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
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
