2019 AMC 10C Problems/Problem 22

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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2019 AMC 10C Problems/Problem 22

Problem

Solution

Solution 2 (Correct Version)