What is the units digit of 23^14 + 17^12 ?

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 4

Cyclicity 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
\(3^2\) = 9

Divide 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 1

Add the units digits
1 + 9 = 10

The units digit of the sum of the
terms in the prompt is 0

Answer A

Since we only care about units digits, we can rewrite the expression as:

3^14 + 7^12

Let’s start by evaluating the pattern of the units digits of units digits of 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the units digit of 3 raised to each power.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The pattern of the units digit of powers of 3 repeats every 4 exponents. The pattern is 3-9-7-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

3^12 has a units digit of 1.

3^13 has a units digit of 3.

3^14 has a units digit of 9

Next, we can evaluate the pattern of the units digits of 7^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are ONLY concerned with the units digit of 7 raised to each power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digit of powers of 7 repeats every 4 exponents. The pattern is 7-9-3-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

7^12 has a units digit of 1.

Therefore, 3^14 + 7^12 has a units digit of 0 since 9 + 1 = 10.

Answer: A

\(3^{14} = 3{4*3+2} =\) Units digit 9

\(7^{12}=7^{3*4}=\) Units digit 1

Thus, the units digit of \(23^{14} + 17^{12}\) will be 9 + 1 = 0, Answer must be A. 0