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What is the greatest prime factor of 1+2+3+ . . . + 40?

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by Vincen » Sun May 13, 2018 10:52 am
Gmat_mission wrote:What is the greatest prime factor of 1 + 2 + 3 + . . . + 40?

A. 17
B. 29
C. 31
D. 37
E. 41

[spoiler]OA=E[/spoiler].

What is the fastest way to solve this PS question? Could someone explain this question to me? Thanks.
Hello Gmat_mission.

Here, a useful tool that we can use is that the sum of the first n positive integers is equal to n*(n+1)/2.

Therefore, here we have: $$1+2+3+\cdots+40=\frac{40\cdot\left(40+1\right)}{2}=\frac{40\cdot41}{2}=20\cdot41=2\cdot10\cdot41=2^2\cdot5\cdot41.$$ This shows that the greatest prime factor of the given sum is 41.

Hence, the correct answer is the option E.

I hope it helps.
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by [email protected] » Sun May 13, 2018 11:24 am
Hi Vincen,

We're asked for the greatest prime factor of 1 + 2 + 3 + . . . + 40. This question has a number of built-in 'shortcuts' that you can use to avoid doing lots of calculations.

To start, the answers are all prime numbers - so we know that one of those numbers IS the greatest prime factor of that sum. Second, since we're dealing with 40 consecutive integers, we can use 'bunching' to spot a pattern:
1 + 40 = 41
2 + 39 = 41
3 + 38 = 41
Etc.

Since there are 40 terms, there will be 20 "pairs" of 41. By extension, 41 will divide evenly into that sum - and since it's the biggest prime number in the answer choices, it must be the answer.

Final Answer: E

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Rich
Contact Rich at [email protected]
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by Jeff@TargetTestPrep » Wed May 16, 2018 10:11 am
Gmat_mission wrote:What is the greatest prime factor of 1 + 2 + 3 + . . . + 40?

A. 17
B. 29
C. 31
D. 37
E. 41
The average of the consecutive integers from 1 to 40 inclusive is 41/2 and the number of integers is 40, so the sum is 40 x 41/2 = 20 x 41.

Thus, we see that the largest prime factor is 41.

Answer: E

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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