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# Let a be the sum of x consecutive positive integers. Let b be the sum

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Re: Let a be the sum of x consecutive positive integers. Let b be the sum [#permalink]
rajshreeasati wrote:
PyjamaScientist thanks for this approach. Its much more efficient than my approach. I used the more algebraic approach and then by taking the Sum in an AP sequence, but it took a lot of time.
Also, just a quick tip which I read in Kaplan- when going through options, start backwards from E to A. More often than not, the GMAT hides the Answer to "the following" type questions in these options. Majority of times, it turns out to be D or E.
Anyways, thanks!

Glad that you liked my solution. And thanks for your advice. Though my method is to start from choice (C) and then proceeding from there to other choices depending on the result I get from (C).
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Re: Let a be the sum of x consecutive positive integers. Let b be the sum [#permalink]
Tough one! But not impossible. Here's how I thought of it:

a = sum of x consecutive POSITIVE integers; b = sum of y consecutive POSITIVE integers.

Honestly, I don't think this can be done in time without trial and error from the options. So, here we go:

x = 2; y = 6

x--> 1+2 = 3, 2+3 = 5, 3+4 = 7 (i'm noticing a difference of 2 between numbers, so the sequence is sort of like 3,5,7,9,.....)
y--> 1 + 2 + 3 + 4 + 5 + 6 = 21 (Do this with another and we see the sequence is 21,27,33....)

in Option A, x and y will match at 27-27. Hence Eliminated.

I tried the sme for option B, where I got 21-21 as same. Hence eliminated.

Option C is where I couldn't find a pair and therefore picked that. I didn't look at the other options because it may have taken time.

Bunuel any thoughts on this approach?
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Re: Let a be the sum of x consecutive positive integers. Let b be the sum [#permalink]
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Re: Let a be the sum of x consecutive positive integers. Let b be the sum [#permalink]
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