To get the unit digits of integers, we only need to do the computation of the unit digit of the powers of the original integer and look out for patterns

3^1 = 1
3^2 = 3
3^3 = 7
3^4 = 1 We can see that the unit digit of 23 power repeats after every 3 iteration.
Therefore the unit digit of 23^2 = 23^5 = 23^8 = 23^11 = 23^14 = 9

For 17 we follow the same method.

7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7 We can see that the unit digit of 17 power repeats after every 4 iteration.
Therefore the unit digit of 17^4 = [17^8 = 17^2} = 1

Hence the unit digit of 23^14 + 17^12 = The unit digit of (9+1) = The unit digit of 10 = 0

Hence option A = 0 is the answer.

As we are dealing with unit digits we have to know the pattern of unit 3 and 7.

pattern for 3: 3-9-7-1
pattern for 7: 7-9-3-1

Now \(23^{14}\) unit digits will be 9 and \(17^{12}\) unit digit will be 1. total =9+1=10. So ultimate unit digit will be 0, which is option A.

To understand better one must know the cyclicity of unit digits.

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 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
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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
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