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"

(Remove extra problem section)
(→‎Solution 5)
 
(19 intermediate revisions by 9 users not shown)
Line 1: Line 1:
−
== Problem 11 ==
+
== Problem ==
−
Ms. Math's kindergarten class has 16 registered students. The classroom has a very large number, ''N'', of play blocks which satisfies the conditions:
+
Ms. Math's kindergarten class has <math>16</math> registered students. The classroom has a very large number, <math>N</math>, 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
+
(a) If <math>16</math>, <math>15</math>, or <math>14</math> 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 <math>0 < x < y < z < 14</math> such that when <math>x</math>, <math>y</math>, or <math>z</math> students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
 
(b) There are three integers <math>0 < x < y < z < 14</math> such that when <math>x</math>, <math>y</math>, or <math>z</math> 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.
+
Find the sum of the distinct prime divisors of the least possible value of <math>N</math> satisfying the above conditions.
  
 +
==Solution 1==
 +
<math>N</math> must be some multiple of <math>\text{lcm}(14, 15, 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>.
  
−
== Solution 1 ==
+
<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>
−
''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>.
 
−
 
 
−
1, 2, 3, 4, 5, 6, 7, 8, 10, and 12 all divide <math>k</math>, so <math>x, y, z = 9, 11, 13</math>
 
  
 
We have the following three modulo equations:
 
We have the following three modulo equations:
  
−
<math>nk\equiv 3 \pmod{9}</math>
+
<math>nk\equiv 3\pmod{9}</math>
  
−
<math>nk\equiv 3 \pmod{11}</math>
+
<math>nk\equiv 3\pmod{11}</math>
  
−
<math>nk\equiv 3 \pmod{13}</math>
+
<math>nk\equiv 3\pmod{13}</math>
  
 
To solve the equations, you can notice the answer must be of the form <math>9\cdot 11\cdot 13\cdot m + 3</math> where <math>m</math> is an integer.  
 
To solve the equations, you can notice the answer must be of the form <math>9\cdot 11\cdot 13\cdot m + 3</math> where <math>m</math> is an integer.  
  
−
This must be divisible by LCM<math>(14, 15, 16)</math>, which is <math>560\cdot 3</math>.  
+
This must be divisible by <math>lcm</math> <math>(14, 15, 16)</math>, which is <math>560\cdot 3</math>.  
  
−
Therefore, <math>(9\cdot 11\cdot 13m + 3)/(560\cdot 3) = q</math>, which is an integer. Factor out 3 and divide to get <math>(429m+1)/(560) = q</math>.  
+
Therefore, <math>\frac{9\cdot 11\cdot 13\cdot m + 3}{560\cdot 3} = q</math>, which is an integer. Factor out <math>3</math> and divide to get <math>\frac{429m+1}{560} = q</math>.  
−
Therefore, <math>429m+1=560q</math>. We can use Bezout's Identity or a Euclidean Algorithm bash to solve for the least of <math>m</math> and <math>q</math>.  
+
Therefore, <math>429m+1=560q</math>. We can use [[Bezout's Lemma|Bezout's Identity]] or a [[Euclidean algorithm]] bash to solve for the least of <math>m</math> and <math>q</math>.  
  
 
We find that the least <math>m</math> is <math>171</math> and the least <math>q</math> is <math>131</math>.  
 
We find that the least <math>m</math> is <math>171</math> and the least <math>q</math> is <math>131</math>.  
  
−
Since we want to factor <math>1680 \cdot q</math>, don't multiply; we already know that the prime factors of <math>1680</math> are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>, and since <math>131</math> is prime, we have <math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.
+
Since we want to factor <math>1680\cdot q</math>, don't multiply: we already know that the prime factors of <math>1680</math> are <math>2</math>, <math>3</math>, <math>5</math>, and <math>7</math>, and since <math>131</math> is prime, we have <math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.
  
 
==Solution 2==
 
==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>.
+
Note that the number of play blocks is a multiple of the LCM of <math>16</math>, <math>15</math>, and <math>14</math>. The value of this can be found to be <math>(16)(15)(7) = 1680</math>. This number is also divisible by <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>, 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.
+
Thus, <math>1680k\equiv 3</math> when taken mod <math>9</math>, <math>11</math>, <math>13</math>. Since <math>1680</math> is congruent to <math>6</math> mod <math>9</math> and <math>3</math> mod <math>13</math>, and congruent to <math>8</math> mod <math>11</math>, 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 <math>11</math>.
  
