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Difference between revisions of "2017 AMC 8 Problems/Problem 1"
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==Solution 1==
 
==Solution 1==
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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>- SBose
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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{\textbf{(A) } 2+0+1+7}</math>- SBose
  
 
==Solution 2==
 
==Solution 2==
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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{\textbf{(A) } 2+0+1+7}</math> is your answer.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 09:22, 6 January 2022

Problem

Which of the following values is largest?

$\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7$


Solution 1

We will compute each expression. A) $2+0+1+7=10$. B) $2 \times 0+1+7=8$ C) $2+0\times 1+7=9$ D) $2+0+1\times 7=9$. E)$2\times 0\times 1 \times 7 =0$ Ordering these we get $10,8,9,9,0$ Out of these $10$ is the largest number. Option $(A)$ adds up to ten. Therefore, the answer is $\boxed{\textbf{(A) } 2+0+1+7}$- SBose

Solution 2

We immediately see that every one of the choices, except for A and D, has a number multiplied by $0$. This will only make the expression's value smaller. We are left with A and D, but in D, $1$ is multiplied by $7$ to get $7$, whereas in answer choice A, we get $8$ out of $7$ and $1$, instead of $7$. Therefore, $\boxed{\textbf{(A) } 2+0+1+7}$ is your answer.

Video Solution

https://youtu.be/cY4NYSAD0vQ

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

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