Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AIME I Problems/Problem 11"
(→‎Solution)
Line 9: Line 9:
  
  
−
== Solution ==
+
== Solution 1 ==
 
''N'' must be some multiple of the LCM of 14, 15, and 16 = <math>2^{4} \cdot 3 \cdot 5 \cdot 7</math> ; this LCM is hereby denoted <math>k</math> and <math>N = qk</math>.
 
''N'' must be some multiple of the LCM of 14, 15, and 16 = <math>2^{4} \cdot 3 \cdot 5 \cdot 7</math> ; this LCM is hereby denoted <math>k</math> and <math>N = qk</math>.
  
Line 35: Line 35:
  
 
<math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.
 
<math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.
 +
 +
==Solution 2==
 +
Note that the number of play blocks is a multiple of the LCM of 16, 15, and 14. The value of this can be found to be <math>(16)(15)(7) = 1680</math>. This number is also divisible by 1, 2, 3, 4, 5, 6, 7, 8, 10, and 12, thus, the three numbers <math>x, y, z</math> are <math>9, 11, 13</math>.
 +
 +
Thus, <math>1680k \equiv 3</math> when taken mod 9, 11, 13. Since <math>1680</math> is congruent to 6 mod 9 and 3 mod 13, and congruent to 8 mod 11, the number <math>k</math> must be a number that is congruent to <math>1</math> mod <math>13</math>, <math>2</math> mod <math>3</math> (because <math>6</math> is a multiple of <math>3</math>, which is a factor of <math>9</math> that can be divided out) and cause <math>8</math> to become <math>3</math> when multiplied under modulo 11.
 +
 +
Looking at the last condition shows that <math>k \equiv 10</math> mod 11 (after a bit of bashing) and is congruent to <math>1</math> mod <math>13</math> and <math>2</math> mod <math>3</math> as previously noted. Listing out the numbers congruent to 10 mod 11 and 1 mod 13 yield the following lists:
 +
10 mod 11: 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, 131...
 +
1 mod 13: 14, 27, 40, 53, 66, 79, 92, 105, 118, 131, 144, 157, 170...
 +
Both lists contain <math>x</math> elements where <math>x</math> is the mod being taken, thus, there must be a solution in these lists as adding <math>11(13)</math> to this solution yields the next smallest solution. In this case, <math>131</math> is the solution for <math>k</math> and thus the answer is <math>1680(131)</math>. Since <math>131</math> is prime, the sum of the prime factors is <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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:09, 20 March 2017

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 1

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 $N = qk$.

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:

$nk\equiv 3 \pmod{9}$

$nk\equiv 3 \pmod{11}$

$nk\equiv 3 \pmod{13}$

To solve the equations, you can notice the answer must be of the form $9\cdot 11\cdot 13\cdot m + 3$ where $m$ is an integer.

This must be divisible by LCM$(14, 15, 16)$, which is $560\cdot 3$.

Therefore, $(9\cdot 11\cdot 13m + 3)/(560\cdot 3) = q$, which is an integer. Factor out 3 and divide to get $(429m+1)/(560) = q$. Therefore, $429m+1=560q$. We can use Bezout's Identity or a Euclidean Algorithm bash to solve for the least of $m$ and $q$.

We find that the least $m$ is $171$ and the least $q$ is $131$.

Plug it into $9\cdot 11\cdot 13m + 3$ and factor to get that the distinct prime divisors are $2,3,5,7$ and $131$.


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

Solution 2

Note that the number of play blocks is a multiple of the LCM of 16, 15, and 14. The value of this can be found to be $(16)(15)(7) = 1680$. This number is also divisible by 1, 2, 3, 4, 5, 6, 7, 8, 10, and 12, thus, the three numbers $x, y, z$ are $9, 11, 13$.

Thus, $1680k \equiv 3$ when taken mod 9, 11, 13. Since $1680$ is congruent to 6 mod 9 and 3 mod 13, and congruent to 8 mod 11, the number $k$ must be a number that is congruent to $1$ mod $13$, $2$ mod $3$ (because $6$ is a multiple of $3$, which is a factor of $9$ that can be divided out) and cause $8$ to become $3$ when multiplied under modulo 11.

Looking at the last condition shows that $k \equiv 10$ mod 11 (after a bit of bashing) and is congruent to $1$ mod $13$ and $2$ mod $3$ as previously noted. Listing out the numbers congruent to 10 mod 11 and 1 mod 13 yield the following lists: 10 mod 11: 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, 131... 1 mod 13: 14, 27, 40, 53, 66, 79, 92, 105, 118, 131, 144, 157, 170... Both lists contain $x$ elements where $x$ is the mod being taken, thus, there must be a solution in these lists as adding $11(13)$ to this solution yields the next smallest solution. In this case, $131$ is the solution for $k$ and thus the answer is $1680(131)$. Since $131$ is prime, the sum of the prime factors is $2 + 3 + 5 + 7 + 131 = \boxed{148}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_11&oldid=84811"