How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers?
How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers?
6
7
8
9
Solution
If 1186 divided by a natural number $d$ leaves remainder 31, then $d$ divides $1186 - 31 = 1155$ and $d > 31$. $1155 = 3 \times 5 \times 7 \times 11$, which has $2^4 = 16$ divisors. Those greater than 31 are: 33, 35, 55, 77, 105, 165, 231, 385, 1155 — that is 9 numbers.