LHS is divisible by $ a! $ so as RHS from that a=1 Now if $ 3 \leq b,c $ it is impossible (mod3) so we have $ b\leq3 $ or $ c\leq3 $. Everything else are cases. I think it is easy now.

gobathegreat wrote: LHS is divisible by $ a! $ so as RHS from that a=1 Now if $ 3 \leq b,c $ it is impossible (mod3) so we have $ b\leq3 $ or $ c\leq3 $. Everything else are cases. I think it is easy now. So now comes the cases: $a=1$ then $1+b!+c!=3^d$ $b=1$ then $2+c!=3^d$ but mod $3$ implies $c=1$ for solution $(a,b,c,d)=(1,1,1,1)$ $b=2$ then $3+c!=3^d$ mod $3$ implies $3 \le c$ but if $6 \le c $ taking mod $9$ you get a contradiction ($9 | c! $ and $9 \le c!$) so $c=3$ for solution $(a,b,c,d)=(1,2,3,2)$ $c=4$ for solution $(a,b,c,d)=(1,2,4,3)$ $c=5$ no solution

I also did exactly what mszew did in the exam

Note that if $a > 1$ then the left-hand side is $even$, and therefore $a = 1$. If $b > 2$ then 3 divides $b! +c!$ and hence $3$ does not divide the left-hand side. Therefore$b = 1$ or $b = 2$. If $b = 1$ then $c! + 2 = 3^d$ , so $c < 2$ and hence$d = 1$. If $b = 2$ then $ c!$=$3^d-3$. Note that $d = 1$ does not give any solution. If $d > 1$ then $9 $ does not divide $c!$, so $c < 6$. By checking the values for$ c = 2, 3, 4, 5 $we see that $c = 3 $and $c = 4$ are the only $two$ $solutions$. Thus $(a, b, c, d) = (1, 1, 1, 1),(1, 2, 3, 2) or (1, 2, 4, 3)$. QED

nice solution sriva

My Solution. Let us consider that $a\ge 2.$ Then $b,c\ge 2$ as $a\le b\le c.$ As all of them are greater than or equal to $2,$ we can say that $2|a!,b!,c!.$ Or, $2|a!+b!+c!.$ Or, $2|3^d,$ which contradicts our assumption. So $1\le a<2.$ Which gives us the value of $a,$ which turns out to be $1.$ Now we can say that $b!+c!=3^d-1.$ Let us assume that $b\ge 3.$ Then $c\ge 3$ as $b\le c.$ Then $3|b!,c!.$ Or, $3|b!+c!.$ Or, $3|3^d-1,$ which contradicts our assumption because $d>0.$ Thus we get $1\le b<3.$ Which gives us the values of $b,$ which turns out to be $1$ and $2.$ Now, let us consider $2$ cases. $\text{Case I.}$ In this case, we assume that $b=1.$ Then $c!=3^d-2.$ Clearly, $c<3$ as for $c\ge 3,3$ will divide $3^d-2,$ which is not possible. Hence, $c=1$ or $c=2.$ If $c=2$ then $1!+1!+2!=1+1+2\neq 3^d~\forall~d\in \mathbb{N}.$ And if $c=1,$ then $1!+1!+1!=3^1\Rightarrow \boxed{(a,b,c,d)=(1,1,1,1)}~.$ Now, let's proceed to the $2^{\text{nd}}$ case. $\text{Case II.}$ In this case, we assume that $b=2.$ Then $c!=3^d-3=3(3^{d-1}-1).$ Which means $3|c!,$ but $3^2\nmid c!.$ So $3\le c<6,$ as $3^2|6!.$ If $c=3,$ then $1!+2!+3!=1+2+6=9=3^2\Rightarrow \boxed{(a,b,c,d)=(1,2,3,2)}~.$ If $c=4,$ then $1!+2!+4!=1+2+24=27=3^3\Rightarrow \boxed{(a,b,c,d)=(1,2,4,3)}~.$ If $c=5,$ hen $1!+2!+5!=1+2+120=123\neq 3^d~\forall~d\in \mathbb{N}.$ Thus, our required integer solutions are $\boxed{(a,b,c,d)=(1,1,1,1),(1,2,3,2),(1,2,4,3)}~.$ This completes the solution. $\Square$

RMO 2013 R3 P2 wrote: Find all $4$-tuples $(a,b,c,d)$ of natural numbers with $a \le b \le c$ and $a!+b!+c!=3^d$ Solution: $a \le b \le c$ and, $$a!+b!+c!=3^d$$If $a=1$ $\implies$ $1+b!+c!=3^d$ $$b!+c!=3^d-1$$Obviously, $b \geq 3 $ $\implies$ $b=1,2$. If $b=1$ $\implies$ $c!=3^d-2$. For $c \geq 3$ $\implies$ $3 \mid \text{ LHS}$ but $ 3 ~ \cancel{ \mid} ~ $ $\text{RHS}$ $\implies$ $c=1$ $\implies$ $(a,b,c,d)$ $\equiv$ $(1,1,1,1)$ And, $c=2$ $\implies$ Contradiction! If $b=2$ $$\implies c!=3^d-3$$Since, $9 ~ \cancel{\mid} ~ \text{RHS}$ $\implies$ $c \le 5 $. Now, $c=2,5$ Don't work! For, $c=2,3$ $\implies$ $(1,2,3,2)$ and $(1,2,4,3)$. For $a >1 $ $\implies$ $2 ~ \cancel{\mid} ~ \text{RHS}$ $\boxed{(a,b,c,d) \equiv (1,1,1,1), (1,2,3,2), (1,2,4,3)} $ $\qquad \blacksquare$

The only trick is to notice is $x!|y!$ if $x<y$ Rest is easy for mathlovers.

Since $a!\mid\text{LHS}$, we have that $a=1$ or $a!$ is a power of $3$. For $a\ge2$, it's always true that $2\mid a!$, so this leaves $a=1$. If $b\ge3$, then we have a contradiction$\pmod3$. Then we have two cases. Case 1: $b=1$ We have $2+c!=3^d$, and similarly $c<3$. If $c=1$ we have the solution $(a,b,c,d)=\boxed{(1,1,1,1)}$. If $c=2$ there is no solution. Case 2: $b=2$ Then $3+c!=3^d$. Also, $d=v_3(3+c!)$, so if $c\ge6$ we will have $d=1$, which doesn't work. Then we can manually check the remaining cases (noting that $c\ge3$) to get the solutions $(a,b,c,d)=\boxed{(1,2,3,2)},\boxed{(1,2,4,3)}$.