Difference between revisions of "2018 AMC 10B Problems/Problem 13"

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==Solution==
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== Video Solution ==
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>
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https://youtu.be/zfChnbMGLVQ?t=5122
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~ pi_is_3.14
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== Solution 1 ==
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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.
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It follows that <math>10^n\equiv -1\pmod{101}</math> if and only if <math>n\equiv 2\pmod 4</math>.
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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>\dfrac{2020}{4}=\boxed{\textbf{(C) } 505}</math> numbers in that list.
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<br>
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Minor fixes by fasterthanlight
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More minor fixes by BakedPotato66
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==Solution 2==
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Note that <math>10^{2k}+1</math> for some odd <math>k</math> will suffice <math>\mod {101}</math>. Each <math>2k \in \{2,6,10,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math>
 
(AOPS12142015)
 
(AOPS12142015)
  
==Solution 2==
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==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>.
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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. Looking at our pattern of four numbers from above, the first number is divisible by <math>101</math>. This means that the first of the <math>2</math> left over will be divisible by <math>101</math>, so our answer is <math>\boxed{\textbf{(C) } 505}</math>.
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==Solution 4==
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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>.
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-googleghosh
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==See Also==
 
==See Also==

Latest revision as of 02:29, 17 January 2021

Problem

How many of the first $2018$ numbers in the sequence $101, 1001, 10001, 100001, \dots$ are divisible by $101$?

$\textbf{(A) }253 \qquad \textbf{(B) }504 \qquad \textbf{(C) }505 \qquad \textbf{(D) }506  \qquad \textbf{(E) }1009 \qquad$

Video Solution

https://youtu.be/zfChnbMGLVQ?t=5122

~ pi_is_3.14

Solution 1

The number $10^n+1$ is divisible by 101 if and only if $10^n\equiv -1\pmod{101}$. We note that $(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}$, so the powers of 10 are 4-periodic mod 101.

It follows that $10^n\equiv -1\pmod{101}$ if and only if $n\equiv 2\pmod 4$.

In the given list, $10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1$, the desired exponents are $2,6,10,\dots,2018$, and there are $\dfrac{2020}{4}=\boxed{\textbf{(C) } 505}$ numbers in that list.

Minor fixes by fasterthanlight

More minor fixes by BakedPotato66

Solution 2

Note that $10^{2k}+1$ for some odd $k$ will suffice $\mod {101}$. Each $2k \in \{2,6,10,\dots,2018\}$, so the answer is $\boxed{\textbf{(C) } 505}$ (AOPS12142015)

Solution 3

If we divide each number by $101$, we see a pattern occuring in every 4 numbers. $101, 1000001, 10000000001, \dots$. We divide $2018$ by $4$ to get $504$ with $2$ left over. Looking at our pattern of four numbers from above, the first number is divisible by $101$. This means that the first of the $2$ left over will be divisible by $101$, so our answer is $\boxed{\textbf{(C) } 505}$.

Solution 4

Note that $909$ is divisible by $101$, and thus $9999$ is too. We know that $101$ is divisible and $1001$ isn't so let us start from $10001$. We subtract $9999$ to get 2. Likewise from $100001$ we subtract, but we instead subtract $9999$ times $10$ or $99990$ to get $11$. We do it again and multiply the 9's by $10$ to get $101$. Following the same knowledge, we can use mod $101$ to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is ${0,-9,-99 ( 2),11, 0, ...}$. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide $2017$ by four to get $504$ remainder $1$. Thus the answer is $504$ plus the 1st term or $\boxed{\textbf{(C) } 505}$.

-googleghosh



See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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

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