Factorial magic

Algebra Level 3

n 6 6 ! + n 5 5 ! + n 4 4 ! + n 3 3 ! + n 2 2 ! + n + 1 \frac {n^6}{6!} + \frac {n^5}{5!} + \frac {n^4}{4!} + \frac {n^3}{3!} + \frac {n^2}{2!} + n + 1

For which values of n n between 2018 2018 and 2099 2099 such that the expression above is an integer?

2040, 2070, 2034, 2064, 2094 2040, 2070, 2034, 2039, 2095 2040, 2099, 2025, 2094, 2064 2018, 2020, 2084, 2034, 2094 2033, 2090, 2035, 2064, 2074

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

All the numbers in any option must be divisible by 3 3

Only the first option satisfies this.

Why must they be divisible by 3?

Pi Han Goh - 9 months, 2 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...