Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the greatest prime factor of 1+2+3+....+36?

A. 2
B. 3
C. 9
D. 31
E. 37

$$?\,\,\,:\,\,\,{\rm{greatest}}\,\,{\rm{prime}}\,\,{\rm{factor}}$$
$$1 + 2 + 3 + \ldots + 36\,\,\mathop = \limits^{\left( * \right)} \,\,\left( {{{1 + 36} \over 2}} \right) \cdot 36 = 18 \cdot 37\,\,\,\,\mathop \Rightarrow \limits^{2,3\,\,{\rm{or}}\,\,37} \,\,\,\,? = 37$$
(*) The parcels 1, 2, 3, ..., 36 constitute a finite arithmetic sequence, hence their average is equal to the average of the first and last terms (or any two symmetric to the median).
Besides that, the sum 1+2+3+...+36 is equal to the average (explained above) times the number of parcels (36), due to the homogeneity nature of the average.

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Since 1 + 2 + 3 + ... + n = n(n+1)/2, we have 1 + 2 + 3 + ... + 36 = (36*37)/2 = 18*37 = 2*32*37.
Thus, 37 is the greatest prime factor of 2*3^2*37.

Therefore, the answer is E.
Answer: E