Difference between revisions of "2017 AIME I Problems/Problem 9"

Latest revision as of 12:54, 2 November 2024

Problem 9

Let $a_{10} = 10$ , and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$ . Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$ .

Solution 1

Writing out the recursive statement for $a_n, a_{n-1}, \dots, a_{10}$ and summing them gives $a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10$ Which simplifies to $a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)$ Therefore, $a_n$ is divisible by 99 if and only if $\frac{1}{2}(n+10)(n-9)$ is divisible by 99, so $(n+10)(n-9)$ needs to be divisible by 9 and 11. Assume that $n+10$ is a multiple of 11. Writing out a few terms, $n=12, 23, 34, 45$ , we see that $n=45$ is the smallest $n$ that works in this case. Next, assume that $n-9$ is a multiple of 11. Writing out a few terms, $n=20, 31, 42, 53$ , we see that $n=53$ is the smallest $n$ that works in this case. The smallest $n$ is $\boxed{045}$ .

Note that we can also construct the solution using CRT by assuming either $11$ divides $n+10$ and $9$ divides $n-9$ , or $9$ divides $n+10$ and $11$ divides $n-9$ , and taking the smaller solution.

Another way to get the quadratic

Writing out the first couple of terms modulo $99$ , we find $a_n = a_{n-1}+n$ so we have $a_{10}=10,a_{11}=21,a_{12}=33,...$ We can compute their finite differences, $10,21,33,46,...$ $11,12,13,...$ $1,1,...$ Since there are 3 rows we know it is a quadratic and we can continue, by finding the quadratic passing through $(10,10),(11,21),(12,33)$ to get $\frac{(n^2+n-90)}{2}.$ -mathkiddus

Solution 2

$a_n \equiv a_{n-1} + n \pmod {99}$ By looking at the first few terms, we can see that $a_n \equiv 10+11+12+ \dots + n \pmod {99}$ This implies $a_n \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}$ Since $a_n \equiv 0 \pmod {99}$ , we can rewrite the equivalence, and simplify $0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99}$ $0 \equiv n(n+1) - 90 \pmod {99}$ $0 \equiv 4n^2+4n+36 \pmod {99}$ $0 \equiv (2n+1)^2+35 \pmod {99}$ $64 \equiv (2n+1)^2 \pmod {99}$ The only squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$ , so $2n+1 \equiv -8, 8, 19, \text{or } {-19} \pmod {99}$ $2n+1 \equiv -8 \pmod {99}$ yields $n=45$ as the smallest integer solution.

$2n+1 \equiv 8 \pmod {99}$ yields $n=53$ as the smallest integer solution.

$2n+1 \equiv -19 \pmod {99}$ yields $n=89$ as the smallest integer solution.

$2n+1 \equiv 19 \pmod {99}$ yields $n=9$ as the smallest integer solution. However, $n$ must be greater than $10$ .

The smallest positive integer solution greater than $10$ is $n=\boxed{045}$ .

Question

Is there an efficient way to notice that the only squares that are congruent to $64 \pmod {99}$ are $(\pm 8)^2$ and $(\pm 19)^2$ ?

Answer: Yes.

We will solve the question by looking for solutions for $m\in[-44,-43,\cdots,-1,0,1,\cdots,43,44]$ . Let the square be $m$ , thus satisfying $m^2-64\equiv 0\pmod{99}$ and $m^2-64=(m+8)(m-8)=99k$ for some integer $k$ . Checking the first factor, $(m+8)$ , for factors of $11$ yields:

$m+8=0\Rightarrow m-8=-16$

$m+8=11\Rightarrow m-8=-5$

$m+8=22\Rightarrow m-8=6$

$m+8=33\Rightarrow m-8=17$

$m+8=44\Rightarrow m-8=28$

In the first case, the product does divide $99$ , so $m=-8\equiv91$ is a solution. For the others, since the first factor already divides $11$ , the second factor must divide $9$ (or $3$ in the case of $m+8=33$ , which already has a factor of $3$ ) in order for the product to divide $99$ . Here, that is not the case, so $m=-8$ is the only possible solution.

