Calculate the units digit of the following expression:

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Gmat_mission wrote:
Sun Jan 17, 2021 11:16 am
Calculate the units digit of the following expression.
\(
1!^1+2!^2+3!^3+4!^4+5!^5+\cdots+10!^{10}
\)

A) 1
B) 3
C) 5
D) 7
E) 9

Answer: D

Solution:

Recall that if n ≥ 5, then the units digit of n! will always be 0. Furthermore, the units digit of any power of a number with a units digit 0 is also 0. Therefore, the units digits of 5!^5, …, 10!^10 are all zero. In other words, we need to concentrate on only the first 4 terms of the sum. Simplifying those 4 terms by first expanding each factorial, we have:

1^1 + (2 x 1)^2 + (3 x 2 x 1)^3 + (4 x 3 x 2 x 1)^4

1^1 + 2^2 + 6^3 + 24^4

1 + 4 + 216 + 24^4

Since 24^4 has the same units digit as 4^4 and 4^4 = 256, we see that 24^4 has a units digit of 6.

Now, adding the units digit of these 4 terms, we have 1 + 4 + 6 + 6 = 17, which means the units digit of the entire sum is 7.

Answer: D

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