What is the remainder when the product of 335, 608 and 853 is divided by 13?

To find the remainder when the product $335\times 608\times 853$ is divided by $13$, we can find the remainder of each individual number when divided by $13$ and then multiply these remainders together.

According to the properties of modular arithmetic:
$(a\times b\times c)\mathrm{mod}n=\left[\right(a\mathrm{mod}n)\times (b\mathrm{mod}n)\times (c\mathrm{mod}n\left)\right]\mathrm{mod}n$

Step 1: Find the remainder of each number when divided by 13

For $335$:
$335=13\times 25+10$
So, the remainder is $10$, meaning $335\equiv 10\left(\mathrm{mod}13\right)$.

For $608$:
$608=13\times 46+10$
So, the remainder is $10$, meaning $608\equiv 10\left(\mathrm{mod}13\right)$.

For $853$:
$853=13\times 65+8$
So, the remainder is $8$, meaning $853\equiv 8\left(\mathrm{mod}13\right)$.

Step 2: Multiply the individual remainders together

Now, we calculate the product of these individual remainders:
$10\times 10\times 8=800$

Step 3: Find the remainder of this product when divided by 13

We divide $800$ by $13$ to find the final remainder:
$800=13\times 61+7$
So, $800\equiv 7\left(\mathrm{mod}13\right)$.

Therefore, the remainder when the product of 335, 608, and 853 is divided by 13 is 7.