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

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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>.
 
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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==Solution 3(Video solution using [The Apple Method])==
 
https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo
 
https://www.youtube.com/watch?v=rz86M2hlOGk&feature=emb_logo
 
https://youtu.be/NRdOxPUDngI
 
https://youtu.be/NRdOxPUDngI
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~ pi_is_3.14
 
~ pi_is_3.14
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==Video Solution by Daily Dose of Math==
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https://youtu.be/X1pdpIGSFRU
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~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==

Latest revision as of 08:57, 1 September 2024

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

Video Solution by Daily Dose of Math

https://youtu.be/X1pdpIGSFRU

~Thesmartgreekmathdude

See Also

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 AMC 10 Problems and Solutions

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