Current Events > A trio of similar number theory equations. Can you handle it?

(+) Yeah! (+)

What are all of the positive integer solution pairs (n, k) for the following equations? From easiest to hardest, IMO...

n! + 7 = 2^k

n! + 10 = 2^k

n! + 8 = 2^k

All these equations are identical except for the constant term, which means they require slightly different strategies. It's easy to inspect solutions for small values of n & k, but the tougher part is proving there is a bound on what n can be.

Can you handle it?

No.

do I wanna?

2^k is even, which means n! must be odd for the first one. since n! is even for n >= 2 that seems like only (1, 3) is viable

or I could be completely wrong >_>

teep_ posted...

2^k is even, which means n! must be odd for the first one. since n! is even for n >= 2 that seems like only (1, 3) is viable

you can do a similar sort of thing with the second one

when n is 5 or higher, n! will be divisible by 5, which means n! + 10 will be divisible by 5, and 2^k is never divisible by 5 (it's only divisible by powers of 2). that means the only options for n are 1-2-3-4 and you can pretty easily just inspect them all to find (3, 4) as the only pair.

The final one I think has something to do with breaking up the n! into 4 * (the rest) so you can add the 8 and get 4 * (the rest + 2), but I'm having trouble defining a concrete sort of answer out of this

Sariana21 posted...

Can you handle it?

No.

CobraGT posted...

do I wanna?

Also no

kirbymuncher posted...

The final one I think has something to do with breaking up the n! into 4 * (the rest) so you can add the 8 and get 4 * (the rest + 2), but I'm having trouble defining a concrete sort of answer out of this

24 + 8 = 32 is a solution of the final case (n=4, k=5).

1) n=1,2,3 is ruled out by inspection

2) n! = 120 for n= 5, so 120 + 8 =128 is another solution (n = 5, k=7)

3) If n is 6 or more, n! is divisible by 16, which means that n!/8 is an even number. Dividing through by 8 (k would clearly be 4 or more for any solution with n > 5) then gives even + odd = even which is impossible, ruling out any other solutions.

FLOUR posted...

n! + 7 = 2^k

Well, 1 and 3 was pretty easy to figure out. You've got nothing else where k is less than 16, just with a quick glance at a factorial table. I'm aware that's not how we're supposed to solve the problem, but still.

FLOUR posted...

n! + 10 = 2^k

Oh, I see these are all pretty much the same. Using the same technique, we have 3 and 4 and no more where k is less than 16.

FLOUR posted...

n! + 8 = 2^k

4 and 5. 5 and 7.

ReturnOfDevsman posted...

I'm aware that's not how we're supposed to solve the problem, but still.

yeah we both know this isn't how it's supposed to work, but at the same time it kind of is how it works.

I'm really curious if there's ever been a question like this (with a finite number of answers) where the options are like, k = 3 or k = 4 or k = 12362432