Peter wrote: Given that \[34! = 95232799cd96041408476186096435ab000000_{(10)},\] determine the digits $a, b, c$, and $d$. First, there is a typo and the number should read \[34! = 295232799cd96041408476186096435ab000000\] Now starting with b, we must examine how many zeros 34! has at the end of the number. To do this, it is best to factor it using \[n! = \prod_{p\le n}p_{i}^{\sum^{\infty}_{r=1}\lfloor \frac{n}{p_{i}^{r}}\rfloor}\] which gives $34! = 2^{32}\cdot3^{15}\cdot5^{7}\cdot7^{4}\cdot11^{3}\cdot13^{2}\cdot17^{2}\cdot19\cdot23\cdot29\cdot31$ The number of zeros at the end is limited by the powers of 5 and 2, but namely 5 since there are always less of them. So this implies that there are 7 zeros and since we see 6 of them implying b=0. To find a, we divide $10^{7}$ leaving us with $\frac{34!}{10^{7}}= 2^{25}\cdot3^{15}\cdot7^{4}\cdot11^{3}\cdot13^{2}\cdot17^{2}\cdot19\cdot23\cdot29\cdot31$ Now a is the ones digit of this number, so if we look at the conguence of this product (mod10) we will know a. $a = 2^{25}\cdot3^{15}\cdot7^{4}\cdot11^{3}\cdot13^{2}\cdot17^{2}\cdot19\cdot23\cdot29\cdot31 \equiv 2\cdot7\cdot1\cdot1\cdot9\cdot9\cdot9\cdot3\cdot9\cdot1 \equiv 2\cdot3^{9}\cdot7 \equiv2\cdot3\cdot7 \equiv 2 (mod10)$ Hence a = 2. Now for c,d. These two were a bit more troublesome, but remembering a few tricks from algebra about finding if a number is divisible by 3, 9, and 11 by merely looking at its digits help. Recall, a number is divisible by 3 if and only if the sum of its digits is divisible by three, this implies $34!=\sum^{k}_{i=0}a_{i}\cdot10^{i}=141+c+d \equiv c+d \equiv 0 (mod3)$ Similarly, a number is divisible by 9 if the sum of its digits are divisible by 9, implying $34!=\sum^{k}_{i=0}a_{i}\cdot10^{i}=141+c+d \equiv 0 (mod9)$ Numbers divisible by 11 have a more complicated rule. a number is divisible by 11 if and only if 11 divides $\sum^{k}_{i=0}(-1)^{i}\cdot a_{i}$ This leads to $\sum^{k}_{i=0}(-1)^{i}\cdot a_{i}= 19-c+d \equiv 0 (mod11)$ So we have three constraints on (c,d) $c+d \equiv 0 (mod3)$ $c+d \equiv 0 (mod9)$ $19-c+d \equiv 0 (mod11)$ After testing all the combinations of (c,d) this leaves (c,d) = (0,3) as the unique solution. Summing it up, {a,b,c,d} = {2,0,0,3}

34!=295... (the first digit is two, not nine) b=0 because there are seven five (factor 5, 10, 15, 20, 25, 30) that divide 34! (and, of course also seven two), so the last seven digits are all zeros.

$34!=295232799039604140847618609643520000000\implies c=0,d=3,a=0$

Peter wrote: Given that \[34! = 95232799cd96041408476186096435ab000000_{(10)},\] determine the digits $a, b, c$, and $d$. $v_p(34!)=[\frac{34}{p}]+[\frac{34}{p^2}]+...$. $p=5, v_5(34!)=7\to b=0,v_2(34!)>7\to a-$even and ${a\mod 5=(1*2*3*4)^7*(\frac 55*\frac{10}{2}*\frac{15}{5}*\frac{20}*5}*\frac{30}{5}*\frac{25}{25}=(-1)^7*(-1)=1$. Therefore $a=6$ $9|34!\to c+d=1\mod 9\to c+d=10 or (cd)=(1,0),(0,1), 11|34!\to c-d-13=0\mod 11\to c-d=2\mod 11\to (c,d)=(0,9),(d+2,d)$. It give $c=6,d=4.$

Not sure how you got those numbers.....but if you use a powerful calculator you can actually verify the numbers. Even microsoft windows calculator will show you the right answers for all but digit b.