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Difference between revisions of "2019 AMC 10C Problems/Problem 22"

(Created page with "===Problem=== <math>100</math> chicks are sitting in a circle. The first chick in the circle says the number <math>1</math>, then the chick <math>2</math> seats away from the...")
 
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The <math>n^{th}</math> number is said by the chick <math>\frac{n(n+1)}{2}-1\pmod{100}</math> away from the first chick. This expression is <math>0</math> when <math>n=1001</math>, so we are looking when <math>n(n+1)\equiv 2\pmod{4}</math> and <math>n(n+1)\equiv 2\pmod{25}</math>. The first congruence is satisfied when <math>n\equiv 1,2\pmod{4}</math> and the second congruence is satisfied when <math>n\equiv 1\pmod{25}</math>, so the answer is <math>21</math>.
 
The <math>n^{th}</math> number is said by the chick <math>\frac{n(n+1)}{2}-1\pmod{100}</math> away from the first chick. This expression is <math>0</math> when <math>n=1001</math>, so we are looking when <math>n(n+1)\equiv 2\pmod{4}</math> and <math>n(n+1)\equiv 2\pmod{25}</math>. The first congruence is satisfied when <math>n\equiv 1,2\pmod{4}</math> and the second congruence is satisfied when <math>n\equiv 1\pmod{25}</math>, so the answer is <math>21</math>.
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Edit: This solution is flawed, although it does lead to the correct answer.
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===Solution 2 (Correct Version)===
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The <math>n^{th}</math> number is said by chick <math>\frac{n(n+1)}{2} \pmod{100}</math>. Note that when <math>n=1001</math>, the first chick said the number. Thus, we are looking for <math>\frac{n(n+1)}{2}\equiv{2} \pmod{100}</math>, or <math>\frac{(n+2)(n-1)}{2}\equiv{0} \pmod{100}</math>. This is equivalent to stating that <math>(n+2)(n-1)\equiv{0} \pmod{8}</math> and <math>(n+2)(n-1)\equiv{0} \pmod{25}</math>. It is clear that there are <math>4</math> numbers that the first chick says every <math>200</math> numbers said by all chicks, so our answer is <math>4\cdot{1000/200}+1=\boxed{21}</math>

Revision as of 19:15, 5 November 2022

Problem

$100$ chicks are sitting in a circle. The first chick in the circle says the number $1$, then the chick $2$ seats away from the first chick says the number $2$, then the chick $3$ seats away from the chicken that said the number $2$ says the number $3$, and so on. The process will always go clockwise. Some of the chicks in the circle will say more than one number while others might not even say a number at all. The process stops when the $1001$th number is said. How many numbers would the chick that said $1001$ have said by that point (including $1001$)?


$\mathrm{(A) \ } 20\qquad \mathrm{(B) \ } 21\qquad \mathrm{(C) \ } 30\qquad \mathrm{(D) \ } 31\qquad \mathrm{(E) \ } 41$

Solution

The $n^{th}$ number is said by the chick $\frac{n(n+1)}{2}-1\pmod{100}$ away from the first chick. This expression is $0$ when $n=1001$, so we are looking when $n(n+1)\equiv 2\pmod{4}$ and $n(n+1)\equiv 2\pmod{25}$. The first congruence is satisfied when $n\equiv 1,2\pmod{4}$ and the second congruence is satisfied when $n\equiv 1\pmod{25}$, so the answer is $21$.

Edit: This solution is flawed, although it does lead to the correct answer.

Solution 2 (Correct Version)

The $n^{th}$ number is said by chick $\frac{n(n+1)}{2} \pmod{100}$. Note that when $n=1001$, the first chick said the number. Thus, we are looking for $\frac{n(n+1)}{2}\equiv{2} \pmod{100}$, or $\frac{(n+2)(n-1)}{2}\equiv{0} \pmod{100}$. This is equivalent to stating that $(n+2)(n-1)\equiv{0} \pmod{8}$ and $(n+2)(n-1)\equiv{0} \pmod{25}$. It is clear that there are $4$ numbers that the first chick says every $200$ numbers said by all chicks, so our answer is $4\cdot{1000/200}+1=\boxed{21}$

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