Difference between revisions of "2013 AIME I Problems/Problem 11"

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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} * 3 * 5 * 7 * 131</math>
 
Solving these equations, we find <math>n = 131 \pmod{9*11*13}</math> , so the smallest possible value of N is <math>2^{4} * 3 * 5 * 7 * 131</math>
 +
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<math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2013|n=I|num-b=10|num-a=12}}
 
<math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>
 

Revision as of 20:47, 16 March 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 = $2^{4} * 3 * 5 * 7$ ; 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 $n = 131 \pmod{9*11*13}$ , so the smallest possible value of N is $2^{4} * 3 * 5 * 7 * 131$

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

See also

2013 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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