How many digits does (n!)^3 have for n=1 to 99?

[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.

 

[Hurkyl]

[?]

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

 

[PrudensOptimus]

The problem is:

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

How many digits are in the resulting sum?

 

[Hurkyl]

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?

 

[PrudensOptimus]

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

 

[PrudensOptimus]

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

 

[Hurkyl]

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

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

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

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.

 

[PrudensOptimus]

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

 

[NateTG]

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.

 

[Hurkyl]

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

 

[PrudensOptimus]

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

Isn't it true that:

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

 

[NateTG]

Originally posted by PrudensOptimus
Isn't it true that:

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

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

 

[Hurkyl]

\Sigma^N_{n=1} n_N

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