2^n+n=m!

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2^n+n=m!

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Source: Turkey National Olympiad Second Round 2013 P4

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1345 posts

#1 h Nov 28, 2013, 9:44 PM • 5 Y

Y by Mathematicalx, highvalley, Mathuzb, Adventure10, Mango247

Find all positive integers $m$ and $n$ satisfying $2^n+n=m!$ .

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1510 posts

mathuz

#2 h Nov 29, 2013, 6:01 AM • 9 Y

Y by Ashutoshmaths, Pirkuliyev Rovsen, Mathematicalx, highvalley, SupermemberVN, AlastorMoody, Adventure10, Mango247, Tellocan

We have $n>1$ . If $m\in \{1,2,3,4,5,6 \}$ then solution is only $m=3$ , $n=2$ . So we solve at $m\ge 7$ and $n\ge 2$ .
(1). Suppose that $n=2^b(b\ge 1)$ . Then since $7|m!=2^n+n=2^x+2^y\equiv \{ 1,2,4 \} +\{ 1,2,4 \}$ we get contradiction.
(2). Suppose that $n=2^b\cdot k, (b\ge 1,(k>1) - odd)$ .
If $k\le m$ then $k|m!$ and $k|2^n$ - contradiction.
If $k\ge m$ then since $a=v_2(m!)$ and $\lfloor \frac{m}{2} \rfloor \le a \le m$ and $m\ge 7$ we get $2^n >2^{m^2}>m^m>m!$ and contradiction.

Answer: $(m,n)=(3,2)$ .

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867 posts

#3 h Dec 13, 2013, 4:27 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I didn't understood the part $k\ge m$ , why $n>m^2$ ?

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867 posts

#4 h Dec 18, 2013, 3:23 PM • 1 Y

Y by Adventure10

why $n>m^2$ ?

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mathuz

#5 h Dec 18, 2013, 5:02 PM • 2 Y

Y by highvalley, Adventure10

sorry, ok @rightways;
We have $v(2^n+n)=b\ge \lfloor \frac{m}{2} \rfloor$

Those not $m^2$ $\rightarrow m\cdot \lfloor \frac{m}{2} \rfloor$ . But, it's correct, because $m\ge 7.$

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mathuz

#6 h Dec 18, 2013, 5:10 PM • 3 Y

Y by rightways, highvalley, Adventure10

of course, we have $2^{\lfloor \frac{m}{2}\rfloor}\ge m$ for any $m\ge 7$ .

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sayantanchakraborty

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sayantanchakraborty

#7 h Jan 21, 2014, 7:04 AM • 1 Y

Y by Adventure10

Assume that $m\ge4$ and $n\ge2$ .First note that n must be even.If n is of form 4k+2 then
$2^n+n \equiv 2\bmod{4}$ .
But $4|m!$ and hence there is no solution.
Hence $4|n$
Case-1

$n \equiv 0\bmod{12}$
In this case,
$2^n+n=2^{12k}+12k\equiv 1\bmod{3}$
But $3|m!$ .No solution.

Case-2
$n\equiv 4\bmod{12}$
In this case,
$2^n+n=2^{12k+4}+12k+4\equiv 2\bmod{3}$
No solution.

There's only one case left, when $n\equiv 8\mod{12}$
I will be grateful to the person who solves this case.Please try.

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537 posts

#8 h Feb 22, 2014, 3:32 PM • 2 Y

Y by Adventure10, Mango247

Sayantanchakraborty, sorry to say this but your solution is nothing. Because we obviously know $n=p-1(modp)$ will satisfy $(modp)$ so you cant get something from mods.
(but i noticed that after exam:()

This post has been edited 1 time. Last edited by Mathematicalx, Jul 2, 2014, 7:26 PM

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TNOT

#9 h Apr 12, 2018, 6:53 AM • 2 Y

Y by Adventure10, Mango247

Find all positive integer (x, y) such that
2(x^3-x) =y^3-y

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111 posts

#10 h Aug 2, 2018, 9:12 AM • 4 Y

Y by top1csp2020, ThisNameIsNotAvailable, Adventure10, Mango247

Try the base case when m<=6 the solution is (m,n)=(3,2)
We have a lemma: v2(m!)=m-s2(m) (s2 is the sum of digit of n written in binary). Note that 2^m+m<m! for m>=4 which can prove easily by induction m. From the condition of the problem we have n>m and from v2(2^n)=n so it equivalent to v2(m!-n)=n but v2(m!)=m-s2(m)<m-1 by the lemma
Written n=2^(m-s2(m)).k (k is odd) then if k<m then m!-n divisible by k and 2^n divisible by k (contradiction). So k>=m hence n<m which is false because n>m

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548 posts

#11 h Feb 7, 2022, 10:02 AM • 1 Y

Y by tuymurod2005

If $m \geq 2$ $\implies n$ is even number. $n=2k$ .
$2^{2k} + 2k \geq (2^k + 1)^2 =2^{2k}+ 2^{k+1} + 1$ $2k \geq 2^{k+1}+1$ Remaining easy. $\blacksquare$

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661 posts

#12 h Jul 5, 2024, 10:43 AM

Y by

Answer: $(m,n)=(3,2)$ .
$\textbf{Claim:} \ 2^{2^x}>(x+1)!$
Proof: It's true for $x=1,2$ . If inequality holds for $x=k,$ then $2^{2^{k+1}}=(2^{2^k})^2>(k+1)!^2>(k+2)!$ since $(k+1)!\geq k(k+1)>(k+2)$ for $k\geq 2$ . $\square$

Denote $n=2^\alpha k$ where $k$ is odd. Let $m=2^c+b$ where $b\in [0,2^c)$ . Suppose that $k\geq 3$ .
If $p|k$ and $p<m,$ then $p|m!=2^\alpha(2^{2^\alpha k-\alpha}+k)\implies p|2$ which is impossible. Thus, $p|k\implies p>m$ so $k>m$ .
$\alpha=v_2(m!)\geq m-\lceil log_2(m) \rceil\geq 2^c+b-c-1$

$(2^c+b)!=2^{2^\alpha k}+2^\alpha k>2^{2^\alpha m}\geq 2^{2^{2^c+b-c-1} m}\geq 2^{2^{2^c+b-c-1}.2^c}=2^{2^{2^{c}+b-1}}>(2^c+b)!$ Which gives a contradiction. Thus, $k=1$ .
$2^{\alpha}(2^{2^\alpha-\alpha}+1)=m!$ If $m\geq 5,$ then $5|2^{2^\alpha-\alpha}+1\implies 2^\alpha-\alpha\equiv 2(mod \ 4)\implies \alpha$ is even.
$3|2^{2^\alpha-\alpha}+1\implies \alpha$ is odd.
These two contradict hence $m\leq 4$ . There is no solution for $m=1,2,4$ and $m=3$ gives $n=2$ as desired. $\blacksquare$

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