Question

A five-digit number divisible by $3$ is to be formed using the numericals $0,1,2,3,4and5$ without repetition. The total number of ways this can be done is

• A. $216$
• B. $240$
• C. $600$
• D. $3125$

Answer 1

Answer: (A)

$216$

We know that a number is divisible by $3$ if the sum of its digits is divisible by $3$

Now the sum of the digits $1,2,3,4$ is $15$ which is divisible by $3$

$\therefore$ All the five digit numbers formed by the digits $1,2,3,4,5$ are divisible by $3$ and their number $=5!=120$

When we include $0$, the four other digits whose sum is divisible by $3$ are $1,2,4$ and $5$

$\therefore$ The number of numbers in this case $=4\times 4!=4\times 24=96$

Hence, the total number of ways $=96+120=216$