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

Revision as of 12:51, 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

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 $i$ and $-i$ . Thus $x^2+1 \mid x^n+1$ iff both $i$ and $-i$ are roots of $x^n+1$ , i.e., \[ i^n=-1 \quad \text{and} \quad (-i)^n=-1. \] 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 $\boxed{\textbf{(C)}}$ .

@@ Line 32: / Line 32: @@
 ==Solution 6==
-Let <math>x = 10</math>. We can rewrite the nth term in this sequence as <math>x^n + 1</math> for all <math>2\leq n \leq 2017</math>. For <math>101</math> to divide <math>100...001</math>, we have that the kth term must satisfy <math>x^2 + 1 | x^k + 1</math>. Notice that the roots of <math>x^k + 1</math> can be expressed
+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
+<cmath>x^2+1 \mid x^n+1</cmath>
+Over <math>\mathbb{C}</math>, <math>x^2+1</math> has roots <math>i</math> and <math>-i</math>. Thus <math>x^2+1 \mid x^n+1</math> iff both <math>i</math> and <math>-i</math> are roots of <math>x^n+1</math>, i.e.,
+\[
+i^n=-1 \quad \text{and} \quad (-i)^n=-1.
+\]
+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
+\[
+n \equiv 2 \pmod{4}.
+\]
+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
+<math>2,6,10,\dots,2018</math>.
+This is an arithmetic progression with common difference <math>4</math>; the count is
+<math>\frac{2018-2}{4}+1=\frac{2016}{4}+1=504+1=505</math>.
+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>.
 ==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
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