Difference between revisions of "2017 AMC 8 Problems/Problem 1"

(Problem 1)
Line 11: Line 11:
 
==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.
 +
 +
==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?

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


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 $\boxed{(A)  2+0+1+7}$- 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, $\boxed{(A) 2+0+1+7}$ is your answer.

Video Solution

https://youtu.be/cY4NYSAD0vQ

~ icematrix2

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png