Productive! Topic: Algebra/NT. Level: AIME.

Problem: (2006 Bellevue BATH Team) Evaluate $ \displaystyle \sum_{a=1}^{10} \sum_{b=1}^{10} \sum_{c=1}^{10} \sum_{d=1}^{10} dcab $.

Solution: Note that we can write the sum as

$ \displaystyle \sum_{b=1}^{10} \sum_{c=1}^{10} \sum_{d=1}^{10} (1 \cdot bcd+ 2 \cdot bcd + \cdots + 10 \cdot bcd) = \sum_{b=1}^{10} \sum_{c=1}^{10} \sum_{d=1}^{10} (1+2+\cdots+10)bcd$.

Using the same idea, we can write it as

$ (1+2+\cdots+10)(1+2+\cdots+10)(1+2+\cdots+10)(1+2+\cdots+10) = 55^4 $.

QED.

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Comment: Evaluating multiple summations this way is very effective. Other examples include factoring the harmonic series

$ \displaystyle 1+\frac{1}{2}+\frac{1}{3}+\cdots = \left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right) \left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots\right) \left(1+\frac{1}{5}+\frac{1}{5^2}+\cdots\right) = \prod \sum_{i=0}^\infty \frac{1}{p^i} $

or, more generally, any integer value of the Riemann Zeta Function

$ \displaystyle \zeta(s) = 1^s+\frac{1}{2^s}+\frac{1}{3^s}+\cdots = \prod \sum_{i=0}^\infty \frac{1}{p^is} $.

Furthermore, since

$ \displaystyle \sum_{i=0}^\infty \frac{1}{p^is} $

is an infinite geometric series of common ratio $ \frac{1}{p^s} $, we can say that

$ \zeta(s) = \prod \frac{1}{p^s-1} $,

where the product is taken over all primes $ p $.