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Source: JBMO 2023 Problem 1

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#1 h Jun 26, 2023, 7:14 AM • 4 Y

Y by GeoKing, Sedro, ItsBesi, lian_the_noob12

Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$ .

Nikola Velov, North Macedonia

This post has been edited 2 times. Last edited by Orestis_Lignos, Jun 26, 2023, 4:02 PM

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#2 h Jun 26, 2023, 7:36 AM • 4 Y

Y by GeoKing, ItsBesi, FeridMath, Yassperx

WLOG assume that $a \geq b$ . We split into two cases.

Case 1: $a=b$ . Then, $a!+a=5^k$ with $k \geq 1$ . If $k=1$ then obiously $(a,b)=(5,5)$ works. If $k \geq 2$ , then $(a-1)!+1 \equiv 1 \pmod 5$ , and since $5^k=a!+a=a((a-1)!+1),$ we must have $(a-1)!+1=1,$ a contradiction.

Case 2: $a>b$ . Then, $a=b+s$ with $s \geq 1$ . Let $a!+b=5^k$ with $k \geq 1$ . Therefore,

$b((b-1)! \cdot (b+1) \cdots (b+s)+1)=5^k$ .

Thus, $b=5^\ell$ with $\ell \geq 0$ , and if $b-1 \geq 5$ then $(b-1)! \cdot (b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction. Moreover, if $s \geq 5$ , then $5 \mid (b+1) \cdots (b+5) \mid (b+1) \cdots (b+s),$ and so $(b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction.

Therefore, $b \leq 5$ and $s \leq 4$ .

Taking all cases, we obtain $(b,s)=(1,3),$ that is $(a,b)=(4,1)$ .

To sum up, we obtain the solutions $(a,b)=(5,5), (4,1)$ and $(1,4)$ .

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ismayilzadei1387

#3 h Jun 26, 2023, 11:20 AM • 2 Y

Y by Aoxz, FriIzi

Orestis_Lignos wrote:

WLOG assume that $a \geq b$ . We split into two cases.

Case 1: $a=b$ . Then, $a!+a=5^k$ with $k \geq 1$ . If $k=1$ then obiously $(a,b)=(5,5)$ works. If $k \geq 2$ , then $(a-1)!+1 \equiv 1 \pmod 5$ , and since $5^k=a!+a=a((a-1)!+1),$ we must have $(a-1)!+1=1,$ a contradiction.

Case 2: $a>b$ . Then, $a=b+s$ with $s \geq 1$ . Let $a!+b=5^k$ with $k \geq 1$ . Therefore,

$b((b-1)! \cdot (b+1) \cdots (b+s)+1)=5^k$ .

Thus, $b=5^\ell$ with $\ell \geq 0$ , and if $b-1 \geq 5$ then $(b-1)! \cdot (b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction. Moreover, if $s \geq 5$ , then $5 \mid (b+1) \cdots (b+5) \mid (b+1) \cdots (b+s),$ and so $(b+1) \cdots (b+s)+1 \equiv 1 \pmod 5,$ a contradiction.

Therefore, $b \leq 5$ and $s \leq 4$ .

Taking all cases, we obtain $(b,s)=(1,3),$ that is $(a,b)=(4,1)$ .

To sum up, we obtain the solutions $(a,b)=(5,5), (4,1)$ and $(1,4)$ .

ohh Same solution

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#4 h Jun 29, 2023, 2:59 PM

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We claim that the only solutions are $(1,4),(4,1)$ , and $(5,5)$ .
Suppose that we have integers $a$ and $b$ such that $a!+b=5^r$ and $b!+a=5^t$ for some $r,t\geq 1$ .
If $a\geq 5$ , we have that $5\mid a!$ .
Since $5\mid a!$ and $5\mid 5^r$ , we have $5\mid b$ .
This means that $b\geq 5$ . By the same argument, we have $5\mid a$ .
So $a=5x$ and $b=5y$ for some positive integers $x,y$ .

