Difference between revisions of "2021 AMC 12B Problems/Problem 10"

Revision as of 19:44, 11 February 2021

The sum of the first $37$ integers is given by $n(n+1)/2$ , so $37(37+1)/2=703$ .

Therefore, $703-x-y=xy$

Rearranging, $xy+x+y=703$

$(x+1)(y+1)=704$

Looking at the possible divisors of $704 = 2^6*11$ , $22$ and $32$ are within the constraints of $0 < x <= y <= 37$ so we try those:

$(x+1)(y+1) = 22 * 32$

$x+1=22, y+1 = 32$

$x = 21, y = 31$

Therefore, the difference $y-x=31-21=10$ , choice E). ~ SoySoy4444

Revision as of 19:43, 11 February 2021 (view source) Soysoy4444 (talk \| contribs) (Created page with " The sum of the first <math>37</math> integers is given by <math>n(n+1)/2</math>, so <math>37(37+1)/2=703</math>. Therefore, <math>703-x-y=xy</math> Rearranging, <math>xy+x+...")		Revision as of 19:44, 11 February 2021 (view source) Soysoy4444 (talk \| contribs) m Newer edit →
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	Therefore, the difference <math>y-x=31-21=10</math>, choice E).		Therefore, the difference <math>y-x=31-21=10</math>, choice E).
		+	~ SoySoy4444