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Bunuel
What is the units digit of \(13^{2003}\)?

(A) 1
(B) 3
(C) 7
(D) 8
(E) 9

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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Bunuel
What is the units digit of \(13^{2003}\)?

(A) 1
(B) 3
(C) 7
(D) 8
(E) 9

\(13^{2003}\)

13^2000 *13^3
cyclicty = 13 is 4 so 13
13^3 = 7
IMO C
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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
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Bunuel
What is the units digit of \(13^{2003}\)?

(A) 1
(B) 3
(C) 7
(D) 8
(E) 9

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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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

Why are we considering the last two in this case? And how do we know to?
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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
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\({13}^{2003}\): 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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