Difference between revisions of "2017 AMC 8 Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
| − | We immediately see that every one of the choices, except for A, has a number multiplied by <math>0</math>. This will only make the expression's value smaller. Therefore, <math>\boxed{\textbf{(A) }2+0+1+7}</math> is your answer | + | We immediately see that every one of the choices, except for A and D, has a number multiplied by <math>0</math>. This will only make the expression's value smaller. We are left with A and D, but in D, <math>1</math> is multiplied by <math>7</math> to get <math>7</math>, whereas in answer choice A, we get <math>8</math> out of <math>7</math> and <math>1</math> instead of <math>7</math>. Therefore, <math>\boxed{\textbf{(A) }2+0+1+7}</math> is your answer. |
==See Also== | ==See Also== | ||
Revision as of 02:43, 10 August 2018
Contents
Problem 1
Which of the following values is largest?
Solution 1
We compute each expression individually according to the order of operations. We get 
, 
, 
, 
, and 
. Since 
is the greatest out of these numbers, 
is the answer.
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.
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. 
