Are you finding the last nonzero digit of that product?

the answer is 4 right ?

@Nguyen No. Correct answer is 2

How do you solve this?

Maybe find the number of zeros that $A$ has... call that $x$. You could do $\frac{A}{10^x} \pmod{10}$.. Not sure, of any other easy way to do this. I am not sure on how to use, $A \equiv 1 \pmod{3}$. Working on the actual solution, I will post it if I can get it

Certainly it is just CRT on $2$ and $5$, and since $v_2(A)>v_5(A)$ we know that it is even. At that point we partition the sequence into subsequences based on the $5$-adic valuation (since $2014$ is not too big), and bash the product using periodicities.

ar03 wrote: Maybe find the number of zeros that $A$ has... call that $x$. You could do $\frac{A}{10^x} \pmod{10}$.. Not sure, of any other easy way to do this. I am not sure on how to use, $A \equiv 1 \pmod{3}$. Working on the actual solution, I will post it if I can get it it's just to know that the terms of the product are on arithmetic sequence, $a_i = 1+(i-1)3$ so $A= 10^{67}*k$

Wait, why $10^{67}$ Wouldn't be $10^{670}$? (I am probably just missing something.)

@above $a_{672}= 2014$ and $a_i \equiv 0 (mod 10) \iff i \equiv 4 (mod 10)$ so it happens $ \frac {672}{10}=67.2 \rightarrow 67 $ times but im not considering the cases where $a_i * a_j \equiv 0 (mod 10)$ so it's very bashy to do this

Answer is 4 We see that by periodicity this reduces to $$ (1*4*7*0*6*9*2*5*8)^{67} *1*4$$We see that the number of occurrences of $5$ is less than $2$ therefore we can ignore the $0$ and $5$ in the above .Simplifying we get the desired result.

It must be (1×4×7×0×3×6×9×2×5×8)^67×1×4

Here's my solution.

Attachments:
Digits.docx (13kb)

@VicKmath7 after you get (1*4*7*0*3*6*9*2*5*8)^67 *1*4 how you reach the final answer which is 4???

I mean... I am trying to figure it out but i cant!

Notice that \[ A=\prod_{k=1}^{672}3(k-1)+1 \]Clearly $\nu_2(A)>\nu_5(A)$. To find trailing zeros of $A$, sufficient to find how many $5$'s are in $A$ i.e. how many $5, 5^2, 5^3, \ldots$ terms are in $A$. Since $k$ varies from $1$ to $672$, \begin{align*} 3(k-1)+1&=5m \implies m \text{ has } 134 \text{ solutions} \\ 3(k-1)+1&=5^2m \implies m \text{ has } 27 \text{ solutions} \\ 3(k-1)+1&=5^3m \implies m \text{ has } 5 \text{ solutions} \\ 3(k-1)+1&=5^4m \implies m \text{ has } 1 \text{ solutions} \end{align*}Hence $\nu_5(A)=134+27+5+1=167$. So now we are interested in $\frac{A}{10^{167}} \pmod{10}$. Let $B=\frac{A}{10^{167}}$. Clearly \[ B\equiv 0\pmod 2.\tag{1} \]And \begin{align*} 2^{167}\cdot B&\equiv (1\cdot 2\cdot 3\cdot 4)^{134}\cdot 4\cdot (1\cdot 2\cdot 3\cdot 4)^{26}\cdot 2\cdot 3\cdot (1\cdot 2\cdot 3\cdot 4)^5\cdot 1\cdot 4\cdot (1\cdot 2\cdot 3\cdot 4)^1 \\ &\equiv 1\pmod 5 \end{align*}Since $2^{167}\equiv 3\pmod 5$, we have \[ B\equiv 2 \pmod 5.\tag{2} \]Combining congruence $(1), (2)$ (or applying CRT) we have $B\equiv 2\pmod{10}$. So answer is $\boxed{2}$.

$1,4,7,10,13,16,19,22,25,28$ $31,34…$ $2011.2014$ Take mod 10 with getting rid of 5’s $(1.4.7.2.3.6.9.2.8)^{67} \times 1.4 \equiv 2(\bmod 10)$ hence the answer is 2

well it's cycling and it's not to much long it comes from there