Difference between revisions of "2017 AMC 8 Problems/Problem 1"
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<math>\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7</math> | <math>\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7</math> | ||
− | + | We will compute each expression. A) <math>2+0+1+7=10</math>. B) <math>2 \times 0+1+7=8</math> C) <math>2+0\times 1+7=9</math> D) <math>2+0+1\times 7=9</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. Option <math>(A)</math> adds up to ten. Therefore, the answer is <math>\boxed{(A) 2+0+1+7}</math> | |
− | We will compute each expression. A) <math>2+0+1+7=10</math>. B) <math>2 \times 0+1+7=8</math> C) <math>2+0\times 1+7=9</math> D) <math>2+0+1\times 7=9</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. Option <math>(A)</math> adds up to ten. Therefore, the answer is <math>\boxed{( | ||
==Solution 2== | ==Solution 2== |
Revision as of 22:48, 14 June 2020
Problem 1
Which of the following values is largest?
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
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.