Bunuel
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)
---------------------
\(3^5 = 243\) (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