if a 3-digit number is randomly chosen, what is the probability that either the number itself or some permutation of the number ( which is a 3-digit number) is divisible by 4 and 5 ?

Options

A. `1/45`
B. `29/180`
C. `11/60`
D. `1/4`

Answer

b

Explanation

We need 3-digit number which is divisible by 4 & 5 both.
i.e their last two diigits are
00, 20, 40, 60 & 80
Now, ending with 00 and '9'
{ 100, 200,…….,900}
if digit repeat other than '0' then they are
{ 220, 440, 660, 880, }
but 220 numbers can be permuted according to the condition as { 220,202}
so , there are '8' other favorable cases.
if the number have no digit reperated like 320 .
(320) can be permuted in 4 ways.
( 302, 230, 320,203)
so, such numbers are `8 xx 4xx 4 =128`
Total favorable = 9 + 8 + 128 =145
so, required prob= `145/900 = 29/180`