the units digit of n^100?

Question

There is some sort of pattern... I can't find it!

1^1001
2^100 1.2676506 × 1030
3^100 5.15377521 × 1047
4^1001.60693804 × 1060
5^1007.88860905 × 1069
6^1006.53318624 × 1077
7^1003.23447651 × 1084
8^1002.03703598 × 1090
9^1002.65613989 × 1095
10^1001.0 × 10100
11^1001.37806123 × 10104
12^1008.28179745 × 10107
13^1002.47933511 × 10111
14^1004.10018609 × 10114
15^1004.06561178 × 10117
16^1002.58224988 × 10120
17^1001.10889937 × 10123
18^1003.36705732 × 10125
19^1007.50516242 × 10127
20^1001.2676506 × 10130
21^1001.66697648 × 10132
22^1001.74690015 × 10134
23^1001.48861915 × 10136

Answer 1

You are asking what the units digit would be for any number raised to the power of 100, right?

I'm gonna start thinking out loud here to see if I can reason this out:

If n ended with 1, then n^100 would end with 1 (since 1 * 1 = 1, you can only get 1 in the digit field when you start with a number ending in 1).

Similarly if n ended with a 0 then n^100 would end with 0 (since 0 * 0 = 0)

Similarly if n ended with a 5, then n^100 would end with 5 (since 5 * 5 = 25)

Similarly if n ended with a 6, then n^100 would end with 6 (since 6 * 6 = 36)

So let's look at some other numbers to see the patterns:

2^1 = 2 (last digit only)
2^2 = 4 (last digit only)
2^3 = 8 (last digit only)
2^4 = 6 (last digit only)
2^5 = 2 (last digit only)

So the pattern for numbers ending with 2 is 2 4 8 6 then repeats. When the power is a mutiple of 4, the last digit is a 6.

Therefore, n^100 has a last digit of 6 when n has a last digit of 2.

Similarly, find a pattern with the other numbers (I will omit a lot of my work here for brevity, but the idea is the same as what I demonstrated for n=2).

The patterns show that if the number n ends in the digit "X" (first column below) then n^100 ends in the digit "Y" (second column below):

X Y
--------
0 0
1 1
2 6
3 1
4 6
5 5
6 6
7 1
8 6
9 1

So, hmmmmm, is there a pattern here or a formula you can state?

If n ends in an even digit other than 0, then n^100 ends in 6.

If n ends in an odd digit other than 5, then n^100 ends in 1.

If n ends in 0, then n^100 ends in 0.

If n ends in 5, then n^100 ends in 5.

Not sure you can state much more than this.

Answer 2

Why are you computing the whole number? You're only interested in the units digit. In case you're confused, that's the _last_ digit, not the first. Note that the only thing that affects the last digit is the last digit of n, so we can compute the units digit of any of these numbers just by considering the units digit of 0-9 to the 100th power.

0^100 ≡ 0 mod 10
1^100 ≡ 1 mod 10
For higher powers, just consider the units digit as you multiply by the first few powers and look for a pattern:
2^1 ≡ 2 mod 10
2^2 ≡ 4 mod 10
2^3 ≡ 8 mod 10
2^4 ≡ 6 mod 10
2^5 ≡ 2 mod 10

So the cycle repeats every 4 powers. 2^8 ≡ 6 mod 10, 2^12 ≡ 6 mod 10, and so on. Since 100 is a multiple of 4, you have:

2^100 ≡ 6 mod 10

Similar logic can be used to find the units digit of 3^100:

3^1 ≡ 3 mod 10
3^2 ≡ 9 mod 10
3^3 ≡ 7 mod 10
3^4 ≡ 1 mod 10
3^5 ≡ 3 mod 10

Again, this cycle has a priod of 4, so 3^8 ≡ 1 mod 10, as is 3^12, 3^16... and 3^100. So 3^100 ≡ 1 mod 10

For 4:

4^1 ≡ 4 mod 10
4^2 ≡ 6 mod 10
4^3 ≡ 4 mod 10

So this one has a period of 2, and 4^100 ≡ 6 mod 10. By a similar manner, you can find:

5^100 ≡ 5 mod 10
6^100 ≡ 6 mod 10
7^100 ≡ 1 mod 10
8^100 ≡ 6 mod 10
9^100 ≡ 1 mod 10


And for higher numbers, just refer to the list already computed.

BTW, in case you're wondering what the "mod 10" means, it just means that these numbers have the same remainder when divided by 10 - i.e. they have the same last digit.

Here's the full list:

0^100 ≡ 0 mod 10
1^100 ≡ 1 mod 10
2^100 ≡ 6 mod 10
3^100 ≡ 1 mod 10
4^100 ≡ 6 mod 10
5^100 ≡ 5 mod 10
6^100 ≡ 6 mod 10
7^100 ≡ 1 mod 10
8^100 ≡ 6 mod 10
9^100 ≡ 1 mod 10