What is the units digit of 13^2003 ?

Look for a pattern

13^1 = 13
13^2 = (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) = ----7
13^4 = (13)(13^3) = (13)(---7) = ----1
13^5 = (13)(13^4) = (13)(---1) = ----3

NOTICE that we're back to where we started.
13^5 has units digit 3, and 13^1 has units digit 3
So, 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 = --3
13^2 = ---9
13^3 = ----7
13^4 = ----1
13^5 = ----3
13^6 = ---9
13^7 = ----7
13^8 = ----1
13^9 = ----3
13^10 = ----9
etc.

Notice that when the exponent is a MULTIPLE of 4 (4, 8, 12, 16, ...), the units digit will be 1
Since 2000 is a MULTIPLE of 4, we know that the units digit of 13^2000 will be 1
Continuing with the pattern:
13^2001 = --3
13^2002 = ---9
13^2003 = ----7


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

Given

• The number \(13^{2003}\)

To Find

• Unit digits of the number.

Approach and Working Out

• We just need to consider the last digit of the base and the last two digits of the exponent.
• It makes the number \(3^{03}\)

o The unit digit of this number is 7.

Correct Answer: Option C
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Units digit of 3 has a cycle of 4

Hence units digit of \(13^{2003}\) ==> \(3^{2000}\)*\(3^{3}\)==>7

Hence C
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\(13^{2003}\)

13^2000 *13^3
cyclicty = 13 is 4 so 13
13^3 = 7
IMO C

Units digit of 3 has a cycle of 4 ; 3,9,7,1

Hence units digit of 13^2003 ==> Unit digit of 13^3 = 7

Hence C

Solution:

The units digit of 13^2003 is the same as the units digit of 3^2003. Recall that the units digit pattern of powers of 3 is 3-9-7-1. Thus, when the exponent is a multiple of 4, the units digit is 1. Therefore, the units digit of 3^2004 is 1 and going backward one step in the pattern, the units digit of 3^2003 is 7.

Answer: C

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Why are we considering the last two in this case? And how do we know to?

In solving units digit problems, we just need to determine the value of the units digit of the base raise to the exponent. In this case, however, we have a large-value exponent. Since we know that the patterns for the units digit has maximum of 4 numbers, we can use only the last two digits of the exponent.

Simplified:
Units digit of 13^2003 = units digit of 3^03 = 7
CORRECT ANSWER = C

\({13}^{2003}\): For unit digit consider \(3^{2003}\)

=> \((3^4)^{500} * {3^3}\)

=> \(3^4 * 3^3 => 3^5 * 3^2\)

Every number repeats itself at unit digit when raised to power '5'.

=> \(3 * 3^2\) = 27

Answer C

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