​If a, b, c, d are natural numbers, then how many possible remainders are there when 1^a + 2^b + 3^c + 4^d is divided by 10?​

Correct option is C

Solution:

Unit place of 1^a = 1

Possible unit place of 2^b = 2, 4, 6, 8

Possible unit place of 3^d= 3, 9, 7, 1

Possible unit place of 4^d = 4, 6

Sum of all the above unit places will always be an even number because

sum of two odd and two even numbers will always be even.

Possible even unit digits are = 0, 2, 4, 6, 8

We have to find the combination to get all these numbers for conformation,

1 + 2 + 3 + 4 = 10 (0 at unit place)

1 + 4 + 3 + 4 = 12 (2 at unit place)

1 + 2 + 7 + 4 = 14 (4 at unit place)

1 + 2 + 9 + 4 = 16 (6 at unit place)

1 + 2 + 1 + 4 = 8 (8 at unit place)

All the five unit digit can be found using these combinations.

∴ The correct answer is 5.