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

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E) <math>2 \times 0 \times 1 \times 7 = 0</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 and option <math>(A)</math> adds up to <math>10</math>. Therefore, the answer is <math>\boxed{\textbf{(A) } 2+0+1+7}</math>- SBose
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Ordering these, we get <math>10, 8, 9, 9, 0</math>. Out of these, <math>10</math> is the largest number and option <math>(A)</math> adds up to <math>10</math>. Therefore, the answer is <math>\boxed{\textbf{(A) } 2+0+1+7}</math>.
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- SBose
  
 
==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{\textbf{(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.
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https://youtu.be/S1gAyiyQWYo
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
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https://youtu.be/S1gAyiyQWYo
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~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Latest revision as of 17:45, 15 June 2024

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 and option $(A)$ adds up to $10$. 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.


https://youtu.be/S1gAyiyQWYo

~Education, the Study of Everything

Video Solution

https://youtu.be/cY4NYSAD0vQ

https://youtu.be/o5Aus-6w1vs

~savannahsolver

https://youtu.be/S1gAyiyQWYo

~Education, the Study of Everything

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

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