Suppose that $x>1$ . Then $v_5(a!)>1$ .
Since $a!+b=5^r$ , we have $v_5(b)=v_5(a!).$
However, as $v_5(b)>1$ we have that $v_5(b!)>v_5(b)$ . Hence $v_5(b!+a)=\min\{v_5(b!),v_5(a)\}=v_5(a)$ , but this is clearly a contradiction as $t>v_5(a)$ .
We thus have that $x=1$ . By symmetry we get that $y=1$ which gives us the pair $(a,b)=(5,5)$ which is a solution.
Now it remains to consider $a<5$ .
This implies $b<5$ due to the reasoning above. We check the following cases:
\begin{itemize}
\item If $a=1$ , then $1!+b\equiv 0\pmod 5$ so $b=4$ . This pair is a solution.
\item If $a=2$ , then $2!+b\equiv 0\pmod 5$ so $b=3$ . This pair is not a solution.
\item If $a=3$ , then $3!+b\equiv 0\pmod 5$ so $b=4$ . This pair is not a solution.
\item If $a=4$ , then $4!+b\equiv 0\pmod 5$ so $b=1$ . This pair is a solution.
\end{itemize}

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#5 h Jul 1, 2023, 2:37 PM

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The following solution was proposed together with the problem:

The condition is symmetric so we can assume that $b \leq a$ .

The first case is when $a=b$ . In this case, $a!+a=5^m$ for some positive integer $m$ . We can rewrite this as $a \cdot \left ( (a-1)!+1 \right)=5^{m}$ . This means that $a=5^{k}$ for some integer $k \geq 0$ . It is clear that $k$ cannot be $0$ . If $k \geq 2$ , then $(a-1)!+1 = 5^{l}$ for some $l \geq 1$ , but $a-1=5^{k}-1 > 5$ , so $5|(a-1)!$ , which is not possible because $5|(a-1)!+1$ . This means that $k=1$ and $a=5$ . In this case, $5!+5 = 125$ , which gives us the solution $(5,5)$ .

Let us now assume that $1 \leq b<a$ . Let us first assume that $b=1$ . Then $a+1=5^x$ and $a!+1=5^y$ for integers $x,y \geq 1$ . If $x \geq 2$ , then $a=5^x-1 \geq 5^2-1>5$ , so $5|a!$ . However, $5|5^y=a!+1$ , which leads to a contradiction. We conclude that $x=1$ and $a=4$ . From here $a!+b=25$ and $b!+a=5$ , so we get two more solutions: $(1,4)$ and $(4,1)$ .

Now we focus on the case $1 < b < a$ . Then we have $a!+b = 5^x$ for $x \geq 2$ , so $b \cdot \left ( \frac{a!}{b}+1 \right ) = 5^x$ , where $b|a!$ because $a>b$ . Because $b|5^x$ and $b>1$ , we have $b=5^z$ for $z \geq 1$ . If $z \geq 2$ , then $5<b<a$ , so $5|a!$ , which means that $\frac{a!}{b}+1$ cannot be a power of $5$ . We conclude that $z=1$ and $b=5$ . From here $5!+a$ is a power of $5$ , so $5|a$ , but $a>b=5$ , which gives us $a \geq 10$ . However, this would mean that $25|a!$ , $5|b$ and $25 \nmid b$ , which is not possible, because $a!+b=5^x$ and $25|5^x$ .

We conclude that the only solutions are $(1,4)$ , $(4,1)$ and $(5,5)$ .

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gnoka

#6 h Jul 2, 2023, 2:29 PM

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If $a<5$ , it is obvious that $5$ does not divide $a$ , so $5$ does not divide $b!$ . This implies that $b<5$ . By checking all possible values, we find that $(a,b)=(1,4)$ satisfies the condition.

If $a\geq 5$ , it is obvious that $5$ divides $a!$ , which means $5$ divides $b$ . In turn, this implies that $5$ divides $b!$ , and therefore $5$ divides $a$ . When $a=b=5$ , the condition is satisfied.

If $a>5$ , we will prove by induction that for any $k\in \mathbb{N}^*$ , $5^k$ divides $a$ .

(i) From $a>5$ and $5|a$ , we have $a\geq 10$ .

