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Problem: Given that $34!=295232799cd96041408476186096435ab000000$. Find $a, b, c, d$. $a, b, c, d$ are single digits.

I am able to find $a$ and $b$ but cant find $c, d$. I did the prime factorisation of $34!$ using De-Polignac's formula. I got $\frac{34!}{10^7}=2^{25}\times3^{15}\times7^4\times11^3\times13^2\times17^2\times19\times23\times29\times31$

So, i got $b=0$ and $a=2$ since $2^{25}\times3^{15}\times7^4\times11^3\times13^2\times17^2\times19\times23\times29\times31 \equiv 2 \mod 10$.

So, please help me find $c$ and $d$


marked as duplicate by Jyrki Lahtonen Sep 4 '15 at 19:42

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  • 7
    $\begingroup$ Hint: you know the thing is divisible by $9$, hence the digit sum is divisible by $9$. It is also divisible by $11$, hence the digit sum with alternating signs is divisible by $11$. $\endgroup$ – lulu Aug 31 '15 at 11:21
  • 1
    $\begingroup$ Thanks! I shuld have thought of it. $\endgroup$ – Subham Jaiswal Aug 31 '15 at 11:23
  • 1
    $\begingroup$ The problem has been discussed here: bmoc.maths.org/home/bmomarking.pdf $\endgroup$ – wythagoras Aug 31 '15 at 16:23
  • $\begingroup$ Based on the hot network question link, I thought you were really excited about the intermediate digits of a two digit number and I was thoroughly confused! $\endgroup$ – corsiKa Aug 31 '15 at 23:02
  • $\begingroup$ A related recent easier version with 23 in place of 34. $\endgroup$ – Jyrki Lahtonen Sep 1 '15 at 11:15

HINT: The number is a multiple of $9$ and a multiple of $11$.


You know that $34!$ is divisible by $9$, so, because the sum of all other digits is $141$, you know that $c+d=3$ or $c+d=12$.

Now do the same for divisibility by $11$ (remember the alternating sums criterion?).


Here $34!$ contain $2^{32}$ and $5^7$. So $2^{25}$ remaining and $2^{7}\cdot 5^7 = (10)^7$ form $7$ zeros at the end.

So $34! = 295232799cd9604140809643ab \times 10^7$ now after deleting $7$ zero,s ,

So we get $34! = 295232799cd9604140809643ab$

Now we use divisibility test for last $7$ digits using $2^7$.

Here $(a,b)\in \{0,1,2,3,4,5,6,7,8,9\}$

So Divisiblilty by $2^2$, we get

$00,04,08,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96$ ect

So Divisibility by $2^3$


In a similar manner

at last when divisible by $2^{7},$ we get $ab = 52$

Now, since these are digits, $0 \leq c, d \leq 9$, so $-9 \leq c-d \leq 9$ and $0 \leq c+d \leq 18$.

Use the fact that $34!$ is a multiple of $9,$ to tell you the value of $c+d$. We get that $c+d = 3$ or $12$.

Use the fact that $34!$ is a multiply of $11,$ to tell you the value of $c-d$.

We get that $c-d = -3$ or $8$. Since $2c$ is an even number from $0$ to $18,$

we conclude that $c=0, d=3$.

So $(a,b,c,d) = (5,2,0,3)$


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