# Factorial Digits Sum

A natural number is a factorion if it is the sum of the factorials of each of its decimal digits. For example, $$145$$ is a factorion because $$145 = 1! + 4! + 5!$$. Find every $$3$$-digit number which is a factorion.

My solution:

We can obviously only have factorials $$1$$$$6$$.

• If we have a $$6$$ factorial, our number would need to start with $$6$$ to be the biggest possible, but that is still too small. Doesn't work.
• If we have a $$5$$ factorial, since that's $$120$$, we need our number to start with $$1$$. $$5!+1!=121$$. We need a $$1\_\_$$. If after the $$1$$ we have a $$5$$, we can do the casework.

• $$151$$ does not work.
• Neither does $$152$$.
• Neither does $$153$$.
• Neither does $$154$$.
• But we did the computations for $$151$$, $$152$$, $$153$$, and $$154$$ to find this, and we notice that $$154$$'s answer is $$145$$. So, $$145$$ works, and we quickly see there are no others, so only $$145$$ works.

How can we generalize this?

• What exactly do you mean by "how to generalize"? What kind of generalization do you have in mind? Do you know that $153$ is the sum of the cubes of its digits? Nov 17, 2019 at 22:05
• You're jumping ahead a little when you say "we need our number to start with $1$". You need to say why it cannot start with $2$, i.e. it cannot have two digits equal to $5$ (or even $3$). That's easy of course, but should be done. Also, as Gerry asked, what kind of generalisation, different number of digits, different function applied to the digits, both? Nov 17, 2019 at 22:08

Brainstorming:

As you pointed out. If one of the digits of $$N$$ is $$7$$ then $$7! > 1000> N$$ which is impossible.

If one of the digits is $$6$$ then $$N > 6! = 720$$ so one of the digits is at least $$7$$ which we just showed was impossible.

Also if all three digits, call them $$a,b,c$$, are all $$< 5$$ then $$a!, b!,c! \le 4! =24$$ and $$N = a! + b!+c! \le 3*24 = 72 < 100$$ so that's impossible. So there must be at least one digit that is equal to $$5$$.

So one of the digits is $$5$$. Then if the other two are $$a,b$$ then $$a,b \le 5$$ and $$N = a! + b! + 5! \le 5! + 5! + 5! = 360$$. So the first digit is $$3$$ or less.

So $$N = a! + b! +5! \le 3! + 5! + 5! = 246$$ so the first digit is $$2$$ or less.

If the first digit, let's assume the first digit is $$a$$ is equal $$2$$ then $$200 < N = 2! + b! + 5! = b! + 122$$ so $$b! > 78$$ so $$b >4$$ but $$b \le 5$$ so $$b=5$$ and $$N = 255$$ but $$2! + 5! + 5! = 242$$ so $$255$$ is not factorial. So the first digit is less than $$2$$.

But $$N > 100$$ so the first digit is $$1$$.

So $$N = 1! + b! + 5! = 126 +b!$$. Where $$b = 0,1,2,3,4,5$$ and I guess at this point we can do trial and error.

But we have $$100 < N < 200$$ and $$N = 126+b!$$ so $$b! < 74$$ for $$b \le 4$$. So $$N=126 +b! \le 126 + 24 = 150$$. And $$1! + 5! + 0! = 122\ne 150$$ so $$150$$ is not factorial so $$N \le 149$$. but as one of the digits is $$5$$ that must be the last digit and $$N = 100 + 10b + 5$$.

So $$N = 105 + 10b = 1! + b! + 5! = 121 + b!$$ so $$10b = b! + 16$$. And we know $$N>121$$ so $$b \ge 2$$ and $$b \le 4$$ and $$b! + 16 \equiv 0 \pmod {10}$$ and.... how much further can I go without saying, look, $$b$$ has got to be $$4$$; that's the only option?

Well a little further.... just to be masochistic. $$10b = 20,30$$ or $$40$$ so $$b! = 4,14$$ or $$24$$ and $$b = 2,3,4$$ so either $$2! = 4$$, $$3!=14$$ or $$4! = 24$$ and ....

That's as far as I can go.

$$b =4$$ and $$N = 100 + 40 + 5 = 1! + 4! +5!$$.

One could generalize this approach, but the analysis of cases will become quite more boring, and complicated. However, one can easily prove that every factorion has seven digits or less, since for $$k\geq8$$, $$9!\cdot k<10^k.$$ A quick computer check can then do all the dirty work, and verify that the complete list of factorions (base $$10$$) is $$\boxed{1,2,145,40585.}$$