Maths Olympiad Prep

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February, 2017

Algebra Difficulty 4.9 AIME Prove it United States

Problem:
Find the value of
∑1≤a<b<c12a3b5c \sum_{1 \leq a < b < c} \frac{1}{2^{a} 3^{b} 5^{c}}
(i.e. the sum of 12a3b5c\frac{1}{2^{a} 3^{b} 5^{c}} over all triples of positive integers ( a,b,ca, b, c ) satisfying a<b<ca < b < c )

Solution

Solution:
Let x=b−ax = b - a and y=c−by = c - b so that b=a+xb = a + x and c=a+x+yc = a + x + y. Then
2a3b5c=2a3a+x5a+x+y=30a15x5y 2^{a} 3^{b} 5^{c} = 2^{a} 3^{a + x} 5^{a + x + y} = 30^{a} 15^{x} 5^{y}
and a,x,ya, x, y are any positive integers. Thus
∑1≤a<b<c12a3b5c=∑1≤a,x,y130a15x5y=∑1≤a130a∑1≤x115x∑1≤y15y=129⋅114⋅14=11624 \begin{aligned} \sum_{1 \leq a < b < c} \frac{1}{2^{a} 3^{b} 5^{c}} & = \sum_{1 \leq a, x, y} \frac{1}{30^{a} 15^{x} 5^{y}} \\ & = \sum_{1 \leq a} \frac{1}{30^{a}} \sum_{1 \leq x} \frac{1}{15^{x}} \sum_{1 \leq y} \frac{1}{5^{y}} \\ & = \frac{1}{29} \cdot \frac{1}{14} \cdot \frac{1}{4} \\ & = \frac{1}{1624} \end{aligned}

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