Nashan randomly chooses 6 positive integers a, b, c, d, e, f. Find the probability that 2 ^a + 2^b + 2^c + 2^d + 2^e + 2^f is divisible by 5.

Answer: 0.20

Step-by-step explanation:

we have that:

a number is only divisible by 5 if:

the unit digit of the number is 0 or 5.

As we have:

2^a + 2^b + 2^c + 2^d + 2^e + 2^f = X

we can never have 5 as the last digit, so we only can have a zero as the last digit.

Now, we know that

2^a + 2^b + 2^c + 2^d + 2^e + 2^f can only end with even numbers, if we assume the same probability for all of them, we have 0, 2, 4, 6, 8

5 even numbers, and the sum will be divisible by 5 only if the last number is a zero, then the probability having a zero in the last number is equal to the number of times that zero appears in that set, divided the total number of elements in that set.

p = 1/5 = 0.20