Thread: number theroy

Determine all positive integers n so that n^2 equals the sum of the factorial of the digit of n.

by "the sum of factorials of n" i mean n! + (n - 1)! + (n - 2)! + ... + 2! + 1

Well, factorials grow at a larger rate than exponentials for large values of n. In fact, for n>=4, n, the sum of the factorials will always be larger than their exponents, so you can just do the other cases by hand. To prove that it is always larger, you can do induction, starting with n =4
4^2 =- 16
\sum_{1 4} n! = 24 + 6 + 2 +1 = 33
and show for 5, assume true for n, and show true for n +1.

Then you can look at 1, 2, 3.
1! = 1^2 = 1
2! = 2+1=3 != 2^2 = 4
3! = 6 + 2 + 1=9 = 3^2 = 9

So only 1 and 3.

Originally Posted by terr13

Well, factorials grow at a larger rate than exponentials for large values of n. In fact, for n>=4, n, the sum of the factorials will always be larger than their exponents, so you can just do the other cases by hand. To prove that it is always larger, you can do induction, starting with n =4
4^2 =- 16
\sum_{1 4} n! = 24 + 6 + 2 +1 = 33
and show for 5, assume true for n, and show true for n +1.

Then you can look at 1, 2, 3.
1! = 1^2 = 1
2! = 2+1=3 != 2^2 = 4
3! = 6 + 2 + 1=9 = 3^2 = 9

So only 1 and 3.

what do u mean by \sum_{1 4} n! = 24 + 6 + 2 +1 = 33?? Cant tell what you did here

I'm taking the sum of that expression from 1 to 4, so 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33, which is larger than 4^2 = 16. This is always true for n>=4.

The sum of the factorial of digit of 1000 is 4 but 10^3^2=10^6. It is not about the sum of factorial from 1 to n.

I know that 71 works.
71^2 = 1! + 7! = 5041
But I don't know how to list these solutions

Originally Posted by vincisonfire

I know that 71 works.
71^2 = 1! + 7! = 5041
But I don't know how to list these solutions

71 does not work
it would have to be 71^2=1! + 2! +3! + 4!+....71! 71! it self is just bigger then 71^2

The question says that n^2 should be equal to the sum of the factorial of the DIGIT of n.
n = a1*10^k+...+an.
You search n such that (a1*10^k+...+an)^2 = a1! + ...+ an!

Originally Posted by vincisonfire

The question says that n^2 should be equal to the sum of the factorial of the DIGIT of n.
n = a1*10^k+...+an.
You search n such that (a1*10^k+...+an)^2 = a1! + ...+ an!

This is the idea:
n = sum a_k * 10^k, 0 <= k <= m
with a_k in {0, 1, 2, .., 9} and a_m > 0. When is

n^2 = sum (a_k)!.

For example,
7! + 1! = 71^2.

I'm asking to find (with proof) all such examples.

Does this help?