2009 AIME I Problems/Problem 14

Problem

For $t = 1, 2, 3, 4$ , define $S_t = \sum_{i = 1}^{350}a_i^t$ , where $a_i \in \{1,2,3,4\}$ . If $S_1 = 513$ and $S_4 = 4745$ , find the minimum possible value for $S_2$ .

Solution

Because the order of the $a$ s doesn't matter, we simply need to find the number of $1$ s $2$ s $3$ s and $4$ s that minimize $S_2$ . So let $w, x, y,$ and $z$ represent the number of $1$ s, $2$ s, $3$ s, and $4$ s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $a$ s, we know that $w + x + y + z = 350$ . We also know that $w + 2x + 3y + 4z = 513$ and $w + 16x + 81y + 256z = 4745$ . We can now solve these down to two variables: $w = 350 - x - y - z$ Substituting this into the second and third equations, we get $x + 2y + 3z = 163$ and $15x + 80y + 255z = 4395.$ The second of these can be reduced to $3x + 16y + 51z = 879.$ Now we substitute $x$ from the first new equation into the other new equation. $x = 163 - 2y - 3z$ $3(163 - 2y - 3z) + 16y + 51z = 879$ $489 + 10y + 42z = 879$ $5y + 21z = 195$ Since $y$ and $z$ are integers, the two solutions to this are $(y,z) = (39,0)$ or $(18,5)$ . If you plug both these solutions in to $S_2$ it is apparent that the second one returns a smaller value. It turns out that $w = 215$ , $x = 112$ , $y = 18$ , and $z = 5$ , so $S_2 = 215 + 4*112 + 9*18 + 16*5 = 215 + 448 + 162 + 80 = \boxed{905}$ .

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
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2009 AIME I Problems/Problem 14

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