Carcass wrote:
What is the units digit of \(13^{2003}\)?
(A) 1
(B) 3
(C) 7
(D) 8
(E) 9
Kudos for the right answer and explanation
Look for a
pattern13^1 = 1
313^2 = (13)(13) = ---
9 [aside: we need not determine the other digits. All we care about is the units digit]
13^3 = (13)(13^2) = (13)(---9) = ----
713^4 = (13)(13^3) = (13)(---7) = ----
113^5 = (13)(13^4) = (13)(---1) = ----
3NOTICE that we're back to where we started.
13^5 has units digit
3, and 13^1 has units digit
3So, at this point, our pattern of units digits keep repeating
3, 9, 7, 1, 3, 9, 7, 2, . . . We say that we have a "cycle" of 4, which means the digits repeat every 4 powers.
So, we get:
13^1 = --
313^2 = ---
9 13^3 = ----
713^4 = ----
113^5 = ----
313^6 = ---
9 13^7 = ----
713^8 = ----
113^9 = ----
313^10 = ----
9 etc.
Notice that when the exponent is a MULTIPLE of 4 (4, 8, 12, 16, ...), the units digit will be
1Since 2000 is a MULTIPLE of 4, we know that the units digit of 13^2000 will be
1Continuing with the pattern:
13^2001 = --
313^2002 = ---
9 13^2003 = ----7
Answer: C
Here's an article I wrote on this topic (with additional practice questions):
https://www.greenlighttestprep.com/arti ... big-powers Cheers,
Brent