Question

Five digit numbers divisible by 3 are formed sing 0, 1, 2, 3, 4, 5, 6 and 7 without repetition. The total number of such numbers is

• A. 312
• B. 3125
• C. 120
• D. 216

Answer 1

The correct option is B 3125

We should determine which 5 digits from given 8 could from a 5 digit no. divisible by 3.
Sum of all digits = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
Hence the largest sum divisible by 3 = 27
Least sum = 0 + 1 + 2 + 3 + 4 = 10
$\therefore$ Hence the smallest sum divisibly by 3 = 12
Now we need to find combinations of 5 no's
where sum is 12, 15, 18, 21, 24 and 27.
Let us take 12:
$12\to 0,1,2,3,6\to$ first number cannot be zero.
So we have 4 first numbers and rest 4 arranged in 4! ways
$\Rightarrow 4\times 4!$ ways
$12\to 0,1,2,4,5$ $\to 4\times 4!$ ways
Vect 15
15 : 0, 1, 2, 5, 7
1, 2, 3, 4, 5
0, 1, 3, 4, 7
0, 1, 3, 5, 6
The above itself comes to $4\times 4!\times 5cards+5!$
$=20\times 4!+120$
$=480+120=600$
Similarly you can work out for 18, 21, 24, and 27
$\Rightarrow$ The only option correct is 3125