Now we check the second factor, $(m-8)$ :

$m-8=0\Rightarrow m-8=16$

$m-8=11\Rightarrow m-8=27$

$m-8=22\Rightarrow m-8=38$

$m-8=33\Rightarrow m-8=49$

$m-8=44\Rightarrow m-8=60$

We immediately see that $m=8$ is a solution. Using a similar argument as above for the others, we notice that $m=19$ is the only solution in this group. Using the property that $ab\equiv(-a)(-b)\mod99$ , it is clear that $m=-19$ is also a solution. As a result, $m=\pm8$ and $m=\pm19$ are the only solutions. (This process takes a much shorter time than it seems.)

~eevee9406

Solution 3

$a_n=a_{n-1} + n \pmod{99}$ . Using the steps of the previous solution we get up to $n^2+n \equiv 90 \pmod{99}$ . This gives away the fact that $(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}$ so either $n$ or $n+1$ must be a multiple of 9.

Case 1 ( $9 \mid n$ ): Say $n=9x$ and after simplification $x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}$ .

Case 2: ( $9 \mid n+1$ ): Say $n=9a-1$ and after simplification $(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}$ .

As a result $a$ must be a divisor of $10$ and after doing some testing in both cases the smallest value that works is $x=5 \implies \boxed{045}$ .

~First

Solution 4 (not good, risky)

We just notice that $100 \equiv 1 \pmod{99}$ , so we are just trying to find $10 + 11 + 12 + \cdots + n$ modulo $99$ , or $\dfrac{n(n+1)}{2} - 45$ modulo $99$ . Also, the sum to $44$ is divisible by $99$ , and is the first one that is. Thus, if we sum to $45$ the $45$ is cut off and thus is just a sum to $44$ .

Without checking whether there are other sums congruent to $45 \pmod{99}$ , we can just write the answer to be $\boxed{045}$ .

Solution 5

Let $b_n = 2a_{n+10}$ . We can find a formula for $b_n$ :

$b_n = (20+n)(n+1)$ .

Notice that both can't have a factor of 3. Thus we can limit our search range of n to $n \equiv 7,8 \pmod{9}$ . Testing values for n in our search range (like 7,8,16,17,25,26...), we get that 35 is the least n. But, don't write that down! Remember, $b_n = 2a_{n+10}$ , so, the 35th term in b corresponds to the 45th term in a. Thus our answer is $\boxed{045}$ .

-AlexLikeMath

Solution 6 (bash, slower, but safer)

The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. $10, 1011, 101112, \dots$ . Now the fact that the sequence starts at $10$ can be completely discarded for this solution. Just consider $a(10)$ then same as $a(1)$ , and we can add nine to the answer at the end.

The second step is to split $99$ as $9$ and $11$ and solve for divisibility rules individually. Let's start with $11$ because it gives us the most information to continue.

In any number generated, if the numbers don't go beyond $20$ , then the highest number we can get is $10111213141516171819$ , with every odd digit being $1$ . This is a little risky because we are assuming that it doesn't exceed $20$ . If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by $11$ . The expression being $\frac{((n-1) + 0)\cdot n)}{2}-n$ .

The concept here is to add $0$ to the $n-1^{th}$ term altogether, then subtract the number of ones in it, which is $n$ . Simplify to $\frac{n(n-3)}{2}$ congruent to $0 \pmod{11}$ . Now notice the divide by two can be discarded because one of $n$ or $(n-3)$ will be even. So if $n$ or $n-3$ is to be divisible by $11$ , we can make a simple list.

$n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, \dots$

Now we test each $n$ for divisibility by $9$ . This is done by making a list that ultimately calculates the sum of every digit in the large number. $n(1)$ to $n(10)$ has the first digit $1$ . $n(11)$ to $n(20)$ has the first digit $2$ , and so on. The necessary thing to realize is that the sum of all digits $0-9$ is divisible by $9$ , so we only have to solve for the sum of the first digits, and then the short list of second digits.

For example, let's test $n=25$ .

So we know that $25$ include both $1-10$ and $11-20$ , so that's $10 + 20$ right away. $21-25$ contains $5$ numbers that have the first digit $3$ , so $+15$ . Then we add $0-4$ together, which is $10$ . $10+20+15+10=55$ , which is not divisible by $9$ , so it is not the answer.

Do this for just a minute you get that $36$ sums to $99$ , a multiple of nine! So $n(36)$ is the answer, right? Don't forget we have to add $9$ because we translated $n(10)$ to $n(1)$ at the very beginning! Finally, after a short bash, we get $\boxed{045}$ .

