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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>\text{a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5}</math>. What is <math>\text{a + b + c + d}</math>?
+Let <math>a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5</math>. What is <math>a + b + c + d</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==
+==Solution 1==
-Let <math>\text{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. <math>\text{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>\text{3(a+b+c+d)=-10}</math>, so <math>\text{a+b+c+d}=\frac{-10}{3}</math>. Thus, our answer is <math>\boxed{\text{(B)}\ -10/3}</math>.
+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>.
+==Solution 2==
+Take
+<math>a + 1 = a + b + c + d + 5</math>.
+Now we can clearly see:
+<math>-4 = b + c + d</math>.
+Continuing this same method with <math>b + 2, c + 3</math>, and <math>d + 4</math> we get:
+<math> -4 = b + c + d</math>,
+<math> -3 = a + c + d</math>,
+<math> -2 = a + b + d</math>, and
+<math> -1 = a + b + c</math>,
+Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \boxed{\frac{-10}{3}}</math>.
+==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==
@@ Line 12: / Line 39: @@
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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