Difference between revisions of "2017 AIME I Problems/Problem 9"
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==Problem 9== | ==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== | ==Solution 1== | ||
− | Writing out the recursive statement for <math>a_n, a_{n-1}, \dots, a_{10}</math> and summing them gives <cmath>a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{ | + | Writing out the recursive statement for <math>a_n, a_{n-1}, \dots, a_{10}</math> and summing them gives <cmath>a_n+\dots+a_{10}=100(a_{n-1}+\dots+a_{10})+n+\dots+10</cmath> |
Which simplifies to <cmath>a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)</cmath> | Which simplifies to <cmath>a_n=99(a_{n-1}+\dots+a_{10})+\frac{1}{2}(n+10)(n-9)</cmath> | ||
− | Therefore, <math>a_n</math> is divisible by 99 if and only if <math>\frac{1}{2}(n+10)(n-9)</math> is divisible by 99, so <math>(n+10)(n-9)</math> needs to be divisible by 9 and 11. Assume that <math>n+10</math> is a multiple of 11. Writing out a few terms, <math>n=12, 23, 34, 45</math>, we see that <math>n=45</math> is the smallest <math>n</math> that works in this case. Next, assume that <math>n-9</math> is a multiple of 11. Writing out a few terms, <math>n=20, 31, 42, 53</math>, we see that <math>n=53</math> is the smallest <math>n</math> that works in this case. The smallest <math>n</math> is <math>\boxed{ | + | Therefore, <math>a_n</math> is divisible by 99 if and only if <math>\frac{1}{2}(n+10)(n-9)</math> is divisible by 99, so <math>(n+10)(n-9)</math> needs to be divisible by 9 and 11. Assume that <math>n+10</math> is a multiple of 11. Writing out a few terms, <math>n=12, 23, 34, 45</math>, we see that <math>n=45</math> is the smallest <math>n</math> that works in this case. Next, assume that <math>n-9</math> is a multiple of 11. Writing out a few terms, <math>n=20, 31, 42, 53</math>, we see that <math>n=53</math> is the smallest <math>n</math> that works in this case. The smallest <math>n</math> is <math>\boxed{045}</math>. |
+ | |||
+ | 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. | ||
==Solution 2== | ==Solution 2== | ||
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<cmath>0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} </cmath> | <cmath>0 \equiv \frac{n(n+1)}{2} - \frac{10*9}{2} \pmod {99} </cmath> | ||
<cmath>0 \equiv n(n+1) - 90 \pmod {99} </cmath> | <cmath>0 \equiv n(n+1) - 90 \pmod {99} </cmath> | ||
− | |||
<cmath>0 \equiv 4n^2+4n+36 \pmod {99} </cmath> | <cmath>0 \equiv 4n^2+4n+36 \pmod {99} </cmath> | ||
<cmath>0 \equiv (2n+1)^2+35 \pmod {99} </cmath> | <cmath>0 \equiv (2n+1)^2+35 \pmod {99} </cmath> | ||
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The smallest positive integer solution greater than <math>10</math> is <math>n=\boxed{045}</math>. | The smallest positive integer solution greater than <math>10</math> is <math>n=\boxed{045}</math>. | ||
+ | |||
+ | ==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>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>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>. | ||
+ | |||
+ | ~First | ||
+ | |||
+ | ==Solution 4 (not good, risky)== | ||
+ | We just notice that <math>100 \equiv 1 \pmod{99}</math>, so we are just trying to find <math>10 + 11 + 12 + \cdots + n</math> modulo <math>99</math>, or <math>\dfrac{n(n+1)}{2} - 45</math> modulo <math>99</math>. Also, the sum to <math>44</math> is divisible by <math>99</math>, and is the first one that is. Thus, if we sum to <math>45</math> the <math>45</math> is cut off and thus is just a sum to <math>44</math>. | ||
+ | |||
+ | Without checking whether there are other sums congruent to <math>45 \pmod{99}</math>, we can just write the answer to be <math>\boxed{045}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>b_n = 2a_{n+10}</math>. We can find a formula for <math>b_n</math>: | ||
+ | |||
+ | <math>b_n = (20+n)(n+1)</math>. | ||
+ | |||
+ | Notice that both can't have a factor of 3. Thus we can limit our search range of n to <math>n \equiv 7,8 \pmod{9}</math>. 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, <math>b_n = 2a_{n+10}</math>, so, the 35th term in b corresponds to the 45th term in a. Thus our answer is <math>\boxed{045}</math>. | ||
+ | |||
+ | -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. <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 <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 <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>. | ||
+ | |||
+ | 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 <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. | ||
+ | <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 <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 <math>n=25</math>. | ||
