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

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<math>N</math> must be some multiple of the LCM of <math>14</math>, <math>15</math>, and <math>16 = 2^{4}\cdot 3\cdot 5\cdot 7</math> ; this <math>lcm</math> is hereby denoted <math>k</math> and <math>N = qk</math>.
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<math>N</math> must be some multiple of the <math>lcm</math> of <math>14</math>, <math>15</math>, and <math>16 = 2^{4}\cdot 3\cdot 5\cdot 7</math> ; this <math>lcm</math> is hereby denoted <math>k</math> and <math>N = qk</math>.
  
 
<math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>10</math>, and <math>12</math> all divide <math>k</math>, so <math>x, y, z = 9, 11, 13</math>
 
<math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>10</math>, and <math>12</math> all divide <math>k</math>, so <math>x, y, z = 9, 11, 13</math>

Revision as of 16:53, 7 August 2020

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

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, $\frac{9\cdot 11\cdot 13m + 3}{560\cdot 3} = q$, which is an integer. Factor out $3$ and divide to get $\frac{429m+1}{560} = q$. 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 $131$.

Since we want to factor $1680\cdot q$, don't multiply: we already know that the prime factors of $1680$ are $2$, $3$, $5$, and $7$, and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = \boxed{148}$.

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

Solution 3

It is obvious that $N=a\cdot 2^4 \cdot 3\cdot 5\cdot 7$ and so the only mod $3$ number of students are $9, 11, 13$. Therefore, $N=1287\cdot k+3$. Try some approaches and you will see that this one is one of the few successful ones:

Start by setting the two $N$ equations together, then we get $1680a=1287k+3$. Divide by $3$. Note that since the RHS is $1 (mod 3)$, and since $560$ is $2 (mod 3)$, then $a=3b+2$, where $b$ is some nonnegative integer, because $a$ must be $2 (mod 3)$.

This reduces to $560 \cdot 3b + 1119 = 429k$. Now, take out the $11!$ With the same procedure, $b=11c-1$, where $c$ is some nonnegative integer.

You also get $c=13d+4$, at which point $k=171+560d$. $d$ cannot be equal to $0$. Therefore, $c=4, b=43, a=131$, and we know the prime factors of $N$ are $2, 3, 5, 7, 131$ so the answer is $\boxed{148}$.

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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