−
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:
+
Looking at the last condition shows that <math>k\equiv 10</math> mod <math>11</math> (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 <math>10</math> mod <math>11</math> and <math>1</math> mod <math>13</math> yield the following lists:
  
−
10 mod 11: 21, 32, 43, 54, 65, 76, 87, 98, 109, 120, 131...
+
<math>10</math> mod <math>11</math>: <math>21</math>, <math>32</math>, <math>43</math>, <math>54</math>, <math>65</math>, <math>76</math>, <math>87</math>, <math>98</math>, <math>109</math>, <math>120</math>, <math>131</math>...
  
−
1 mod 13: 14, 27, 40, 53, 66, 79, 92, 105, 118, 131, 144, 157, 170...
+
<math>1</math> mod <math>13</math>: <math>14</math>, <math>27</math>, <math>40</math>, <math>53</math>, <math>66</math>, <math>79</math>, <math>92</math>, <math>105</math>, <math>118</math>, <math>131</math>, <math>144</math>, <math>157</math>, <math>170</math>...
  
 
Both lists contain <math>x</math> elements where <math>x</math> is the modulo 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>.
 
Both lists contain <math>x</math> elements where <math>x</math> is the modulo 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>.
  
−
==Modulus Solution==
+
==Solution 3==
−
It is obvious that <math>N=a*2^4*3*5*7</math> and so the only <math>(mod 3)</math> number of students are <math>9, 11, 13</math>. Therefore, <math>N=1287*k+3</math>. Try some approaches and you will see that this one is one of the few successful ones:
+
It is obvious that <math>N=a\cdot 2^4 \cdot 3\cdot 5\cdot 7</math> and so the only mod <math>3</math> number of students are <math>9, 11, 13</math>. Therefore, <math>N=1287\cdot k+3</math>. Try some approaches and you will see that this one is one of the few successful ones:
 +
 
 +
Start by setting the two <math>N</math> equations together, then we get <math>1680a=1287k+3</math>. Divide by <math>3</math>. Note that since the RHS is <math>1\pmod{3}</math>, and since <math>560</math> is <math>2\pmod{3}</math>, then <math>a=3b+2</math>, where <math>b</math> is some nonnegative integer, because <math>a</math> must be <math>2\pmod{3}</math>.
 +
 
 +
This reduces to <math>560 \cdot 3b + 1119 = 429k</math>. Now, take out the <math>11!</math> With the same procedure, <math>b=11c-1</math>, where <math>c</math> is some nonnegative integer.
 +
 
 +
You also get <math>c=13d+4</math>, at which point <math>k=171+560d</math>. <math>d</math> cannot be equal to <math>0</math>. Therefore, <math>c=4, b=43, a=131</math>, and we know the prime factors of <math>N</math> are <math>2, 3, 5, 7, 131</math> so the answer is <math>\boxed{148}</math>.
 +
 
 +
==Solution 4 ==
 +
 
 +
We start by noticing that <math>N = a\textbf{lcm}(14, 15, 16) = 1680a</math> for some integer <math>a</math> in order to satisfy the first condition.
 +
 
 +
Next, we satisfy the second condition. Since <math>x<y<z < 14</math> must leave a remainder when dividing <math>1680a</math>, they are not divisors of <math>1680x</math>. Thus, we can eliminate all <math>y \le 14</math> s.t. <math>\gcd(y, 1680x) \ne 1</math> which leaves <math>(x, y, z) = (9, 11, 13)</math>. Thus, <math>N = 1680a \equiv 3 \pmod 9 \equiv 3 \pmod {11} \equiv 3 \pmod {13}</math>. Now, we seek to find the least <math>a</math> which satisfies this set of congruences.
 +
 
 +
By Chinese Remainder Theorem on the first two congruences, we find that <math>a \equiv 32 \pmod {33}</math> (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on <math>a \equiv 32 \pmod {33}</math> and <math>1680a \equiv 3 \pmod {13}</math> we find that <math>a \equiv 131 \pmod {429}</math>.
 +
 
 +
Thus, the minimal value of <math>N</math> is possible at <math>a = 131</math>. The prime factorization of this minimum value is <math>2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131</math> and so the answer is <math>2 + 3 + 5 + 7 + 131 = \boxed{148}</math>.
 +
 
