How do I find the remainder of $23! +29!$ divided by $13! +19!$?

Question

What is the remainder of $23! +29!$ divided by $13! +19!$?

Attempt

$$23!+29!=23![1+(29×28×27×26×25×24)]$$ and

$$13!+19!=13![1+(19×18×17×16×15×14)]$$

Now how do I proceed..I am stuck at this initial stage.

Please help.

Answer 1

Partial answer

$$\begin{align}\frac{23!+29!}{13!+19!}&=\frac{23!(1+24\cdot25\cdot26\cdot27\cdot28\cdot29)}{13!(1+14\cdot15\cdot16\cdot17\cdot18\cdot19)}\\&=\frac{23!}{13!}\cdot\frac{2\cdot3\cdot7\cdot2+(2\cdot24)\cdot(3\cdot25)\cdot(7\cdot26)\cdot(2\cdot27)\cdot28\cdot29}{13\cdot5\cdot3\cdot3+(13\cdot14)\cdot(5\cdot15)\cdot(3\cdot16)\cdot17\cdot(3\cdot18)\cdot19}\cdot\frac{13\cdot5\cdot3\cdot3}{2\cdot3\cdot7\cdot2}\\&=\frac{23!}{13!}\cdot\frac{195}{28}\cdot\frac{84+48\cdot75\cdot182\cdot54}{585+48\cdot75\cdot182\cdot54}\cdot\frac{812}{323}\\&=(23\cdot22\cdot21\cdot20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14)\cdot\frac{3\cdot5\cdot13\cdot29}{17\cdot19}\cdot\frac{84+48\cdot75\cdot182\cdot54}{585+48\cdot75\cdot182\cdot54}\\&=(3\cdot5\cdot13\cdot14\cdot15\cdot16\cdot18\cdot20\cdot21\cdot22\cdot23\cdot29)\left(1-\frac{501+48\cdot75\cdot182\cdot54}{585+48\cdot75\cdot182\cdot54}\right)\\&=(3\cdot5\cdot13\cdot14\cdot15\cdot16\cdot18\cdot20\cdot21\cdot22\cdot23\cdot29)\\&\quad\quad-(14\cdot15\cdot16\cdot18\cdot20\cdot21\cdot22\cdot23\cdot29)\cdot\color{red}{\frac{167+3\cdot75\cdot182\cdot54}{1+16\cdot25\cdot14\cdot18}}\end{align}$$

and I think the next step would be to find what happens to the fractional term.

Using a calculator

The fraction becomes $$\frac{2211467}{108001}\approx 20.476...$$ and unfortunately $108001$ is prime...

Answer 2

Well... as a simple check, we have the common factor of $13!$ which I will ignore initially, then

$$ \begin{align} 14\cdot 15\cdot 16\cdot 17\cdot 18\cdot 19 + 1 &= 19535041 \\ (23!/13!) \equiv 19535040\cdot 20 \cdot 21 \cdot 22 \cdot 23 &\equiv 19322521 \bmod 19535041 \\ (29!/13!) \equiv 19322521\cdot 24 \cdot 25 \cdot 26 \cdot 27 \cdot 28 \cdot 29 &\equiv 5837545 \bmod 19535041 \\ 19322521 + 5837545 &\equiv 5625025 \bmod 19535041 \end{align}$$

so $23! + 29! \equiv 13!\cdot 5625025 \bmod (13!+19!)$