-jackshi2006 ( $\text{[math]}$ by PureSwag)

Solution 7 (Bash Bash Bash)

We will work $\mod 99$ . The recursive formula now becomes $a_n=a_{n-1}+n$ . Now, we will bash. For convenience, everything is taken $\mod 99$ . The sequence is $\begin{align*}a_{10}&=10 \\ a_{11}&=10+11=21 \\ a_{12}&=12+21=33 \\ a_{13}&=13+33=46 \\ a_{14}&=46+14=60 \\ a_{15}&=60+15=75 \\ a_{16}&=75+16=91 \\ a_{17}&=91+17=9 \\ a_{18}&=9+18=27 \\ a_{19}&=27+19=46 \\ a_{20}&=46+20=66 \\ a_{21}&=66+21=87 \\ a_{22}&=87+22=10 \\ a_{23}&=10+23=33 \\ a_{24}&=33+24=57 \\ a_{25}&=57+25=82 \\ a_{26}&=82+26=9 \\ a_{27}&=9+27=36 \\ a_{28}&=36+28=64 \\ a_{29}&=64+29=93 \\ a_{30}&=93+30=24 \\ a_{31}&=24+31=55 \\ a_{32}&=55+32=87 \\ a_{33}&=87+33=21 \\ a_{34}&=21+34=55 \\ a_{35}&=55+35=90 \\ a_{36}&=90+36=27 \\ a_{37}&=27+37=64 \\ a_{38}&=64+38=3 \\ a_{39}&=3+39=42 \\ a_{40}&=42+40=82 \\ a_{41}&=82+41=24 \\ a_{42}&=24+42=66 \\ a_{43}&=66+43=10 \\ a_{44}&=10+44=54 \\ a_{45}&=54+45=0.\end{align*}$ Hence the least number $n$ is $n=45$ .

~typos fixed by lpieleanu

Solution 8

Taking the recurrence mod $99$ , we have $a_n=a_{n-1}+n$ for $a_{10}=10$ . Then, we have $a_n=10+11+12+\cdots+n \implies a_n=\frac{n(n+1)}2-45 \equiv 0 \pmod{99} \implies n(n+1)-90 \equiv 0 \pmod{99} \implies n(n+1)+9 \equiv 0 \pmod{99} \implies n \equiv 0 \pmod{9}.$ Then, we simply can test these values of $n$ to find that $n=\boxed{045}$ produces a value that is also divisible by $11$ .

-A1001

Note

By the way, if you're wondering, $a_{45}$ is the $72$ -digit number $101\text{,}112\text{,}131\text{,}415\text{,}161\text{,}718\text{,}192\text{,}021\text{,}222\text{,}324\text{,}252\text{,}627\text{,}282\text{,}930\text{,}313\text{,}233\text{,}343\text{,}536\text{,}373\text{,}839\text{,}404\text{,}142\text{,}434\text{,}445.$ The prime factorization of $a_{45}$ is $3^{2}~\cdot~5~\cdot~11~\cdot~151~\cdot~1381~\cdot~1559~\cdot~1\text{,}511\text{,}647~\cdot~448\text{,}966\text{,}261\text{,}198\text{,}213\text{,}862\text{,}368\text{,}469~\cdot~925\text{,}800\text{,}120\text{,}162\text{,}193\text{,}934\text{,}310\text{,}647\text{,}599\text{,}013$ and $\frac{a_{45}}{99}$ is the $70$ -digit number $1\text{,}021\text{,}334\text{,}660\text{,}759\text{,}209\text{,}274\text{,}666\text{,}881\text{,}033\text{,}578\text{,}309\text{,}366\text{,}494\text{,}245\text{,}588\text{,}215\text{,}591\text{,}276\text{,}503\text{,}428\text{,}324\text{,}671\text{,}055.$