(ii) If there exists a positive integer $m>1$ such that $5^m|a$ , then $5^{m+1}|a!$ . Since $a!+b$ is a power of $5$ , we have $5^{m+1}|b$ . Hence, $5^{m+1}|b!$ . Furthermore, since $b!+a$ is a power of $5$ , we have $5^{m+1}|a$ .

Based on (i) and (ii), we can conclude that for any $k\in \mathbb{N}^*$ , $5^k|a$ . However, this is clearly impossible.

In conclusion, $(a,b)$ can only be $(1,4)$ or $(4,1)$ or $(5,5)$ .

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dancho

#7 h Jul 5, 2023, 11:35 PM

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WLOG $a\geq b$
We have that $a!+b=5^t$ and $b!+a=5^s$

Let's assume that $a\geq10$
Let $v_5(a!)=x>1$
From $a!+b=5^t$ we have that $v_5(b)=x\implies b\geq5^x$
Now note that $b!=1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot b$ which has $\frac{b}{5}$ numbers divisible by 5. Thus $v_5(b!)\geq\frac{b}{5}\geq\frac{5^x}{5}=5^{x-1}$
From $b!+a=5^s$ and $v_5(b!)\geq5^{x-1}$ we get that $v_5(a)\geq5^{x-1}$
$5^{x-1}>x$ for $x>1$ which is a contradiction with $v_5(a!)\geq v_5(a)$
Hence $a<10$

When we check the remaining possibilities we conclude that $(a,b)=(1,4)$ or $(4,1)$ or $(5,5)$

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#8 h Aug 2, 2023, 12:10 PM

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mathmax12 wanted me to share his solution:
WLOG $a\ge b$ , now we have $2$ cases:

If $a=b$ :
So $a!+a=5^n$ , where $n \ge 1.$ When $n=3$ , we have that $a=b=5$ , works. If $n \ge 3$ we have contradiction.

If $a>b$ :
Let $a=b+x$ , so $s \ge 1$ , we also have $a!+b=5^k$ , we have $b\le 5$ , and $s \le 4$ , otherwise contradiction. If we take all cases, we get that, $(4,1)$ works, so the solutions are $(a,b)=(5,5), (4,1) and (1,4).$

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#9 h Sep 13, 2023, 12:14 PM • 2 Y

Y by PennyLane_31, Mr_GermanCano

After the competition one of the Bulgarian contestants proposed looking at triples $(a,b,p)$ , where $p$ is prime, such that $a! + b$ and $b!+a$ are both powers of $p$ . I just solved this now, here is a solution. (This would have been a great Problem 4 (or maybe 3 if one wants the problem set to be extreme) in my opinion, shame we did not think about it in time.)

Answer. $(a,b,p) = (2, 2, 2), (3, 3, 3), (5, 5, 5)$ , $(1, 1, 2)$ , $(2, 1, 3)$ , $(1, 2, 3)$ , $(4, 1, 5)$ , $(1, 4, 5)$

Without loss of generality treat $a\geq b$ . Suppose firstly that $a \geq 2p$ . In
$a! + b = b \cdot (1 \cdot 2 \cdots (b-1) \cdot b+1 \cdots (a-1) \cdot a + 1)$ both multipliers on the right must be powers of $p$ . The second multiplier is greater than $1$ , while $1 \cdot 2 \cdots (b-1) \cdot b+1 \cdots (a-1) \cdot a$ is divisible by $p$ , since for $b\neq p$ we have $p$ as a multiplier, while for $b=p$ we have $2p$ as a multiplier, hence $(1 \cdot 2 \cdots (b-1) \cdot b+1 \cdots (a-1) \cdot a + 1$ is not divisible by $p$ , which is impossible. Hence $b \leq a \leq 2p-1$ and since $b$ is a power of $p$ , we must have $b=p$ or $b=1$ .

