Fast Way of Finding The Remainder

Hint:

If $$a\equiv b\pmod c$$ And $$d\equiv e\pmod c$$ You can multiply the two to get $$a\times d\equiv b\times e\pmod c$$

Doing this for the numbers separately, then multiplying the expressions will get you the solution very quickly.

You can write each number as a multiple of $17$ plus a small remainder:

$$(17+12)(170\cdot 17+11)(118\cdot 17+11).$$

When you multiply this out (which you don't have to do) every term is a multiple of $17$ except the last one $12\cdot 11\cdot 11$. So you need consider only this last product. We can repeat what we just did with a little factoring. The above $= 3\cdot 11 \cdot 4 \cdot 11 = 33\cdot 44 = (17+16)(2\cdot 17 + 10).$ So you need consider only $16\cdot 10$.

Find remainder after dividing $29 \times 2901 \times 2017$ by $17$.

Work $\mod 17$ !

\begin{align} & 29 \times 2901 \times 2017 \\ & 12 \times 1201 \times 317 \\ & 12 \times 1201 \times 147 \\ & 12 \times 351 \times 62 \\ & 12 \times 11 \times 11 \\ & 1452 \\ & 602 \\ & 92 \\ & 7. \end{align}