$a !+ b !+c != d !,$ where $a, b, c, d$ are all positive integers.

What is the number of possible solutions $(a, b, c, d)?$

0
1
2
3
4

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How can d be less than c?

Mario Diaz
- 2 years, 9 months ago

Very clever. Thanks.

David Salimonu
- 2 years, 9 months ago

Shouldn't each letter represent an individual integer? Because why would variables a,b,c have the same value if its not written somewhere? I got the answer wrong because of that.

Rūdis Rozītis
- 2 years, 9 months ago

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There was no such restriction and what's not forbidden is allowed - although both you and I assumed instinctively that restriction and thus we led ourselves astray.

Sándor Gombai
- 2 years, 9 months ago

No it shouldn't. This is not a cryptarithm.

Sharky Kesa
- 2 years, 9 months ago

I assumed they were all different integers and used your method to prove that there were no solutions...

William Kennedy
- 2 years, 9 months ago

I couldn’t get “How d=2 can’t work since n!>=1)” Plz someone explain.

Nitish Kumar
- 2 years, 9 months ago

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If $d=2$ , then the equation is $2! = a! + b! + c! \implies 2 = a! + b! + c!$ . But since $a!, b!, c! \ge 1$ , the sum has to be at least $3$ , so $d=2$ is not possible.

Zain Majumder
- 2 years, 9 months ago

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Nice simple solution!

Jeremy Galvagni
- 2 years, 9 months ago

Something needs to be said to the effect that d > max(a,b,c). Also the first sentence needs to be generalized for d > 4. My first point might be too nitpicky but I don't think my second one is. (Besides, it takes as little effort write the argument once for all d > 3 as it does to write the argument once for d=4.) ("For d > 3, 3*(d-1)! < d!").

Richard Desper
- 2 years, 9 months ago

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Your second point is absolutely critical. Without it, the argument is wrong.

Chris Maitland
- 2 years, 9 months ago

Let
$\alpha!<\beta!<\gamma!<\delta!$
$\gamma!(\alpha)(\alpha-1)(\alpha-2)...(\alpha-\gamma+1)(\alpha-\gamma)+(\beta)(\beta-1)(\beta-2)...(\beta+-\gamma+1)(\beta-\gamma)+1=\delta!$
.

When
$\alpha!,\beta!,\gamma!$
are equal.
$\gamma!(\alpha)(\alpha-1)(\alpha-2)...(\alpha-\gamma+1)(\alpha-\gamma)+(\beta)(\beta-1)(\beta-2)...(\beta+-\gamma+1)(\beta-\gamma)+1=3\gamma!$
$3\gamma!=\delta!$

Then the only different factor of
$\gamma!$
and
$\delta !$
is
$3$
,When
$\delta! =3!, \gamma!= 2!$
the equality holds.

To prove that no other solutions when
$\delta<3$
or
$\delta>3$

Case 1:
$\delta<3$
We know that 2!=2 since 2 can only be written as 2 positive integers that is 1+1 it will not satisfy the equation. The case for 1! is also similar as above.

Case 2:
$\delta>3$
We know that
$n!=n(n-1)!$
Since n>3 and
$3(n-1)!< n!=n(n-1)!$
we could show that the sum of three factorials that is smaller than
$\delta!$
when
$\delta>3$
is impossible.

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Great! I couldn't do it, but this is definitely helping

Zoe Codrington
- 2 years, 9 months ago

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Don't worry, neither could I. At least not this way.

Alan Laifer
- 2 years, 9 months ago

Without loss of generality let $a\leq b\leq c$ . Then we have $a!+b!+c!\leq 3c!$ . So the extra factor that $d!$ has over $c!$ is 3 or less. Note that there is no integer $k$ such that $k!=0$ . Therefore $c<d$ and so $d=2 \text{ or } 3$ and $c=1 \text{ or } 2$ .

There are only 4 possibilities left to check.

$1!+1!+1!=3$ which is not a factorial.

$1!+1!+2!=4$ which is not a factorial.

$1!+2!+2!=5$ which is not a factorial.

$2!+2!+2!=3\cdot 2!=3!$ which is a valid solution and therefore the only valid solution.

Note: this is pretty much identical to Sharky's solution but I tried to express it more naturally. The "without loss of generality" statement and the ordering of the variables are extremely useful and common tricks that reduce the complexity of the problem. They also make it easy to express the next idea that $d!$ can't be more than 3 times the largest term of the left hand side. Again, this inequality trick is another common way to reduce the scale of the solution space. Now, fortunately the solution space is so small you can use even inefficient and unintelligent ideas to find the solution. In conclusion, there is a standard bag of tricks built by experience that you should always think about using to reduce the complexity and solution space of a problem before you need to try to be clever.

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What if a=b=c=d? Who says they must be different? 🙈

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That would mean that $3d!=d!$ . Since there is no non-negative integer whose factorial is 0 or infinite, this implies $3=1$ or $c\neq d$ .

Chris Maitland
- 2 years, 9 months ago

Solution $2!+2!+2!=3!$ is the only solution in which $a=b=c$ , as every other situation in which $a=b=c$ is such that $a!+b!+c!=3a!$ , which is only a factorial if $a=2$ .

Therefore, every other solution must involve at least two distinct positive integers among $a,b,c$ . Suppose, without loss of generality, that $a$ is the smallest value between a,b,c. Also suppose, without loss of generality, that $b\ge c$ . That being the case, there exist integers $k>0,l \ge 0$ such that $b=a+k$ and $c=a+l$ .

In that way, we have that $a!+b!+c!=a!+(a+k)!+(a+l)!$ , which is only a valid solution if $1+\frac{(a+k)!}{a!}+\frac{(a+l)!}{a!}=\frac{(a+m)!}{a!}$ for some $m>k,l$ .

Note that $1+\frac{(a+k)!}{a!}+\frac{(a+l)!}{a!} \le 2\frac{(a+k)!}{a!}+1$ . Also note that $\frac{(a+m)!}{a!} \ge (a+m)\frac{(a+k)!}{a!}$ , but since $(a+k) >1$ , we have that $(a+m) \ge 3$ , implying that it is impossible for $1+\frac{(a+k)!}{a!}+\frac{(a+l)!}{a!}=\frac{(a+m)!}{a!}$ when $m>k \ge l$ .

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I'm waiting for someone smarter than me to explain it in terms stupid enough for me to understand...

lost zahhak
- 2 years, 9 months ago

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And? I guessed guess. As I said. Algebra 1, 8th Grade. I don’t do this stuff...

Trevor Greenlee
- 2 years, 9 months ago

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yeah, so what i said was I'm with you there... no clue how to solve.

lost zahhak
- 2 years, 9 months ago

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WLOG $a\ \leq b \leq c$ . Note that $d > c$ since factorials is a monotonicaly increasing function. We have $\begin{aligned} d! &= a!+b!+c!\\ &\leq 3c!\\ \dfrac{d!}{c!} &\leq 3\\ \dfrac{d!}{c!} &= 2, 3\\ \implies (d,c) &= (3,2), (2,1), (2,0) \end{aligned}$

We note that the $d=2$ cannot work since $n! \geq 1$ . Thus, the only possible solution is $d=3$ , $c=2$ , which yields $a=b=2$ . Thus, the only solution is $(2,2,2,3)$ .