Let $b=p$ . Since $b!+a$ is a power of $p$ , we must have that $a$ is divisible by $p$ and now $a\leq 2p-1$ implies $a=p$ . In this case $b! + a = p((p-1)! + 1)$ and so it remains to see for which $p$ there is $k$ with $(p-1)! + 1 = p^k$ . For $p=2$ we have $k=1$ , for $p=3$ we have $k=1$ , for $p=5$ we have $k=2$ and we now show that no $p\geq 7$ works. Divide by $p-1$ in $(p-1)! = p^k - 1$ to obtain
$(p-2)! = p^{k-1} + p^{k-2} + \cdots + p + 1.$ Consider the latter modulo $p-1$ . For $p\geq 7$ the left-hand side contains the different multipliers $2$ and $\frac{p-1}{2}$ and so it is divisible by $p-1$ . Each summand on the right gives remainder $1$ when divided by $p-1$ , so the right-hand side is congruent to $k$ . Hence $k$ must be divisible by $p-1$ , but then $k\geq p-1$ and $(p-2)! > p^k \geq p^{p-1}$ , which is impossible.

Let $b=1$ . We want $a+1 = p^m$ for some $m\geq 1$ , i.e. $a = p^m-1$ , as well as $a! + 1 = p^n$ for some $n\geq 1$ , i.e. $(p^m-1)! = p^n - 1$ . If $m\geq 2$ , then the left-hand side is divisible by $p$ , while the right-hand side is not, contradiction. For $m=1$ we get $(p-1)! = p^n - 1$ , thus as in the previous case we obtain only $p=2$ (with $a=1$ ), $p=3$ (with $a=2$ ) and $p=5$ (with $a=4$ ).

This post has been edited 2 times. Last edited by Assassino9931, Sep 13, 2023, 12:18 PM

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#10 h Jan 20, 2024, 5:23 AM • 2 Y

Y by dhfurir, ventic

If $a=b$ , then $a((a-1)!+1)$ is a power of $5$ . In particular, $a$ and $(a-1)!+1$ are powers of $5$ , and if $a\geq 6$ the latter is equivalent to $1$ mod $5$ . Check $1,2,3,4,5$ to get $(5,5)$ .
Otherwise WLOG $a>b$ . First, if $b=1$ then $a!+1$ and $1+a$ are powers of $5$ . In particular, $a< 5$ since otherwise $a!+1\equiv 1 \pmod 5$ . Else, $b>1$ so $b \mid a!+b$ , implying $5\mid b$ . Moreover, $5^x=a!+b=b\left(\frac{a!}{b}+1\right)$ Then, if $a\geq 10$ , we must have $\frac{a!}{b}+1 \equiv 1 \pmod 5$ so therefore $b=5$ . A finite case check yields no solutions.
Therefore, $(a,b)=(4,1),(1,4),(5,5)$

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#11 h Mar 15, 2024, 10:42 AM

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Here is my solution sketch...
Case 1:Let $a,b >5$
It is easy to see that both $a$ and $b$ must be powers of $5$ . Now, WLOG, let $a \geq b$ ,
$a!+b= b(\frac{a!}{b}+1)$
We see that $\frac{a!}{b}+1$ is $1$ modulo $5$ , so this choice of $(a,b)$ fails.
Case 2:Let $a>5$ and $b<5$
We see that $b$ must be odd. So $b=1$ or $b=3$ . However, both these choices of $b$ fail as $a!+b \equiv b \pmod 5$ .
Thus we can't choose such pairs of $(a,b)$ either.
Case 3: Let $a,b<5$ . Note that $(a,b)=(5,5)$ works.
In the third case, if $a,b\geq 2$ , it is easy to conclude that both $a$ and $b$ must be odd. However, $(3,3)$ does not work.
So both $a$ or $b$ must be $1$ , Now checking the last few cases left we see that $(a,b)=(4,1);(1,4)$ work.
We thus have gotten three such pairs of $a$ and $b$ .

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#12 h Mar 19, 2024, 1:52 PM

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Notice that if $x\in\mathbb{Z^*_+}$ , then $x!$ ends in 1 (1!), 2 (2!), 6(3!), 4(4!), or 0 ( $x\geq 5$ )
Let
$\begin{cases} a!+b= 5^x\\ b!+a= 5^y \end{cases}$
Where $x,y\in\mathbb{Z^*_+}$

If $a=1\implies b!+1= 5^y\implies b!\equiv 4 \mbox{ (mod 10)}$
So, $b= 4$ (if $b\geq 5\implies b!\equiv 0 \mbox{ (mod 10)})$
$\therefore \boxed{(a,b)= (1,4)}$ is solution ( $1!+4= 5= 5^1$ )

If $a=2\implies b!+2= 5^y\implies b!\equiv 3\mbox{ (mod 10)}$ , which is impossible.

