Difference between revisions of "2019 AMC 10C Problems/Problem 22"
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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)=== | ||
+ | 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 20:15, 5 November 2022
Problem
chicks are sitting in a circle. The first chick in the circle says the number
, then the chick
seats away from the first chick says the number
, then the chick
seats away from the chicken that said the number
says the number
, 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
th number is said. How many numbers would the chick that said
have said by that point (including
)?
Solution
The number is said by the chick
away from the first chick. This expression is
when
, so we are looking when
and
. The first congruence is satisfied when
and the second congruence is satisfied when
, so the answer is
.
Edit: This solution is flawed, although it does lead to the correct answer.
Solution 2 (Correct Version)
The number is said by chick
. Note that when
, the first chick said the number. Thus, we are looking for
, or
. This is equivalent to stating that
and
. It is clear that there are
numbers that the first chick says every
numbers said by all chicks, so our answer is