Difference between revisions of "2009 AIME I Problems/Problem 14"
(New page: == Problem == For <math>t = 1, 2, 3, 4</math>, define <math>S_t = \sum_{i = 1}^{350}a_i^t</math>, where <math>a_i \in \{1,2,3,4\}</math>. If <math>S_1 = 513</math> and <math>S_4 = 4745</ma...) |
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== Solution == | == Solution == | ||
− | Because the order of the <math>a</math>s doesn't matter, we simply need to find the number of <math>1</math>s <math>2</math>s <math>3</math>s and <math>4</math>s that minimize <math>S_2</math>. So let <math>w, x, y,</math> and <math>z</math> represent the number of <math>1</math>s, <math>2</math>s, <math>3</math>s, and <math>4</math>s respectively. Then we can write three equations based on these variables. Since there are a total of <math>350 a</math>s, we know that <math>w + x + y + z = 350</math>. We also know that <math>w + 2x + 3y + 4z = 513</math> and <math>w + 16x + 81y + 256z = 4745</math>. | + | Because the order of the <math>a</math>'s doesn't matter, we simply need to find the number of <math>1</math>s <math>2</math>s <math>3</math>s and <math>4</math>s that minimize <math>S_2</math>. So let <math>w, x, y,</math> and <math>z</math> represent the number of <math>1</math>s, <math>2</math>s, <math>3</math>s, and <math>4</math>s respectively. Then we can write three equations based on these variables. Since there are a total of <math>350</math> <math>a</math>s, we know that <math>w + x + y + z = 350</math>. We also know that <math>w + 2x + 3y + 4z = 513</math> and <math>w + 16x + 81y + 256z = 4745</math>. We can now solve these down to two variables: |
+ | <cmath>w = 350 - x - y - z</cmath> | ||
+ | Substituting this into the second and third equations, we get | ||
+ | <cmath>x + 2y + 3z = 163</cmath> | ||
+ | and | ||
+ | <cmath>15x + 80y + 255z = 4395.</cmath> | ||
+ | The second of these can be reduced to | ||
+ | <cmath>3x + 16y + 51z = 879.</cmath> | ||
+ | Now we substitute <math>x</math> from the first new equation into the other new equation. | ||
+ | <cmath>x = 163 - 2y - 3z</cmath> | ||
+ | <cmath>3(163 - 2y - 3z) + 16y + 51z = 879</cmath> | ||
+ | <cmath>489 + 10y + 42z = 879</cmath> | ||
+ | <cmath>5y + 21z = 195</cmath> | ||
+ | Since <math>y</math> and <math>z</math> are integers, the two solutions to this are <math>(y,z) = (39,0)</math> or <math>(18,5)</math>. | ||
+ | If you plug both these solutions in to <math>S_2</math> it is apparent that the second one returns a smaller value. It turns out that <math>w = 215</math>, <math>x = 112</math>, <math>y = 18</math>, and <math>z = 5</math>, so <math>S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \boxed{905}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=13|num-a=15}} | {{AIME box|year=2009|n=I|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:12, 16 August 2020
Problem
For , define , where . If and , find the minimum possible value for .
Solution
Because the order of the 's doesn't matter, we simply need to find the number of s s s and s that minimize . So let and represent the number of s, s, s, and s respectively. Then we can write three equations based on these variables. Since there are a total of s, we know that . We also know that and . We can now solve these down to two variables: Substituting this into the second and third equations, we get and The second of these can be reduced to Now we substitute from the first new equation into the other new equation. Since and are integers, the two solutions to this are or . If you plug both these solutions in to it is apparent that the second one returns a smaller value. It turns out that , , , and , so .
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.