If $a=3\implies b!+3= 5^y\implies b!\equiv 2 \mbox{ (mod 10)}\implies b=2$ . But, we also need that $a!+b$ is a power of five, so $3!+2= 8$ is a power of 5, contradiction!

If $a= 4$ , we get the solution $\boxed{(a,b)= (4,1)}$

If $a=5\implies 5!+b= 5^x\implies b!\equiv 5 \mbox{ (mod 100)}$ . Also, we know that $b!+5=5^y\implies b!\equiv 20 \mbox{ (mod 100)}$ .
But, if $b\geq 10, b!\equiv 0 \mbox{ (mod 100)}$ . So, we must have $b<10$ , giving us the solution $\boxed{(a,b)= (5,5)}$

Noooow, consider WLOG $5<a<b$ .

Let $b=a+b_0$ , where $b_0\in\mathbb{Z^*_+}$ (with a quick check, you can verify that $a=b$ gives us only the solution a=b=5)
We know that:

$a!(a+1)(a+2)\dots(a+b_0)+a= 5^y\implies a\mid 5^y$ .

So, $\boxed{\mbox{a is a power of 5}}.$
Call $a= 5^c$ , $c\in\mathbb{Z^*_+}$

We have:
$b!+5^c= 5^y\implies b!= 5^y-5^c= 5^c(5^{y-c}-1)$
$\implies v_5(b!)=c$ , but $b!$ has at least $(c+1)$ factors of 5 (actually much more than this), because he has at least the 5 and $a$ (a has c factors 5). $\implies c= v_5(b!)\geq c+1$ , contradiction!!

Therefore, 3 solutions:
$(a,b)= (1,4); (4,1); (5,5)$

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#13 h Apr 16, 2024, 2:09 AM • 1 Y

Y by Siddharthmaybe

Clearly $v_5(a!)=v_5(b)$ and $v_5(b!)=v_5(a)$ . Since $v_5(a!) \geq v_5(a)$ and $v_5(b!) \geq v_5(b)$ , it follows that $v_5(a)=v_5(b)=v_5(a!)=v_5(b!)$ .
Thus, $v_5((a-1)!)=v_5((b-1)!)=0 \implies a,b \leq 5$ . Checking gives $(a,b)=(1,4),(4,1),(5,5)$ as the only solutions.

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#14 h May 17, 2024, 7:13 PM

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Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$ .

Claim: If $x + y$ is a power of $5$ , then $\nu_5(x) = \nu_5(y)$ .
Proof: Suppose otherwise and WLOG $\nu_5(x) > \nu_5(y)$ . Then $\nu_5(x+y) = \nu_5(y)$ , so $x + y = 5^{\nu_5(y)} < y$ , absurd. $\square$

Hence $\nu_5(a!) = \nu_5(b)$ and $\nu_5(b!) = \nu_5(a)$ .

Claim: For all $a > 5$ , we have $\nu_5(a!) > \nu_5(a)$ .
Proof: This is equivalent to there being a positive integer multiple of $5$ less than $a$ . $\square$

Now for all positive integers $a$ , we have $\nu_5(a!) \ge \nu_5(a)$ since $a \mid a!$ . Thus, $\nu_5(b) = \nu_5(a!) \ge \nu_5(a)$ and $\nu_5(a) = \nu_5(b!) \ge \nu_5(b)$ , meaning $\nu_5(a) = \nu_5(b)$ . Then $\nu_5(a!) = \nu_5(b) = \nu_5(a)$ and $\nu_5(b) = \nu_5(a) = \nu_5(b!)$ , so $a \le 5$ and $b \le 5$ .

Thus, $1 < a! + b \le 5! + 5 = 125$ , so $a! + b \in \{5, 25, 125\}$ .