@@ Line 1: / Line 1: @@
 ==Problem 9==
-Let <math>a_{10} = 10</math>, and for each integer <math>n >10</math> let <math>a_n = 100a_{n - 1} + n</math>. Find the least <math>n > 10</math> such that <math>a_n</math> is a multiple of <math>99</math>.
+Let <math>a_{10} = 10</math>, and for each positive integer <math>n >10</math> let <math>a_n = 100a_{n - 1} + n</math>. Find the least positive <math>n > 10</math> such that <math>a_n</math> is a multiple of <math>99</math>.
 ==Solution 1==
@@ Line 8: / Line 8: @@
 Note that we can also construct the solution using CRT by assuming either <math>11</math> divides <math>n+10</math> and <math>9</math> divides <math>n-9</math>, or <math>9</math> divides <math>n+10</math> and <math>11</math> divides <math>n-9</math>, and taking the smaller solution.
+==Another way to get the quadratic==
+Writing out the first couple of terms modulo <math>99</math>, we find <math>a_n = a_{n-1}+n</math> so we have <math>a_{10}=10,a_{11}=21,a_{12}=33,...</math> We can compute their finite differences, <cmath>10,21,33,46,...</cmath><cmath>11,12,13,...</cmath><cmath>1,1,...</cmath> Since there are 3 rows we know it is a quadratic and we can continue, by finding the quadratic passing through <math>(10,10),(11,21),(12,33)</math> to get <math>\frac{(n^2+n-90)}{2}.</math>
+-mathkiddus
 ==Solution 2==
@@ Line 32: / Line 35: @@
 The smallest positive integer solution greater than <math>10</math> is <math>n=\boxed{045}</math>.
+==Question==
+Is there an efficient way to notice that the only squares that are congruent to <math>64 \pmod {99}</math> are <math>(\pm 8)^2</math> and <math>(\pm 19)^2</math>?
+<b>Answer:</b> Yes.
+We will solve the question by looking for solutions for <math>m\in[-44,-43,\cdots,-1,0,1,\cdots,43,44]</math>. Let the square be <math>m</math>, thus satisfying <math>m^2-64\equiv 0\pmod{99}</math>  and <math>m^2-64=(m+8)(m-8)=99k</math> for some integer <math>k</math>. Checking the first factor, <math>(m+8)</math>, for factors of <math>11</math> yields:
+<math>m+8=0\Rightarrow m-8=-16</math>
+<math>m+8=11\Rightarrow m-8=-5</math>
+<math>m+8=22\Rightarrow m-8=6</math>
+<math>m+8=33\Rightarrow m-8=17</math>
+<math>m+8=44\Rightarrow m-8=28</math>
+In the first case, the product does divide <math>99</math>, so <math>m=-8\equiv91</math> is a solution. For the others, since the first factor already divides <math>11</math>, the second factor must divide <math>9</math> (or <math>3</math> in the case of <math>m+8=33</math>, which already has a factor of <math>3</math>) in order for the product to divide <math>99</math>. Here, that is not the case, so <math>m=-8</math> is the only possible solution.
+Now we check the second factor, <math>(m-8)</math>:
+<math>m-8=0\Rightarrow m-8=16</math>
+<math>m-8=11\Rightarrow m-8=27</math>
+<math>m-8=22\Rightarrow m-8=38</math>
+<math>m-8=33\Rightarrow m-8=49</math>
+<math>m-8=44\Rightarrow m-8=60</math>
+We immediately see that <math>m=8</math> is a solution. Using a similar argument as above for the others, we notice that <math>m=19</math> is the only solution in this group. Using the property that <math>ab\equiv(-a)(-b)\mod99</math>, it is clear that <math>m=-19</math> is also a solution. As a result, <math>m=\pm8</math> and <math>m=\pm19</math> are the only solutions. (This process takes a much shorter time than it seems.)
+~eevee9406
 ==Solution 3==
 <math>a_n=a_{n-1} + n \pmod{99}</math>.  Using the steps of the previous solution we get up to <math>n^2+n \equiv 90 \pmod{99}</math>. This gives away the fact that <math>(n)(n+1) \equiv 0 \pmod{9} \implies n \equiv \{0, 8\} \pmod{9}</math> so either <math>n</math> or <math>n+1</math> must be a multiple of 9.
-Case 1 (<math>n|9</math>): Say <math>n=9x</math> and after simplification <math>x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}</math>.
+Case 1 (<math>9 \mid n</math>): Say <math>n=9x</math> and after simplification <math>x(9x+1) = 10 \pmod{90} \forall x \in \mathbb{Z}</math>.
-Case 2: (<math>n+1|9</math>): Say <math>n=9a-1</math> and after simplification <math>(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}</math>.
