Exponents/Powers

swerve wrote: ↑

Sun Feb 23, 2020 11:21 am

\(17^{27}\) has a units digit of:

A. 1
B. 2
C. 3
D. 7
E. 9

The OA is C

Source: Manhattan Prep

Look for a pattern

17^1 = 17
17^2 = (17)(17) = ---9 [aside: we need not determine the other digits. All we care about is the units digit]
17^3 = (17)(17^2) = (17)(---9) = ----3
17^4 = (17)(17^3) = (17)(---3) = ----1
17^5 = (17)(17^4) = (17)(---1) = ----7

NOTICE that we're back to where we started.
17^5 has units digit 7, and 17^1 has units digit 7
So, at this point, our pattern of units digits keep repeating 7, 9, 3, 1, 7, 9, 3, 1, . . .
We say that we have a "cycle" of 4, which means the digits repeat every 4 powers.

So, we get:
17^1 = --7
17^2 = ---9
17^3 = ----3
17^4 = ----1
17^5 = ----7
17^6 = ---9
17^7 = ----3
17^8 = ----1
17^9 = ----7
17^10 = ----9
etc.

Notice that when the exponent is a MULTIPLE of 4 (4, 8, 12, 16, ...), the units digit will be 1
Since 24 is a MULTIPLE of 4, we know that the units digit of 17^24 will be 1
Continuing on, we get:
17^25 = ----7
17^26 = ---9
17^27 = ----3

Answer: C

Here's an article I wrote on this topic (with additional practice questions): https://www.gmatprepnow.com/articles/un ... big-powers

Cheers,
Brent

swerve wrote: ↑

Sun Feb 23, 2020 11:21 am

\(17^{27}\) has a units digit of:

A. 1
B. 2
C. 3
D. 7
E. 9

The OA is C

Source: Manhattan Prep

The pattern for the units digits of powers of 7 is 7 - 9 - 3 - 1. So, when 7 is raised to a power that is a multiple of 4, the units digit is 1. Thus, the units digit of 7^28 is 1, and so the units digit of 7^27 is 3.

Answer: C