+ | |||
+ | 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 <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=54 \\ a_{29}&=54+29=83 \\ a_{30}&=83+30=14 \\ a_{31}&=14+31=45 \\ a_{32}&=45+32=77 \\ a_{33}&=77+33=1 \\ a_{34}&=11+34=45 \\ a_{35}&=45+35=80 \\ a_{36}&=80+36=17 \\ a_{37}&=17+37=54 \\ a_{38}&=54+38=92 \\ a_{39}&=92+39=32 \\ a_{40}&=32+40=72 \\ a_{41}&=72+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>. | ||
==See also== | ==See also== | ||
{{AIME box|year=2017|n=I|num-b=8|num-a=10}} | {{AIME box|year=2017|n=I|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:35, 16 March 2021
Contents
Problem 9
Let , and for each positive integer let . Find the least positive such that is a multiple of .
Solution 1
Writing out the recursive statement for and summing them gives Which simplifies to Therefore, is divisible by 99 if and only if is divisible by 99, so needs to be divisible by 9 and 11. Assume that is a multiple of 11. Writing out a few terms, , we see that is the smallest that works in this case. Next, assume that is a multiple of 11. Writing out a few terms, , we see that is the smallest that works in this case. The smallest is .
Note that we can also construct the solution using CRT by assuming either divides and divides , or divides and divides , and taking the smaller solution.
Solution 2
By looking at the first few terms, we can see that This implies Since , we can rewrite the equivalence, and simplify The only squares that are congruent to are and , so yields as the smallest integer solution.
yields as the smallest integer solution.
yields as the smallest integer solution.
yields as the smallest integer solution. However, must be greater than .
The smallest positive integer solution greater than is .
Solution 3
. Using the steps of the previous solution we get up to . This gives away the fact that so either or must be a multiple of 9.
Case 1 (): Say and after simplification .
Case 2: (): Say and after simplification .
As a result must be a divisor of and after doing some testing in both cases the smallest value that works is .
~First
Solution 4 (not good, risky)
We just notice that , so we are just trying to find modulo , or modulo . Also, the sum to is divisible by , and is the first one that is. Thus, if we sum to the is cut off and thus is just a sum to .
Without checking whether there are other sums congruent to , we can just write the answer to be .
Solution 5
Let . We can find a formula for :
.
Notice that both can't have a factor of 3. Thus we can limit our search range of n to . 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, , so, the 35th term in b corresponds to the 45th term in a. Thus our answer is .
-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. . Now the fact that the sequence starts at can be completely discarded for this solution. Just consider then same as , and we can add nine to the answer at the end.
The second step is to split as and and solve for divisibility rules individually. Let's start with because it gives us the most information to continue.
In any number generated, if the numbers don't go beyond , then the highest number we can get is , with every odd digit being . This is a little risky because we are assuming that it doesn't exceed . 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 . The expression being .
The concept here is to add to the term altogether, then subtract the number of ones in it, which is . Simplify to congruent to . Now notice the divide by two can be discarded because one of or will be even. So if or is to be divisible by , we can make a simple list.
Now we test each for divisibility by . This is done by making a list that ultimately calculates the sum of every digit in the large number. to has the first digit . to has the first digit , and so on. The necessary thing to realize is that the sum of all digits is divisible by , 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 .
So we know that include both and , so that's right away. contains numbers that have the first digit , so . Then we add together, which is . , which is not divisible by , so it is not the answer.
Do this for just a minute you get that sums to , a multiple of nine! So is the answer, right? Don't forget we have to add because we translated to at the very beginning! Finally, after a short bash, we get .
-jackshi2006 ( by PureSwag)
Solution 7 (Bash Bash Bash)
We will work. The recursive formula now becomes . Now, we will bash. For convenience, everything is taken . The sequence is Hence the least number is .
See also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.