Difference between revisions of "2002 AMC 10A Problems/Problem 16"
(→Solution 2) |
Mathlebron (talk | contribs) m (→Solution 3(Video solution using (The Apple Method)) |
||
| (13 intermediate revisions by 8 users not shown) | |||
| Line 5: | Line 5: | ||
<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== | + | ==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>. | + | 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== | ==Solution 2== | ||
Take | Take | ||
| − | <math>a + 1 = a + b + c + d + 5</math> | + | <math>a + 1 = a + b + c + d + 5</math>. |
Now we can clearly see: | Now we can clearly see: | ||
| − | <math>-4 = b + c + d</math> | + | <math>-4 = b + c + d</math>. |
| − | Continuing this same method with <math>b + 2, c + 3</math>, and <math>d + 4</math> we get | + | 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> -4 = b + c + d</math>, | ||
<math> -3 = a + c + d</math>, | <math> -3 = a + c + d</math>, | ||
<math> -2 = a + b + d</math>, and | <math> -2 = a + b + d</math>, and | ||
<math> -1 = a + b + c</math>, | <math> -1 = a + b + c</math>, | ||
| − | Adding, we see <math> -10 = 3a + 3b + 3c + 3d</math>. Therefore, <math>a + b + c + d = \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>. |
| + | |||
| + | ==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== | ==See Also== | ||
Revision as of 14:32, 13 June 2024
Contents
Problem
Let 
. What is 
?
Solution 1
Let 
. Since one of the sums involves 
and 
it makes sense to consider 
. We have 
. Rearranging, we have 
, so 
. Thus, our answer is 
.
Solution 2
Take
.
Now we can clearly see:
.
Continuing this same method with 
, and 
we get:
,
,
, and
,
Adding, we see 
. Therefore, 
.
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 (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. 
