# 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 $$\mbox{n_N}$$?