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Bunuel
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well, it's easier if you few digits with their powers up till 10 and first 10 factorials and their values.
or let me tell you a simple way
3^7 = 3.3.3.3.3.3.3
so rather than actually multiplying 3 with itself 7 times, you can club it up = 9.9.9.3 = 729*3
i hope this helps.
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n = 4q + 3
possible values of n are 3,7,11,15,19...

The cyclicity of powers of 3 is: 3,9,7,1

By looking at all the possible values of n, you can quickly see that they will always end up being in the third place of the cyclicity of powers of 3 (as you increase every possible value of n by 4 every time)

Therefore, the units digit will always be a 7

therefore the remainder of 3^n + 2 will always be 9/5 = 4
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We can rewrite \(n=4x+3\)
We are then looking for the remainder of 3^(4x+3)+2 / 5

If we observe the remainders of various powers of 3 we notice that:
\(3^1/5=3/5\)
\(3^2/5=9/5\)
\(3^3/5=27/5\)
\(3^4/5=81/5\)
Notice that we only really need the last digit, which is easy to obtain! In this case 81 ends in 1 so the next power will end in \(1*3=3\)
\(3^5/5=243/5\)
\(3^6/5=...9/5\)
\(3^7/5=...7/5\)

We are looking for the remainder of 3^(4x+3)+2 so for example 3^7+2. 3^7 ends in 7, 7+2=9 so 3^7+2 ends in 9 -> remainder 4
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