Sigma math help

1. Nov 23, 2003

PrudensOptimus

Problem:

$$(n!)^3$$ n = {1-99}

How many digit is the resulting $$(n!)^3$$?

Attempted solution:

$$\Sigma^{99}_{n=1} (n!)^3 = (\Sigma^{99}_{n=1} (n!))^3 = 99(\frac{1!^3 - 99!^3}{2}) =~ 470$$ digits. But they say it's wrong. Please help.

2. Nov 23, 2003

Hurkyl

Staff Emeritus
[?]

None of this post made sense... try it again from scratch?

3. Nov 23, 2003

PrudensOptimus

The problem is:

(1)!^3 + (2)!^3 + ... + (99)!^3

How many digits are in the resulting sum?

4. Nov 23, 2003

Hurkyl

Staff Emeritus
Ah, ok, I wasn't sure.

Anyways, I think it's a trick question. (99!)^3 is a lot bigger than the rest of the terms in the sequence.

The formula to compute the number of digits d the number n has is:

$$d = \floor{\log_{10} n} + 1$$

And you can use the properties of logarithms to evaluate this when $$\mbox{n=(99!)^3}$$.

Once you know how many digits there are in (99!)^3, figure out the maximum possible number of digits in the sum of the rest of them and see if you can prove that the number of digits in (99!)^3 is or is not the number of digits in the sum.

Hint: For this second part, it may be simpler to first try and solve this brain teaser:
What is the largest 6 digit number that has the property that if you add a 3 digit number to it, the sum still has 6 digits?

5. Nov 23, 2003

PrudensOptimus

Hurkl, where did you learn all these "mathematic tricks"? They are very valuable, more like "mathematical treasures!!!!"

6. Nov 23, 2003

PrudensOptimus

after solving d, the answer came very close to my answer... 469.

7. Nov 23, 2003

Hurkyl

Staff Emeritus
Bah, the floor function didn't appear in my TeX.

You're supposed to round down in this computation, sorry, so it's 468.

Basically, I've done a lot of math. (It's been a hobby ever since I was like 2) The more math you read and do, the more tricks, facts, et cetera you pick up.

Last edited: Nov 23, 2003
8. Nov 23, 2003

PrudensOptimus

So was there something wrong with my sigma approach? Rounding?

9. Nov 24, 2003

NateTG

Re: Sigma :)

Originally posted by PrudensOptimus

$$\Sigma^{99}_{n=1} (n!)^3 = (\Sigma^{99}_{n=1} (n!))^3$$

Is a pretty huge increase.

There are a bunch of terms in the middle that you skipped which might account for the extra digits.

10. Nov 24, 2003

Hurkyl

Staff Emeritus
The next step doesn't follow either; it is not true that $$\mbox{(\Sigma^{99}_{n=1} (n!))^3 = 99(\frac{1!^3 - 99!^3}{2})}$$

11. Nov 24, 2003

PrudensOptimus

Isn't it true that:

$$\Sigma^N_{n=1} n_N = N(\frac{n_1 + n_N}{2})$$ ?

12. Nov 24, 2003

NateTG

Yeah, but you're dealing with n!, not n.

13. Nov 24, 2003

Hurkyl

Staff Emeritus
What do you mean by $$\mbox{n_N}$$?