Problem: Find the value of 1≤a<b<c∑2a3b5c1 (i.e. the sum of 2a3b5c1 over all triples of positive integers ( a,b,c ) satisfying a<b<c )
Solution
Solution: Let x=b−a and y=c−b so that b=a+x and c=a+x+y. Then 2a3b5c=2a3a+x5a+x+y=30a15x5y and a,x,y are any positive integers. Thus 1≤a<b<c∑2a3b5c1=1≤a,x,y∑30a15x5y1=1≤a∑30a11≤x∑15x11≤y∑5y1=291⋅141⋅41=16241
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