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

Latest revision as of 08:59, 3 November 2024

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

Note that $10^{2k}+1$ for some odd $k$ will satisfy $\mod {101}$ . 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 $\lceil 2018/4 \rceil=\boxed{\textbf{(C) } 505}$

@@ Line 1: / Line 1: @@
+==Problem==
 How many of the first <math>2018</math> numbers in the sequence <math>101, 1001, 10001, 100001, \dots</math> are divisible by <math>101</math>?
@@ Line 8: / Line 9: @@
 \textbf{(E) }1009 \qquad
 </math>
+== 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>\dfrac{2020}{4}=\boxed{\textbf{(C) } 505}</math> numbers in that list.
+<br>
+==Solution 2==
+Note that <math>10^{2k}+1</math> for some odd <math>k</math> will satisfy <math>\mod {101}</math>. Each <math>2k \in \{2,6,10,\dots,2018\}</math>, so the answer is <math>\boxed{\textbf{(C) } 505}</math>
+==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. 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>.
+==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>.
+==Solution 5==
+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>
+==See Also==
+{{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}}
+{{MAA Notice}}

2018 AMC 10B (Problems • Answer Key • Resources)
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