A number when divided by 143 leaves remainder 31. The remainder when the number is divided by 13 is
[a] 0
[b] 1
[c] 3
[d] 5

Hint: Let the number be n. Use Euclid's division lemma with a = n and b = 143. Write 31 as 26+5 and take 13 common from the first two terms. Hence find the remainder obtained on dividing by 13.
Alternatively, you can use the property that if $a\equiv b\bmod m$ and n divides m then $a\equiv b\bmod n$.
Use the fact that if $a\equiv b\bmod m$ thenn$a\equiv b-cm\bmod m$, where c is an integer.
Hence find the remainder on dividing by 13.

Complete step-by-step answer:
We know from Euclid's division lemma if r is the remainder on dividing a by b then
a = bq+r.
Let n be the given number.
Hence n = 143q+31
Hence n = 143q+26+5
Taking 13 common from the first two terms, we get
n = 13(11q+2) +5
i.e. n = 13s+5 where s is an integer.
Since $0\le 5<13$we have
The remainder on dividing n by 13 is 5.
Hence option [d] is correct.

Note: Let n be the given number.
Hence $n\equiv 31\bmod 143$
We know that if $a\equiv b\bmod m$ and n divides m then $a\equiv b\bmod n$.
Since 13 divides 143, using the above property, we get
$\begin{align}
  & n\equiv 31\bmod 13 \\
 & \Rightarrow n\equiv 5\bmod 13 \\
\end{align}$
Hence the remainder obtained on dividing the number by 13 is 5.
Hence option [d] is correct.