 +
 
 +
==Solution 5==
 +
 
 +
As the problem stated, the number of boxes is definitely a multiple of <math>lcm(14,15,16)=1680</math>, so we assume total number of boxes is <math>1680k</math>
 +
 
 +
Then, according to <math>(b)</math> statement, we get <math>1680k \equiv 3 \pmod x \equiv 3 \pmod {y} \equiv 3 \pmod {z}</math>. So we have <math>lcm(x,y,z)+3+m\cdot lcm(x,y,z)=1680k</math>, we just write it to be <math>(1+n)lcm(x,y,z)=1680k-3</math> Which tells that <math>x,y,z</math> must be all odd number. Moreover, we can see <math>(1+m)lcm(x,y,z)</math> can't be a multiple of <math>3,5,7</math>(as <math>1680</math> is a multiple of <math>5,7</math>) which means that <math>lcm(x,y,z)=lcm(9,11,13)=1287</math> We let <math>n=1+m</math>
 +
 
 +
Now, we write <math>1287n+3=1680k, 429n+1=560k</math>
 +
It is true that <math>n\equiv 1 \pmod{10}</math>, let <math>n=10p+1</math>, it has <math>429p+43=56k</math> Then, <math>p</math> must be odd, let <math>p=2q+1</math>, it indicates <math>429q+236=28k</math> Now, <math>q</math> must be even, <math>q=2s</math> tells <math>429s+118=14k</math> Eventually, <math>s</math> must be even, <math>s=2y</math>, <math>858y+59=7k</math>, <math>y=1</math> is the smallest. This time, <math>k=131</math>
 +
 
 +
So the number of balls is <math>1680\cdot 131=2^4\cdot 3\cdot 5\cdot 7 \cdot 131</math>, the desired value is <math>2+3+5+7+131=\boxed{148}</math>
 +
 
 +
~bluesoul
 +
 
  
−
Start by setting the two <math>N</math> equations together, then we get <math>1680a=1287k+3</math>. Divide by 3. Note that since the RHS is <math>1 (mod 3)</math>, and since <math>560</math> is <math>2 (mod 3)</math>, then <math>a=3b+2</math>, where <math>b</math> is some nonnegative integer, because <math>a</math> must be <math>2 (mod 3)</math>.
+
==Solution 6(CRT Bash)==
 +
From part (a), we know that <math>2^4\cdot3\cdot5\cdot7 | N</math>. From part (b), we know that <math>N \equiv 3 \pmod {1287}</math>. We can expand on part (a) by saying that <math>N = 1680k</math> for some <math>k</math>. Rather than taking the three modulos together, we take them individually.
 +
<cmath> 1680k \equiv 6k \equiv 3 \pmod 9</cmath>
 +
<cmath>k \equiv 2^{-1} \pmod 9</cmath>
 +
The inverse of 2 mod 9 is easily seen to be <math>5</math>.
 +
<cmath>k \equiv 5 \pmod 9</cmath>
  
−
This reduces to <math>560 * 3b + 1119 = 429k</math>. Now, take out the 11! With the same procedure, <math>b=11c-1</math>, where <math>c</math> is some nonnegative integer.
+
Now moving to the second modulo which we leave as follows,
 +
<cmath> 1680k \equiv 8k \equiv 3 \pmod {11} </cmath>
 +
Now the last modulo,
 +
<cmath>1680k \equiv 3k \equiv 3 \pmod {13} </cmath>
 +
<cmath>k \equiv 1 \pmod{13} </cmath>
 +
CRT on the first and the third one results in <math>k \equiv 4 \pmod {117}</math>. Now doing the second one and the one we just made, <math>k \equiv 131 \pmod{1287}</math>. Thus, the smallest value that works for <math>k = 131</math>. Thus <math>N = 2^4\cdot3\cdot5\cdot7\cdot131</math> <math>2+3+5+7+131 = \boxed{148}</math>
  
−
You also get <math>c=13d+4</math>, at which point <math>k=171+560d</math>. D CAN BE ZERO!! Therefore, <math>c=4, b=43, a=131</math>, and we know the prime factors of <math>N</math> are <math>2, 3, 5, 7, 131</math> so the answer is <math>\boxed{148}</math>!
+
~YBSuburbanTea
  
−
== 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}}

Latest revision as of 11:48, 20 December 2022

Problem

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 $\text{lcm}(14, 15, 16)= 2^{4}\cdot 3\cdot 5\cdot 7$ ; 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, $\frac{9\cdot 11\cdot 13\cdot m + 3}{560\cdot 3} = q$, 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 $1\pmod{3}$, and since $560$ is $2\pmod{3}$, then $a=3b+2$, where $b$ is some nonnegative integer, because $a$ must be $2\pmod{3}$.

