Difference between revisions of "2017 AMC 8 Problems/Problem 1"
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We compute each expression individually according to the order of operations. We get <math>2 + 0 + 1 + 7 = 10</math>, <math>2 \times 0 + 1 + 7 = 8</math>, <math>2 + 0 \times 1 + 7 = 9</math>, <math>2 + 0 + 1 \times 7 = 9</math>, and <math>2 \times 0 \times 1 \times 7 = 0</math>. Since <math>10</math> is the greatest out of these numbers, <math>\boxed{\textbf{(A) }2+0+1+7}</math> is the answer. | We compute each expression individually according to the order of operations. We get <math>2 + 0 + 1 + 7 = 10</math>, <math>2 \times 0 + 1 + 7 = 8</math>, <math>2 + 0 \times 1 + 7 = 9</math>, <math>2 + 0 + 1 \times 7 = 9</math>, and <math>2 \times 0 \times 1 \times 7 = 0</math>. Since <math>10</math> is the greatest out of these numbers, <math>\boxed{\textbf{(A) }2+0+1+7}</math> is the answer. | ||
− | ==Solution | + | ==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, 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 | ||
Revision as of 02:41, 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, has a number multiplied by . This will only make the expression's value smaller. 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.