The Second of Seven Problems - A Problem for Kids Aged 7 Years And 2 Months

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If this is way too easy,I hope you read the title.(But seriously speaking,if this is too easy,try one of my other problems that doesn't have any sarcasm in the first two sentences)The quantity x is defined as 2 n 1 + 2 n + 2 n + 1 2 m 2 + 2 m 1 + 2 m \frac{{2}^{{n}-{1}}+{2}^{n}+{2}^{{n}+{1}}}{{2}^{m-2}+{2}^{{m}-1}+{2}^{m}} .If x can be written as 2^ (n (fakeoperationlol) m + y(where y is a real number)) find 1 (fakeoperationlol) 2 + 1.


The answer is 0.

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1 solution

Aaryan Vaishya
Oct 3, 2019

Factor out 2^n on the top and the 2^m-1 on the bottom and you have two sequences that cancel out and 2^n/2^m-1 is left which gives 2^(n-m+1) so the fakeoperationlol is subtraction so 1-2+1=0.

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