Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 10C Problems/Problem 22"
 
Line 8: Line 8:
 
===Solution===
 
===Solution===
  
−
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>
+
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>2*2=4</math> numbers that the first chick says every <math>8*25=200</math> numbers said by all chicks, so our answer is <math>4\cdot{1000/200}+1=\boxed{21}</math>

Latest revision as of 19:16, 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 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 $2*2=4$ numbers that the first chick says every $8*25=200$ numbers said by all chicks, so our answer is $4\cdot{1000/200}+1=\boxed{21}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10C_Problems/Problem_22&oldid=180240"