Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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− | ==Solution== | + | == Solution 1 == |
+ | The number <math>10^n+1</math> is divisible by 101 if and only if <math>10^n\equiv -1\pmod{101}</math>. We note that <math>(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}</math>, so the powers of 10 are 4-periodic mod 101. It follows that <math>10^n\equiv -1\pmod{101}</math> if and only if <math>n\equiv 2\pmod 4</math>. | ||
+ | |||
+ | In the given list, <math>10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1</math>, the desired exponents are <math>2,6,10,\dots,2018</math>, and there are <math>2020/4=\boxed{\textbf{(C) } 505}</math> numbers in that list. | ||
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+ | ==Solution 2== | ||
Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,4,6,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math> | Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,4,6,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math> | ||
(AOPS12142015) | (AOPS12142015) | ||
− | ==Solution | + | ==Solution 3== |
If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots</math>. We divide <math>2018</math> by <math>4</math> to get <math>504</math> with <math>2</math> left over. One divisible number will be in the <math>2</math> left over, so out answer is <math>\boxed{\textbf{(C) } 505}</math>. | If we divide each number by <math>101</math>, we see a pattern occuring in every 4 numbers. <math>101, 1000001, 10000000001, \dots</math>. We divide <math>2018</math> by <math>4</math> to get <math>504</math> with <math>2</math> left over. One divisible number will be in the <math>2</math> left over, so out answer is <math>\boxed{\textbf{(C) } 505}</math>. | ||
− | ==Solution | + | ==Solution 4== |
Note that <math>909</math> is divisible by <math>101</math>, and thus <math>9999</math> is too. We know that <math>101</math> is divisible and <math>1001</math> isn't so let us start from <math>10001</math>. We subtract <math>9999</math> to get 2. Likewise from <math>100001</math> we subtract, but we instead subtract <math>9999</math> times <math>10</math> or <math>99990</math> to get <math>11</math>. We do it again and multiply the 9's by <math>10</math> to get <math>101</math>. Following the same knowledge, we can use mod <math>101</math> to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is <math>{0,-9,-99 ( 2),11, 0, ...}</math>. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide <math>2017</math> by four to get <math>504</math> remainder <math>1</math>. Thus the answer is <math>504</math> plus the 1st term or <math>\boxed{\textbf{(C) } 505}</math>. | Note that <math>909</math> is divisible by <math>101</math>, and thus <math>9999</math> is too. We know that <math>101</math> is divisible and <math>1001</math> isn't so let us start from <math>10001</math>. We subtract <math>9999</math> to get 2. Likewise from <math>100001</math> we subtract, but we instead subtract <math>9999</math> times <math>10</math> or <math>99990</math> to get <math>11</math>. We do it again and multiply the 9's by <math>10</math> to get <math>101</math>. Following the same knowledge, we can use mod <math>101</math> to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is <math>{0,-9,-99 ( 2),11, 0, ...}</math>. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide <math>2017</math> by four to get <math>504</math> remainder <math>1</math>. Thus the answer is <math>504</math> plus the 1st term or <math>\boxed{\textbf{(C) } 505}</math>. | ||
Revision as of 22:27, 20 December 2018
Problem
How many of the first numbers in the sequence are divisible by ?
Solution 1
The number is divisible by 101 if and only if . We note that , so the powers of 10 are 4-periodic mod 101. It follows that if and only if .
In the given list, , the desired exponents are , and there are numbers in that list.
Solution 2
Note that for some odd will suffice . Each , so the answer is (AOPS12142015)
Solution 3
If we divide each number by , we see a pattern occuring in every 4 numbers. . We divide by to get with left over. One divisible number will be in the left over, so out answer is .
Solution 4
Note that is divisible by , and thus is too. We know that is divisible and isn't so let us start from . We subtract to get 2. Likewise from we subtract, but we instead subtract times or to get . We do it again and multiply the 9's by to get . Following the same knowledge, we can use mod to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is . Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide by four to get remainder . Thus the answer is plus the 1st term or .
-googleghosh
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.