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- Thread starter jacks
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- Feb 7, 2012

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Would it be better to tackle it the other way round, finding $a$ and $b$ first and then using your clever $9$-and-$11$ idea to get $c$ and $d$?Given $34!=295232799cd9604140847618609643ab0000000.$ Then $(a,b,c,d)$ is

I did not understand how can i calculate $(c,d)$, If I get $(c,d)$,

Then i will easily get $(a,b)$ using divisibility by $9$ and $11$

One way to find $a$ and $b$ would be to start with the prime factorisation $34! = 2^{32}\cdot 3^{15}\cdot 5^7\cdot 7^4\cdot 11^3\cdot 13^2\cdot 17^2\cdot 19\cdot 23\cdot 29\cdot 31$, then divide by $2^7\cdot 5^7$ to eliminate the seven $0$s at the end of $34!$. That leaves you with $2^{25}\cdot 3^{15} \cdot 7^4\cdot 11^3\cdot 13^2\cdot 17^2\cdot 19\cdot 23\cdot 29\cdot 31$. You could find the last two digits of that by explicitly multiplying out that product, reducing$\mod{100}$ as you go – laborious but just about doable with a calculator.

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Thanks opalg got it.

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 $2^3$

$304,312,320,328,336,344,352,368,376,384$

In a similar manner

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

Now Using $9$ and $11$ divisibility test, we get $c$ and $d$

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

Thanks opalg.