Dividing by e

Calculus Level 5

Let $[x]$ denote the closest integer to $x$ , for example $[4.7]=5$ and $[9.1]=9$ .

Find the last 5 digits of $[{\frac{100000!}{e}}]$ .

( $e$ is Euler's constant and $!$ is the factorial function.)

(If the answer begins with one or more zeroes, you don't need to write them, for example if you got 01234 as an answer then you can just write 1234)

The answer is 1.

3 solutions

David Vreken
Feb 22, 2020

By the Taylor Series expansion, $e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} ...$

Therefore, $\frac{100000!}{e} = 100000!e^{-1} = 100000! \cdot 1 - 100000! \cdot 1 + 100000! \cdot \frac{1}{2!} - 100000! \cdot \frac{1}{3!} + 100000! \cdot \frac{1}{4!} - 100000! \cdot \frac{1}{5!} ...$

Since $|100000! \cdot \frac{1}{99999!}| > 99999$ and $\displaystyle 100000! \cdot |\sum_{n=100001}^{\infty} \frac{(-1)^n}{n!}| < \frac{1}{2}$ , the only term that affects the last five digits of $[\frac{100000!}{e}]$ is $100000! \cdot \frac{1}{100000!} = 1$ , so the last five digits round up to $00001$ , which makes the answer $\boxed{1}$ .

Thank you, i had the same solution but I couldn't explain it very well :/

O N - 1 year, 3 months ago

You need to deal with the whole infinite tail of the series, and not just the next term. Since it is an alternating series, however, we can say that $\left|\sum_{n=100001}^\infty \frac{(-1)^n \times 100000!}{n!}\right| \; < \; \frac{1}{100001} \; < \; \frac12$ and we are done.

Mark Hennings - 1 year, 3 months ago

You're right! I edited my solution. Thanks!

David Vreken - 1 year, 3 months ago

Actually, the error term $\sum_{n=100001}^\infty \frac{(-1)^n \times 100000!}{n!}$ has negative first term, so lies strictly between $-\tfrac{1}{100001}$ and $0$ . Thus the last five digits of the integer part, and the first five digits of its fractional part, of $\tfrac{100000!}{e}$ are $00000.99999$ , but that makes the last five digits of the integer closest to this number $00001$ . On the other hand, the last five digits of the integer part of $\tfrac{100000!}{e}$ are $00000$ .

Mark Hennings - 1 year, 3 months ago

@Mark Hennings – I edited my solution. Thanks again!

David Vreken - 1 year, 3 months ago

Mohd Faraz
Feb 24, 2020

Alapan Das
Feb 27, 2020

$\frac{1}{e}=1-1+\frac{1}{2!}-\frac{1}{3!}+.....+(-1)^n\frac{1}{n!}+....$

So, $\frac{n!}{e}=\frac{n!}{2!}-\frac{n!}{3!}+....+(-1)^{n-1}n+(-1)^n{1}+X$

For, $n=100000$ we can easily see $-0.5<<X<<0.5$ .

So, [ $\frac{100000!}{e}$ ]= $A($ say)= $\frac{n!}{2!}-\frac{n!}{3!}+....+(-1)^{n-1}n+(-1)^n{1}$ .

Now, we are asked to find last five digits of $A$ . So, we have to mind

$A(n=10^5)$ modulo(100000) or [ $\frac{n!}{e}$ ]modulo(n).

And obviously, all terms except the last term in $A$ $(-1)^n$ are divisible by $n$ . Hence, $A$ modulo $n$ = $(-1)^n$ .

As, $n=100000, (-1)^n=1$ . Hence the last five terms of the expression is $00001=1$

Dividing by e

The answer is 1.

3 solutions

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