+Case 2: (<math>9 \mid n+1</math>): Say <math>n=9a-1</math> and after simplification <math>(9a-1)(a) = 10 \pmod{90} \forall a \in \mathbb{Z}</math>.
 As a result <math>a</math> must be a divisor of <math>10</math> and after doing some testing in both cases the smallest value that works is <math>x=5 \implies \boxed{045}</math>.
@@ Line 60: / Line 98: @@
 ==Solution 6 (bash, slower, but safer)==
-The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. 10, 1011, 101112, etc. Now the fact that the sequence starts at 10 can be completely discarded for this solution. Just consider a(10) then same as a(1), and we can add nine to the answer at the end.
+The first thing you should realize is that each term after the tenth is another two-digit number chained to the last number. <math>10, 1011, 101112, \dots</math>. Now the fact that the sequence starts at <math>10</math> can be completely discarded for this solution. Just consider <math>a(10)</math> then same as <math>a(1)</math>, and we can add nine to the answer at the end.
-The second step is to split 99 as 9 and 11 and solve for divisibility rules individually. Let's start with 11 because it gives us the most information to continue.
+The second step is to split <math>99</math> as <math>9</math> and <math>11</math> and solve for divisibility rules individually. Let's start with <math>11</math> because it gives us the most information to continue.
-In any number generated, if the numbers don't go beyond 20, then the highest number we can get is 10111213141516171819, with every odd digit being 1. This is a little risky because we are assuming that it doesn't exceed 20. If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by 11. The expression being ((((n-1) + 0) * n)/2)-n. (apologies for latex illiteracy, please add latex if you can) I don't think I need to go in-depth, but the concept here is to add 0 to (nth-1 term) altogether, then subtract the number of ones in it, which is n. Simplify to (n(n-3))/2 congruent to 0 mod 11. Now notice the divide by two can be discarded because one of n or (n-3) will be even. So if n or n-3 is to be divisible by 11, we can make a simple list.
+In any number generated, if the numbers don't go beyond <math>20</math>, then the highest number we can get is <math>10111213141516171819</math>, with every odd digit being <math>1</math>. This is a little risky because we are assuming that it doesn't exceed <math>20</math>. If someone wanted to be absolutely sure they could continue, but this is unnecessary later and a big hassle. Anyways, now we write an equation to check for divisibility by <math>11</math>. The expression being <math>\frac{((n-1) + 0)\cdot n)}{2}-n</math>.
-n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, etc.
+The concept here is to add <math>0</math> to the <math>n-1^{th}</math> term altogether, then subtract the number of ones in it, which is <math>n</math>. Simplify to <math>\frac{n(n-3)}{2}</math> congruent to <math>0 \pmod{11}</math>. Now notice the divide by two can be discarded because one of <math>n</math> or <math>(n-3)</math> will be even. So if <math>n</math> or <math>n-3</math> is to be divisible by <math>11</math>, we can make a simple list.
+<cmath>n = 0, 3, 11, 14, 22, 25, 33, 36, 44, 47, \dots</cmath>
-Now we test each n for divisibility by 9.
+Now we test each <math>n</math> for divisibility by <math>9</math>.
 This is done by making a list that ultimately calculates the sum of every digit in the large number.
-n(1) to n(10) has the first digit 1. n(11) to n(20) has the first digit 2, and so on.
+<math>n(1)</math> to <math>n(10)</math> has the first digit <math>1</math>. <math>n(11)</math> to <math>n(20)</math> has the first digit <math>2</math>, and so on.
-The necessary thing to realize is that the sum of all digits 0-9 is divisible by 9, so we only have to solve for the sum of the first digits, and then the short list of second digits.
+The necessary thing to realize is that the sum of all digits <math>0-9</math> is divisible by <math>9</math>, so we only have to solve for the sum of the first digits, and then the short list of second digits.
-For example, let's test n=25.
