Question

Given that the sum of digits of a 5 digit number is 41. Find the probability that such a number is divisible by 11?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

In order to get the sum as 41, the following 5 digit combinations exist:

99995 → number of 5 digit numbers = 5
99986 → number of 5 digit numbers = 20
99977 → number of 5 digit numbers = 10
99887 → number of 5 digit numbers = 30
98888 → number of 5 digit numbers = 5

Now, 70 such numbers exist.
Now for a 5 digit number of form (pqrst) to be divisible by 11,

(p+r+t)−(q+s) = 11, also (p+r+t)+(q+s) = 41
$\therefore$ p+r+t = 26, q+s = 15

Note that, (p+r+t)−(q+s) = 22 is not possible as it will give fractional values of (p+r+t) and (q+s).

Also, (p+r+t)−(q+s) = 33 is not possible as it will give (p+r+t) = 37 and we know p,r,t are digits.

(p,r,t) = (9,9,8) and (q,s) = (8,7) -------- (i)
or (p,r,t) = (9,9,8) and (q,s) = (9,6) -------- (ii)

Using $1^{st}$ equation we can construct 6 numbers.

Using $2^{nd}$ equation we can construct 6 numbers.

$\therefore$ Number of favourable cases = 12.

Hence, required probability = $\frac{12}{70}$ = $\frac{6}{35}$