Find the units digit of 3^{2018} - 2^{2018}.

C 5
Since exponential power of 3 follows the below pattern for unit digit:
3^1=3, unit digit is 3
3^2=9, unit digit is 9
3^3=27, unit digit is 7
3^4=81, unit digit is 1
3^5=243, unit digit is 3
3^6=729, unit digit is 9
This means cyclicity of exponential power of 3 is 4. Now 2018=2016+2. This means 3^2018 cab be written as 3^2*3^2016. Since 3^2016 will give unit digit as 1 (Since 2016 is completely divisible by 4), 3^2018 will give the unit digit as (3^2)*1= 9
Same way,cyclicity of 2 is 4 (Since 2,4,8,16,32,64,128)
Thus 2^2018 will give same unit digit as (2^2)*(2^2016), which means 4*1=4
So, unit digit of 3^2018 - 2^2018 = 9-4=5

Hence the answer

=>

The units digit is the remainder when 3^{2018} - 2^{2018} is divided by 10.

The remainders when powers of 3 are divided by 10 are
3^1: 3,
3^2: 9,
3^3: 7,
3^4: 1,
3^5: 3,
…
So, the units digits of 3^n have period 4: they form the cycle 3 -> 9 -> 7 -> 1.
Thus, 3^n has the units digit of 9 if n has a remainder of 2 when it is divided by 4.
The remainder when 2018 is divided by 4 is 2, so the units digit of 3^{2018} is 9.

The remainders when powers of 2 are divided by 10 are
2^1: 2,
2^2: 4,
2^3: 8,
2^4: 6,
2^5: 2,
…
So, the units digits of 2^n have period 4: they form the cycle 2 -> 4 -> 8 -> 6.
Thus, 2^n has the units digit of 4 since n has a remainder of 2 when it is divided by 4.
The remainder when 2018 is divided by 4 is 2, so the units digit of 2^{2018} is 4.

3^{2018} - 2^{2018} has remainder 9 – 4 = 5 when it is divided by 10.

Therefore, the answer is C.
Answer: C