Difference between revisions of "2017 AMC 8 Problems/Problem 1"
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E) <math>2 \times 0 \times 1 \times 7 = 0</math> | E) <math>2 \times 0 \times 1 \times 7 = 0</math> | ||
| − | Ordering these, we get <math>10, 8, 9, 9, 0</math>. Out of these, <math>10</math> is the largest number and option <math>(A)</math> adds up to <math>10</math>. Therefore, the answer is <math>\boxed{\textbf{(A) } 2+0+1+7}</math>- SBose | + | Ordering these, we get <math>10, 8, 9, 9, 0</math>. Out of these, <math>10</math> is the largest number and option <math>(A)</math> adds up to <math>10</math>. Therefore, the answer is <math>\boxed{\textbf{(A) } 2+0+1+7}</math>. |
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| + | - SBose | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 04:02, 8 January 2023
Problem
Which of the following values is largest?
Solution 1
We will compute each expression.
A) 
B) 
C) 
D) 
E) 
Ordering these, we get 
. Out of these, 
is the largest number and option 
adds up to . Therefore, the answer is 
.
- SBose
Solution 2
We immediately see that every one of the choices, except for A and D, has a number multiplied by 
. This will only make the expression's value smaller. We are left with A and D, but in D, 
is multiplied by 
to get , whereas in answer choice A, we get 
out of and , instead of . Therefore, is your answer.
Video Solution
~savannahsolver
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
