Difference between revisions of "2002 AMC 10A Problems/Problem 16"

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<math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math>
 
<math>\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5</math>
  
==Solution==
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==Solution 1==
Let <math>x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. Since one of the sums involves a, b, c, and d, it makes sense to consider 4x. We have <math>4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20</math>. Rearranging, we have <math>3(a+b+c+d)=-10</math>, so <math>a+b+c+d=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>.
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Let <math>x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. Since one of the sums involves <math>a, b, c,</math> and <math>d,</math> it makes sense to consider <math>4x</math>. We have <math>4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20</math>. Rearranging, we have <math>3(a+b+c+d)=-10</math>, so <math>a+b+c+d=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>.
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==Solution 2==
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Take
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<math>a + 1 = a + b + c + d + 5</math>.
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Now we can clearly see:
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<math>-4 = b + c + d</math>.
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Continuing this same method with <math>b + 2, c + 3</math>, and <math>d + 4</math> we get:
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<math> -4 = b + c + d</math>,
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<math> -3 = a + c + d</math>,
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<math> -2 = a + b + d</math>, and
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<math> -1 = a + b + c</math>,
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Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \boxed{\frac{-10}{3}}</math>.
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==Solution 3(Video solution using [[The Apple Method]])==
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https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo
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https://youtu.be/NRdOxPUDngI
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== Video Solution by OmegaLearn ==
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https://youtu.be/tKsYSBdeVuw?t=4105
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 03:20, 16 January 2023

Problem

Let $a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. What is $a + b + c + d$?

$\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$

Solution 1

Let $x=a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. Since one of the sums involves $a, b, c,$ and $d,$ it makes sense to consider $4x$. We have $4x=(a+1)+(b+2)+(c+3)+(d+4)=a+b+c+d+10=4(a+b+c+d)+20$. Rearranging, we have $3(a+b+c+d)=-10$, so $a+b+c+d=\frac{-10}{3}$. Thus, our answer is $\boxed{\text{(B)}\ -10/3}$.

Solution 2

Take $a + 1 = a + b + c + d + 5$. Now we can clearly see: $-4 = b + c + d$. Continuing this same method with $b + 2, c + 3$, and $d + 4$ we get: $-4 = b + c + d$, $-3 = a + c + d$, $-2 = a + b + d$, and $-1 = a + b + c$, Adding, we see $-10 = 3a + 3b + 3c + 3d$. Therefore, $a + b + c + d = \boxed{\frac{-10}{3}}$.

Solution 3(Video solution using The Apple Method)

https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo https://youtu.be/NRdOxPUDngI

Video Solution by OmegaLearn

https://youtu.be/tKsYSBdeVuw?t=4105

~ pi_is_3.14

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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