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 $\boxed{148}$.

Solution 4

We start by noticing that $N = a\textbf{lcm}(14, 15, 16) = 1680a$ for some integer $a$ in order to satisfy the first condition.

Next, we satisfy the second condition. Since $x<y<z < 14$ must leave a remainder when dividing $1680a$, they are not divisors of $1680x$. Thus, we can eliminate all $y \le 14$ s.t. $\gcd(y, 1680x) \ne 1$ which leaves $(x, y, z) = (9, 11, 13)$. Thus, $N = 1680a \equiv 3 \pmod 9 \equiv 3 \pmod {11} \equiv 3 \pmod {13}$. Now, we seek to find the least $a$ which satisfies this set of congruences.

By Chinese Remainder Theorem on the first two congruences, we find that $a \equiv 32 \pmod {33}$ (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on $a \equiv 32 \pmod {33}$ and $1680a \equiv 3 \pmod {13}$ we find that $a \equiv 131 \pmod {429}$.

Thus, the minimal value of $N$ is possible at $a = 131$. The prime factorization of this minimum value is $2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131$ and so the answer is $2 + 3 + 5 + 7 + 131 = \boxed{148}$.


Solution 5

As the problem stated, the number of boxes is definitely a multiple of $lcm(14,15,16)=1680$, so we assume total number of boxes is $1680k$

Then, according to $(b)$ statement, we get $1680k \equiv 3 \pmod x \equiv 3 \pmod {y} \equiv 3 \pmod {z}$. So we have $lcm(x,y,z)+3+m\cdot lcm(x,y,z)=1680k$, we just write it to be $(1+n)lcm(x,y,z)=1680k-3$ Which tells that $x,y,z$ must be all odd number. Moreover, we can see $(1+m)lcm(x,y,z)$ can't be a multiple of $3,5,7$(as $1680$ is a multiple of $5,7$) which means that $lcm(x,y,z)=lcm(9,11,13)=1287$ We let $n=1+m$

Now, we write $1287n+3=1680k, 429n+1=560k$ It is true that $n\equiv 1 \pmod{10}$, let $n=10p+1$, it has $429p+43=56k$ Then, $p$ must be odd, let $p=2q+1$, it indicates $429q+236=28k$ Now, $q$ must be even, $q=2s$ tells $429s+118=14k$ Eventually, $s$ must be even, $s=2y$, $858y+59=7k$, $y=1$ is the smallest. This time, $k=131$

So the number of balls is $1680\cdot 131=2^4\cdot 3\cdot 5\cdot 7 \cdot 131$, the desired value is $2+3+5+7+131=\boxed{148}$

~bluesoul


Solution 6(CRT Bash)

From part (a), we know that $2^4\cdot3\cdot5\cdot7 | N$. From part (b), we know that $N \equiv 3 \pmod {1287}$. We can expand on part (a) by saying that $N = 1680k$ for some $k$. Rather than taking the three modulos together, we take them individually. \[1680k \equiv 6k \equiv 3 \pmod 9\] \[k \equiv 2^{-1} \pmod 9\] The inverse of 2 mod 9 is easily seen to be $5$. \[k \equiv 5 \pmod 9\]

Now moving to the second modulo which we leave as follows, \[1680k \equiv 8k \equiv 3 \pmod {11}\] Now the last modulo, \[1680k \equiv 3k \equiv 3 \pmod {13}\] \[k \equiv 1 \pmod{13}\] CRT on the first and the third one results in $k \equiv 4 \pmod {117}$. Now doing the second one and the one we just made, $k \equiv 131 \pmod{1287}$. Thus, the smallest value that works for $k = 131$. Thus $N = 2^4\cdot3\cdot5\cdot7\cdot131$ $2+3+5+7+131 = \boxed{148}$

~YBSuburbanTea

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=184331"