SOLUTION: Suppose 2011 is divided by an integer n. For which integers n is the remainder 9?

Question 420376: Suppose 2011 is divided by an integer n. For which integers n is the remainder 9?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
We have 2011 ≡ 9 (mod n), so if we subtract 9 from both sides, we obtain 2002 ≡ 0 (mod n). Hence, we need to find all factors of 2002 greater than 9 (otherwise 2011 would not be 9 modulo n).

. If you know the algorithm to finding the number of factors of a number, we conclude that the number has 16 factors. Only three of them (1,2,7) are less than 9, so the other 13 factors satisfy for n. I'll leave it to you to find the 13 factors of 2002 greater than 9.
RELATED QUESTIONS
Suppose 2001 is divided by an integer n. For which integers n is the remainder... (answered by robertb)

when 39,500 is divided by an integer n, the quotient is 123 and the remainder is 17. find (answered by ikleyn)

if n is an integer greater then 1, such that n divided by 4 yields a remainder of 0,... (answered by tommyt3rd)

Find all integers n for which (n^2+n+1)/(n-1) is an... (answered by josgarithmetic)

How many positive integers n produce a remainder of 9 when 2009 is divided by n and n > 9 (answered by drk)

For an integer $n,$ let $f(n)$ be the remainder when $n^8 + n^{16}$ is divided by $5.$... (answered by greenestamps)

Find all integers n for which \frac{n^2 + n + 1}{n - 2 + n^3} is an integer. (answered by CPhill,math_tutor2020)

If n is an integer and 4 is the remainder when 4n - 1 is divided by 5, then n could... (answered by RAY100)

When 29500 are divided by and integer n. The quotient is 123 and the remainder is 17.... (answered by Earlsdon)