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Difference between revisions of "2018 AMC 10B Problems/Problem 13"
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==Solution 2==
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==Solution 2 (Similar to solution 1)==
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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>
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Note that <math>10^{2k}+1</math> for some odd <math>k</math> will satisfy <math>10^{2k}+1 \equiv -1\pmod {101}</math>. Each <math>2k \in \{2,6,10,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math>
  
 
==Solution 3==
 
==Solution 3==
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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>.
 
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>.
 
==Solution 5==
 
==Solution 5==
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Note that <math>101=x^2+1</math> and <math>100...0001=x^n+1</math>, where <math>x=10</math>. We have that <math>\frac{x^n+1}{x^2+1}</math> must have a remainder of <math>0</math>. By the remainder theorem, the roots of <math>x^2+1</math> must also be roots of <math>x^n+1</math>. Plugging in <math>i,-i</math> to <math>x^n+1</math> yields that <math>n\equiv2\mod{4}</math>. Because the sequence starts with <math>10^2+1</math>, the answer is <math>\ceil{2018/4}=\boxed{505}</math>
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Note that <math>101=x^2+1</math> and <math>100...0001=x^n+1</math>, where <math>x=10</math>. We have that <math>\frac{x^n+1}{x^2+1}</math> must have a remainder of <math>0</math>. By the remainder theorem, the roots of <math>x^2+1</math> must also be roots of <math>x^n+1</math>. Plugging in <math>i,-i</math> to <math>x^n+1</math> yields that <math>n\equiv2\mod{4}</math>. Because the sequence starts with <math>10^2+1</math>, the answer is <math>\lceil 2018/4 \rceil=\boxed{\textbf{(C) } 505}</math>
  
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==Video Solution==
 
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https://youtu.be/zfChnbMGLVQ?t=5122
 
  
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~ pi_is_3.14
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==Solution 6 (Complex Numbers)==
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The <math>n</math>th term of the sequence is <math>10^n+1</math> for <math>n\ge 2</math> (since <math>10^2+1=101</math>, <math>10^3+1=1001</math>, etc.). We seek those <math>n</math> for which <cmath>101 \mid (10^n+1).</cmath>Let <math>x = 10</math>. Note that <math>101=10^2+1</math>, so this is equivalent to the polynomial divisibility
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<cmath>x^2+1 \mid x^n+1</cmath>
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Over <math>\mathbb{C}</math>, <math>x^2+1</math> has roots <math>i,-i</math>. Thus <math>x^2+1 \mid x^n+1</math> if both <math>i,-i</math> are roots of <math>x^n+1</math>, since,
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<math>i^n=-1 \quad \text{and} \quad (-i)^n=-1.</math>
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Because <math>i^4=1</math>, we have <math>i^n=-1</math> exactly when <math>n\equiv 2 \pmod{4}</math>. For such <math>n</math>, <math>(-i)^n=(-1)^n i^n=i^n=-1</math> as well, so the condition holds precisely when
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<math>n \equiv 2 \pmod{4}.</math>
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Among the first <math>2018</math> terms of the sequence (which correspond to <math>n=2,3,\dots,2019</math>), the values of <math>n</math> congruent to <math>2 \pmod{4}</math> are
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<math>2,6,10,\dots,2018</math>.
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This is an arithmetic progression with common difference <math>4</math>; the count is
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<math>\frac{2018-2}{4}+1=\frac{2016}{4}+1=504+1=505</math>.
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Therefore, exactly <math>\boxed{505}</math> of the first <math>2018</math> numbers are divisible by <math>101</math>, which corresponds to choice <math>\boxed{\textbf{(C)}}</math>.
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~ Aeioujyot
  
 
==See Also==
 
==See Also==

Latest revision as of 15:20, 27 September 2025

Problem

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

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

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.

Solution 2 (Similar to solution 1)

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

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 $\boxed{\textbf{(C) } 505}$.

Solution 5

Note that $101=x^2+1$ and $100...0001=x^n+1$, where $x=10$. We have that $\frac{x^n+1}{x^2+1}$ must have a remainder of $0$. By the remainder theorem, the roots of $x^2+1$ must also be roots of $x^n+1$. Plugging in $i,-i$ to $x^n+1$ yields that $n\equiv2\mod{4}$. Because the sequence starts with $10^2+1$, the answer is $\lceil 2018/4 \rceil=\boxed{\textbf{(C) } 505}$


Solution 6 (Complex Numbers)

The $n$th term of the sequence is $10^n+1$ for $n\ge 2$ (since $10^2+1=101$, $10^3+1=1001$, etc.). We seek those $n$ for which \[101 \mid (10^n+1).\]Let $x = 10$. Note that $101=10^2+1$, so this is equivalent to the polynomial divisibility

\[x^2+1 \mid x^n+1\]

Over $\mathbb{C}$, $x^2+1$ has roots $i,-i$. Thus $x^2+1 \mid x^n+1$ if both $i,-i$ are roots of $x^n+1$, since, $i^n=-1 \quad \text{and} \quad (-i)^n=-1.$ Because $i^4=1$, we have $i^n=-1$ exactly when $n\equiv 2 \pmod{4}$. For such $n$, $(-i)^n=(-1)^n i^n=i^n=-1$ as well, so the condition holds precisely when $n \equiv 2 \pmod{4}.$

Among the first $2018$ terms of the sequence (which correspond to $n=2,3,\dots,2019$), the values of $n$ congruent to $2 \pmod{4}$ are $2,6,10,\dots,2018$. This is an arithmetic progression with common difference $4$; the count is

$\frac{2018-2}{4}+1=\frac{2016}{4}+1=504+1=505$. Therefore, exactly $\boxed{505}$ of the first $2018$ numbers are divisible by $101$, which corresponds to choice $\boxed{\textbf{(C)}}$.

~ Aeioujyot

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

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