RMO Practice Problems - Number theory #1

We note that there is no $n!+5$ a perfect cube for $n<5$ . And for $n \ge 5$ , $n!$ has at least a trailing zero, then the last digit of $n!+5$ is $5$ . And we know that the perfect cube must end with $25$ . Since, $n!$ has at least two tailing zeros for $n\ge 10$ , then the last two digits of $n!+5$ are $05$ , therefore, $4 < n < 10$ . For $4 < n<10$ , there are only three $n!+5$ end with $25$ ; $5!+5=125$ , $6!+5 = 725$ and $8!+5 = 40325$ and only one $n!+5=125$ is a perfect cube. Therefore the answer is $\boxed{5}$ .

If $n$ is at least $7$ , then $n!+5 \equiv 5$ mod $7$ , but the only cubes mod $7$ are $0,1,6$ , so no such $n$ work. It's easy to check the values of $n$ less than $7$ . The only one that works is $n = \fbox{5}$ .

Alternatively, you can use $mod 9$ . Perfect cubes must be $-1, 0 , 1$ . Therefore, if $n! \equiv 0 \text{ (mod 9)}$ , there are no solutions. Thus, we only need to check $n = 5 , 4, 3, 2, 1$ and $n = \boxed{5}$ is the only one that works.

You only need to check up to 9,

Because $n!+5$ , where $n>9$ ,

Then it will end with last digits $805,605,...,005$ ,

For cube of a number ending with 5 it must have its last digits as $625$ or $125$ ,

Clearly,

$n=\boxed{5}$

Well I checked for divisibility by 7... Usually Factorials either have remainders 1,6,4,0 (mod 7) ..We also notice that all numbers after 7 has 0(mod7)... So when we solve n! +5 (mod7) we get n! = 1,3,6,2(mod 7) So we notice that number is less than 7.. Then we try all numbers from 1-6 and get answer as 5