If $a! + b = 5$ , then $a = 1, b = 4$ , which works, or $a = 2, b = 3$ , which doesn't work since $3! + 2$ isn't a power of $5$ .

If $a! + b = 25$ , then $a = 4, b = 1$ , which works.

If $a! + b = 125$ , then $a = b = 5$ , which works.

Hence the answer is $(a,b) \in \{(1,4), (4,1), (5,5)\}$ .

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Sedro

#15 h May 29, 2024, 8:40 PM • 1 Y

Y by sami1618

We claim the only working ordered pairs are $(a,b)=\boxed{(1,4),(4,1),(5,5)}$ . It is easy to see that they work; we now prove they are the only ones.

It is not hard to verify that these are the only solutions when $a,b\le 5$ , so henceforth, FTSOC, assume $a,b>5$ . We must have $a!+b = 5^x$ and $a+b! = 5^y$ , for some positive integers $x$ and $y$ . Adding these two equations together, we have $a((a-1)!+1) + b((b-1)!+1) = 5^x+5^y$ . Note that $v_5(a!), v_5(b!) \ge 1$ , and thus $5\mid a$ , $5\mid b$ . Furthermore, notice that $v_5((a-1)!), v_5((b-1)!) \ge 1$ , and hence $5\nmid (a-1)!+1$ , $5\nmid (b-1)!+1$ . It follows that $v_5(\text{LHS}) = \min\{v_5(a), v_5(b)\}$ . We also have $v_5(\text{RHS}) = \min\{x,y\}$ .

WLOG assume $\min\{v_5(a), v_5(b)\}=v_5(a)$ . Now, suppose $\min\{x,y\}=x$ . Then, $v_5(a)=x$ , and $a\ge 5^x$ . But then $a!+b \ge (5^x)!+b > 5^x$ , contradiction. Thus, $\min\{x,y\}=y$ , and $v_5(a)=y$ . Then, $v_5(b)\ge y$ , and $b\ge 5^y$ . But then $a+b!\ge a+(5^y)! > 5^y$ , absurd. Thus, there are no solutions when $a,b>5$ , and we are done. $\blacksquare$

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#16 h Jun 28, 2024, 6:49 AM

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This was N1 at last year's shortlist.

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#18 h Jan 4, 2025, 1:39 AM

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$(1,4), (4,1), (5,5)$ work.

Case 1: $a=b$ : then $a((a-1)!+1)=5^k$ for some k. clearly a neq 1, so $a$ is a power of 5. $a=5$ works, but if $a$ is 25 or more, $(a-1)!+1)$ isnt a power of 5 since its 1 mod 5. So $(5,5)$ is the only solution here.

Case 2: $a>b$ . Consider b=1 (edge case since we noticed $(1,4)$ works. Then $a!+1=5^k, a+1=5^j.$ So $a=5^j-1$ . If a=4, it works. if a=24 or more, $a! +1$ is 1 mod 5 again. (same technique as before. we will use it again soon). So $(1,4) (4,1)$ works.

Now if $a>b>1$ , then $a!+b=5^k, a+b!=5^j$ . Then factor out b since $a>b$ to get $b(\frac{a!}{b}+1)=5^k$ . So b is a power of 5. Then $a+b!=5^k$ gives $a = 0 \mod 5$ . However, then $(\frac{a!}{b}+1)$ would 1 mod 5 (again) ((a would be the next multiple of 5 after b)). So no more solutions.

This post has been edited 2 times. Last edited by PaixiaoLover, Jan 4, 2025, 1:40 AM

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eg4334

#19 h Jan 4, 2025, 3:42 AM

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The only pairs are $\boxed{(1, 4), (5, 5)}$ in some permutation. If $a=b$ , then we have $a((a-1)!+1)$ to be a power of $5$ , or more specifically $(a-1)!+1$ to be a power of $5$ . This implies $a-1 < 5$ else it will be $1 \pmod{5}$ , and a finite check gives the only solution of $(5, 5)$ as advertised above. Else, WLOG $a>b$ . The key idea is that looking at $a!+b$ , we can factor out a $b$ from $a!$ and hence $\frac{a!}{b}$ cannot be a multiple of $5$ . However, when $a \geq 10$ , we can easily see that $v_5(a!)$ is strictly greater than $v_5(b)$ for all $b < a$ . This is most easily seen by noticing the gaps between powers of $5$ . Therefore, there is a finite check that we need to do with $a \leq 9$ , noting again that when $5 \leq a \leq 9$ then $b=5$ to eliminate the factor of $5$ . This lands us at the other advertised solution above.

