Sum=Sum!

Number Theory Level 3

Find the number of ordered n-tuple of solutions $(a_1, a_2, a_3, ..., a_n)$ of the equation $a_1+a_2+a_3+...+a_n=a_1!+a_2!+a_3!+...+a_n!$

2^n

3^n

1 solution

Finn Hulse
May 28, 2014

Obviously if $a_n=a_n!$ for each $n$ , then the equation will be satisfied. There are $2$ $a_n$ that have this property, $1$ and $2$ . So for any given $a_n$ there are two options, thus there are $\boxed{2^n}$ total solutions.

How do you know that that's the only case?

Nathan Ramesh - 7 years ago

Note that $x \le x!$ is satisfied for all non-negative integers $x$ , where equality case is $x=1,2$ .

Thus, $a_1+a_2+\cdots +a_n\le a_1!+a_2!+\cdots +a_n!$ , where equality case is when $a_i=1,2$ .

Thus, there are $2^n$ different solutions.

Does that satisfy your rigor needs?

Daniel Liu - 7 years ago

Yeah because for all other integers $n$ , $n! > n$ so the whole thing would be larger.

Finn Hulse - 7 years ago

What about 0! = 1?

Jacob Mair - 4 years, 4 months ago

Sum=Sum!

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