Find the sum of all positive integers $n$ and $k$ such that $n! + 8 = 2^k$

The answer is 21.

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Good proof!

Mahdi Raza
- 1 year, 1 month ago

For positive integer n, $n! + 8 =2^k > 8 = 2^3 \implies k> 3$

$n! = 2^k - 8 = 2^3 (2^{k-3} - 1) = \text{Even} \times \text{Odd}$

So, $n!$ contains $8$ only and no move even numbers $\implies 4 \leq n < 6$

$n=4, {1\times \cancel{2} \times 3 \times \cancel{4}} = \cancel{2^3} (2^{k-3} - 1) \implies 2^{k-3} = 4=2^2 \implies k=5$

$n=5, 1\times \cancel{2} \times 3 \times \cancel{4} \times 5 = \cancel{2^3} (2^{k-3} - 1) \implies 2^{k-3} = 16=2^4 \implies k=7$

The answer $= 4+5 +5 +7 =\boxed{21}$

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$n=4 \implies 4! + 8 = 2^{5} \implies k = 5$ $n = 5 \implies 5! + 8 = 2^7 \implies k = 7$

Sum of all values of $n$ and $k$ is: $4 + 5 + 5 + 7 = \boxed{21}$

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This isn't really a solution.

Adhiraj Dutta
- 1 year, 1 month ago

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Uhmm, yes I know. Since I pretty much just checked values till 6 and got the sense that it won't be true for higher values. Though I should've explained

Mahdi Raza
- 1 year, 1 month ago

Also @Adhiraj Dutta , did you find this problem form Michael Penn's channel on YouTube

Mahdi Raza
- 1 year, 1 month ago

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Kind of. There's a Discord server for IMO. It was posted as a daily problem. And turns out they found it from Penn's channel.

Adhiraj Dutta
- 1 year, 1 month ago

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@Adhiraj Dutta – Ok, Can you share that Discord link, Thanks!

Mahdi Raza
- 1 year, 1 month ago

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@Mahdi Raza – Sure, here is the site. You'll love it!

Adhiraj Dutta
- 1 year, 1 month ago

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@Adhiraj Dutta – Thank you so much.. didn't know anything about this. Looks like a great community to be with for maths

Mahdi Raza
- 1 year, 1 month ago

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@Mahdi Raza – You'll love it. There are ex-olympians and medalists as well.

Adhiraj Dutta
- 1 year, 1 month ago

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We mod $16$ on both sides yields $n!+8\equiv 2^k\pmod{16}$ .

If $n>6,2^k>n!=720$ , so $k\ge10$ , which means $n!+8\equiv0\pmod{16}$ , but for $n>6, 16|n!$ . This means $8\equiv0\pmod{16}$ , a contradiction, which means $n\le5$ .

Checking for values $n\le5$ , we get two pairs of solutions: $(n,k)=(5,7),(4,5)$ , which means the answer is $5+7+4+5=\boxed{21}$ .