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Perelman1000000

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Perelman1000000

#20 h Jan 5, 2025, 10:36 AM • 1 Y

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Case 1) $a=b\implies x=y$
$a!+a=5^x$
Case 1.1) $a<5$
$a=1\implies 1+1\neq 5^x$
$a=2\implies 2+2\neq 5^x$
$a=3\implies 6+3\neq 5^x$
$a=4\implies 24+4\neq 5^x$ so this case fails
Case 1.2) $a\geq 5$
$v_5(a!)\geq v_5(a)$
If $v_5(a!)=v_5(a)\implies a=5\implies a!+a=120+5=125=5^3\implies x=3$
If $v_5(a!)>v_5(a)$ ans also we know that $5|a\implies v_5(a!+a)=min(v_5(a!),v_5(a))=v_5(a)=v_5(5^x)=x\implies a\geq 5^x$ contradiction
Case 2) Wlog $a>b>5$
Since $a>b\implies v_5(a!)>v_5(b)\implies v_5(a!+b)=v_5(b)=x\implies b\geq 5^x$ contradiction
Case 3) $5>a>b$
By checking we get $4!+1=25=5^2$
$1!+4=5$ so $(a,b)=(4,1)$ if $b>a\implies (a,b)=(1,4)$
I think someone will ask why not $a>5>b$ because if $a>5$ $5|a!\implies 5|b\implies b>5$ since $a\neq b$ (case 1)
$\boxed{ANSWER:(a,b)=(1,4),(4,1),(5,5)}$

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iStud

#21 h Jan 6, 2025, 3:29 PM

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W.L.O.G. $a\ge b$ .

Case 1. If $a=b$ .
Then we have $a((a-1)!+1)\equiv 0\pmod{5}$ . It's easy to check that any $a\le 4$ doesn't works, so we must have $a\ge 5$ . For $a=5$ , the problem is satisfied. For $a>5$ , we have $(a-1)!+1\equiv 1\pmod{5}$ . But we have $5^t=a((a-1)!+1)$ for some $t\in\mathbb{N}$ , so $(a-1)!+1=1$ , a contradiction. So for this case, the only pair that satisfy is $(a,b)=(5,5)$

Case 2. If $a>b$ .
Let $a=b+k$ for some $k\in\mathbb{N}$ . If $a>b\ge 5$ , we have $5\mid a,b$ . In other side, we have $5^t=b((b+1)(b+2)\dots a+1)$ for some $t\in\mathbb{N}$ , so we must have $(b+1)(b+2)\dots a\equiv -1\pmod{5}$ .This is a contradiction since we must have $(b+1)(b+2)\dots a\equiv 0\pmod{5}$ (since that $5\mid a$ ). So we must have $a,b\le 4$ , which from that we easily get $(a,b)=(4,1)$ .

In the end, the only pairs that satisfy the problem are $(a,b)=(5,5),(4,1),(1,4)$ . $\blacksquare$

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#22 h Mar 24, 2025, 4:38 PM

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We claim $(1,4), (4,1)$ and $(5,5)$ are the only solutions.
We may assume $a\ge b$ and that implies $b\lvert a!$ . As $a!+b=b\left(\frac{a!}{b}+1\right)$ is given to be a power of $5$ , $b$ is necessarily a power of $5$ and is clear that $\gcd(a!/b, 5)=1$ . This forces $b=5^{\nu_5(a!)}$ and is a moment to celebrate as $b$ seems to get larger and larger surpassing $a$ given $a\ge 10$ which is against our assumption. Performing a finite check we get the required. $\square$

This post has been edited 2 times. Last edited by anudeep, Mar 24, 2025, 4:39 PM

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