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Difference between revisions of "2013 AIME I Problems/Problem 11"
m (→‎Solution: changed equal sign into $\equiv$.)
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<math>nk\equiv 3 \pmod{13}</math>
 
<math>nk\equiv 3 \pmod{13}</math>
  
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Solving these equations, we find <math>n = 131 \pmod{9*11*13}</math> , so the smallest possible value of N is <math>2^{4} \cdot 3 \cdot 5 \cdot 7 \cdot 131</math>
+
Solving these equations, we find <math>n \equiv 131 \pmod{9*11*13}</math> , so the smallest possible value of N is <math>2^{4} \cdot 3 \cdot 5 \cdot 7 \cdot 131</math>
  
 
<math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.
 
<math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.

Revision as of 10:11, 16 July 2013

Problem 11

Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, N, of play blocks which satisfies the conditions:

(a) If 16, 15, or 14 students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and

(b) There are three integers $0 < x < y < z < 14$ such that when $x$, $y$, or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.

Find the sum of the distinct prime divisors of the least possible value of N satisfying the above conditions.


Solution

N must be some multiple of the LCM of 14, 15, and 16 = $2^{4} \cdot 3 \cdot 5 \cdot 7$ ; this LCM is hereby denoted $k$ and $N = nk$.

1, 2, 3, 4, 5, 6, 7, 8, 10, and 12 all divide $k$, so $x, y, z = 9, 11, 13$

We have the following three modulo equations:

$nk\equiv 3 \pmod{9}$

$nk\equiv 3 \pmod{11}$

$nk\equiv 3 \pmod{13}$

Solving these equations, we find $n \equiv 131 \pmod{9*11*13}$ , so the smallest possible value of N is $2^{4} \cdot 3 \cdot 5 \cdot 7 \cdot 131$

$2 + 3 + 5 + 7 + 131 = \boxed{148}$.

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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