+For example, let's test <math>n=25</math>.
-So we know that 25 include both 1-10 and 11-20, so that's 10 + 20 right away. 21-25 contains 5 numbers that have the first digit 3, so +15. Then we add 0-4 together, which is 10. 10+20+15+10=55, which is not divisible by 9, so it is not the answer.
+So we know that <math>25</math> include both <math>1-10</math> and <math>11-20</math>, so that's <math>10 + 20</math> right away. <math>21-25</math> contains <math>5</math> numbers that have the first digit <math>3</math>, so <math>+15</math>. Then we add <math>0-4</math> together, which is <math>10</math>. <math>10+20+15+10=55</math>, which is not divisible by <math>9</math>, so it is not the answer.
-Do this for just a minute you get that 36 sums to 99, a multiple of nine! So n(36) is the answer, right? Don't forget we have to add 9 because we translated n(10) to n(1) at the very beginning!
+Do this for just a minute you get that <math>36</math> sums to <math>99</math>, a multiple of nine! So <math>n(36)</math> is the answer, right? Don't forget we have to add <math>9</math> because we translated <math>n(10)</math> to <math>n(1)</math> at the very beginning!
 Finally, after a short bash, we get <math>\boxed{045}</math>.
 -jackshi2006
+(<math>LaTeX</math> by PureSwag)
+==Solution 7 (Bash Bash Bash)==
+We will work<math>\mod 99</math>. The recursive formula now becomes <math>a_n=a_{n-1}+n</math>. Now, we will bash. For convenience, everything is taken <math>\mod 99</math>. The sequence is
+<cmath>\begin{align*}a_{10}&=10 \\ a_{11}&=10+11=21 \\ a_{12}&=12+21=33 \\ a_{13}&=13+33=46 \\ a_{14}&=46+14=60 \\ a_{15}&=60+15=75 \\ a_{16}&=75+16=91 \\ a_{17}&=91+17=9 \\ a_{18}&=9+18=27 \\ a_{19}&=27+19=46 \\ a_{20}&=46+20=66 \\ a_{21}&=66+21=87 \\ a_{22}&=87+22=10 \\ a_{23}&=10+23=33 \\ a_{24}&=33+24=57 \\ a_{25}&=57+25=82 \\ a_{26}&=82+26=9 \\ a_{27}&=9+27=36 \\ a_{28}&=36+28=64 \\ a_{29}&=64+29=93 \\ a_{30}&=93+30=24 \\ a_{31}&=24+31=55 \\ a_{32}&=55+32=87 \\ a_{33}&=87+33=21 \\ a_{34}&=21+34=55 \\ a_{35}&=55+35=90 \\ a_{36}&=90+36=27 \\ a_{37}&=27+37=64 \\ a_{38}&=64+38=3 \\ a_{39}&=3+39=42 \\ a_{40}&=42+40=82 \\ a_{41}&=82+41=24 \\ a_{42}&=24+42=66 \\ a_{43}&=66+43=10 \\ a_{44}&=10+44=54 \\ a_{45}&=54+45=0.\end{align*}</cmath> Hence the least number <math>n</math> is <math>n=45</math>.
+~typos fixed by lpieleanu
+==Solution 8==
+Taking the recurrence mod <math>99</math>, we have <cmath>a_n=a_{n-1}+n</cmath> for <math>a_{10}=10</math>. Then, we have <cmath>a_n=10+11+12+\cdots+n \implies a_n=\frac{n(n+1)}2-45 \equiv 0 \pmod{99} \implies n(n+1)-90 \equiv 0 \pmod{99} \implies n(n+1)+9 \equiv 0 \pmod{99} \implies n \equiv 0 \pmod{9}.</cmath> Then, we simply can test these values of <math>n</math> to find that <math>n=\boxed{045}</math> produces a value that is also divisible by <math>11</math>.
+-A1001
+==Note==
+By the way, if you're wondering, <math>a_{45}</math> is the <math>72</math>-digit number <cmath>101\text{,}112\text{,}131\text{,}415\text{,}161\text{,}718\text{,}192\text{,}021\text{,}222\text{,}324\text{,}252\text{,}627\text{,}282\text{,}930\text{,}313\text{,}233\text{,}343\text{,}536\text{,}373\text{,}839\text{,}404\text{,}142\text{,}434\text{,}445.</cmath> The prime factorization of <math>a_{45}</math> is <cmath>3^{2}~\cdot~5~\cdot~11~\cdot~151~\cdot~1381~\cdot~1559~\cdot~1\text{,}511\text{,}647~\cdot~448\text{,}966\text{,}261\text{,}198\text{,}213\text{,}862\text{,}368\text{,}469~\cdot~925\text{,}800\text{,}120\text{,}162\text{,}193\text{,}934\text{,}310\text{,}647\text{,}599\text{,}013</cmath> and <math>\frac{a_{45}}{99}</math> is the <math>70</math>-digit number <cmath>1\text{,}021\text{,}334\text{,}660\text{,}759\text{,}209\text{,}274\text{,}666\text{,}881\text{,}033\text{,}578\text{,}309\text{,}366\text{,}494\text{,}245\text{,}588\text{,}215\text{,}591\text{,}276\text{,}503\text{,}428\text{,}324\text{,}671\text{,}055.</cmath>
 ==See also==
 {{AIME box|year=2017|n=I|num-b=8|num-a=10}}
+[[Category:Intermediate Number Theory Problems]]
 {{MAA Notice}}

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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