Bunuel wrote:

What is the units digit of \(23^{14} + 17^{12}\) ?

A. 0

B. 1

C. 7

D. 8

E. 9

We care only about units digits / last digits

and cyclicity.

Cyclicity is explained by Bunuel , Last Digit of a Power\(23^{14}\) has the same units digit as \(3^{14}\)

\(17^{12}\) has the same units digit as \(7^{12}\)

To find that last digit,

divide the exponent by the last digit's "

cyclicity number"

Exponents of single-digit integers

form patterns that repeat in cycles.

Cyclicity is the number of numbers

before the pattern repeats.

Find the cyclicity of 3 raised to powers

\(3^1 = 3\)

\(3^2 = 9\)

\(3^3 = 27\) (units digit of 7)

\(3^4 = 81\) (units digit of 1)

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\(3^5 = 243\) (units digit of 3 again)

\(3^6 = xx9\) (units digit of 9 again)

The patterns repeats after four powers of 3

Cyclicity of 3, therefore, is 4Cyclicity of 7 is also 4\(7^1 = 7\)

\(7^2 = 49\) (units digit of 9)

\(7^3 = 343\) (units digit of 3)

\(7^4 = xx1\) (units digit of 1)

\(7^5\) has a units digit of 1*7 = 7

Divide exponent of \(3^{14}\) by cyclicity of 4

\(\frac{14}{4} = 3\), remainder 2

Remainder of 2? Use the units digit of

the SECOND power of 3

\(3^2\) =

9Divide exponent of

\(7^{12}\) by 4

\(\frac{12}{4} = 3\), remainder 0

No remainder? Then the last power in the cycle

has the same units digit as your number.

That is, R = 0? Here, cyclicity of 4?

Use the units digit of the fourth power of 7

\(7^4\) has

units digit of 1Add the units digits

1 + 9 = 10The units digit of the sum of the

terms in the prompt is 0

Answer A

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Never look down on anybody unless you're helping them up.

--Jesse Jackson