Difference between revisions of "2021 AMC 12A Problems/Problem 1"

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So assume that 1 + 1 = 3
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==Problem==
 +
What is the value of<cmath>2^{1+2+3}-(2^1+2^2+2^3)?</cmath><math>\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57</math>
  
Using this logic, prove that:
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==Solution==
looking + at + this + problem = you're a cheater.
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We evaluate the given expression to get that <cmath>2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(A)}}</cmath>
 
 
-jhu08
 
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|before=First problem|num-a=2}}
 
{{AMC12 box|year=2021|ab=A|before=First problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:24, 11 February 2021

Problem

What is the value of\[2^{1+2+3}-(2^1+2^2+2^3)?\]$\textbf{(A) }0 \qquad \textbf{(B) }50 \qquad \textbf{(C) }52 \qquad \textbf{(D) }54 \qquad \textbf{(E) }57$

Solution

We evaluate the given expression to get that \[2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(A)}}\]

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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