 | v; so u = 5w and v = 3x for some positive integers w and x. Therefore n = u^2/5 = 5w^2 and n = v^3/3 = 9x^3. Equating the two expressions for n, we get 5w^2 = 9x^3. Clearly 5 | x and 3 | w; so w = 3y and x = 5z for some positive integers y and z. Thus 5(3y)^2 = 9(5z)^3, which simplies to y^2 = 25z^3. It follows that n = 5w^2 = 5(3y)^2 = 45y^2 = 45(25z^3) = 1125z^3. We are given that p, q, r, s, t < 10000. So 5n < 10000, or 1125z^3 = n < 2000. Therefore z must be 1 and n = 1125. Hence sqrt(p+q+r+s+t) = sqrt(5n) = 75.
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