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 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{(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{(A) 2+0+1+7}</math> is your answer. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/cY4NYSAD0vQ | ||
+ | |||
+ | ~[[User:icematrix2 | icematrix2]] | ||
==See Also== | ==See Also== |
Revision as of 00:44, 3 October 2020
Problem 1
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. Option adds up to ten. 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
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.