Art of Problem Solving

2000 CEMC Gauss (Grade 8) Problems/Problem 1: Difference between revisions

Anabel.disher (talk | contribs)
Created page with "==Problem== The value of <math>2^{5} + 5</math> is <math> \text{ (A) }\ 20 \qquad\text{ (B) }\ 37 \qquad\text{ (C) }\ 11 \qquad\text{ (D) }\ 13 \qquad\text{ (E) }\ 21 </mat..."
 
Anabel.disher (talk | contribs)
No edit summary
 
(One intermediate revision by the same user not shown)
Line 4: Line 4:
<math> \text{ (A) }\ 20 \qquad\text{ (B) }\ 37 \qquad\text{ (C) }\ 11 \qquad\text{ (D) }\ 13 \qquad\text{ (E) }\ 21 </math>   
<math> \text{ (A) }\ 20 \qquad\text{ (B) }\ 37 \qquad\text{ (C) }\ 11 \qquad\text{ (D) }\ 13 \qquad\text{ (E) }\ 21 </math>   
==Solution==
==Solution==
Following [[order of operations]], we have:
<math>2^{5} + 5 = 32 + 5 = \boxed {\textbf {(B) } 37}</math>
<math>2^{5} + 5 = 32 + 5 = \boxed {\textbf {(B) } 37}</math>


~anabel.disher
~anabel.disher
{{CEMC box|year=2000|competition=Gauss (Grade 8)|before=First Problem|num-a=2}}

Latest revision as of 12:38, 20 October 2025

Problem

The value of $2^{5} + 5$ is

$\text{ (A) }\ 20 \qquad\text{ (B) }\ 37 \qquad\text{ (C) }\ 11 \qquad\text{ (D) }\ 13 \qquad\text{ (E) }\ 21$

Solution

Following order of operations, we have:

$2^{5} + 5 = 32 + 5 = \boxed {\textbf {(B) } 37}$

~anabel.disher

2000 CEMC Gauss (Grade 8) (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
CEMC Gauss (Grade 8)