Maths Olympiad Prep

Track / Stage 3 / 236 of 260 #236 of 1964

Problem 236

AMC 10/12, early questions
Number theory Difficulty 3.9 Find the answer China Mathematical Competition (Shaanxi) · China

Let aa, bb, cc, dd be positive integers and logab=32\log_a b = \frac{3}{2}, logcd=54\log_c d = \frac{5}{4}. If ac=9a-c=9, then bd=b-d= ________.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

We have b=a32b = a^{\frac{3}{2}}, d=c54d = c^{\frac{5}{4}} from the assumption. We may assume that a=x2a = x^2, c=y4c = y^4 with xx and yy being positive integers, since aa, bb, cc, dd are all positive integers. Then ac=x2y4=(xy2)(x+y2)=9a-c = x^2 - y^4 = (x-y^2)(x+y^2) = 9. It follows that (xy2,x+y2)=(1,9)(x-y^2, x+y^2) = (1, 9). So we obtain the solution x=5x=5, y=2y=2. That is, bd=x3y5=12532=93b-d = x^3 - y^5 = 125-32=93.

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