What are the last two digits of 5 to the power of 347?

Use arithmetic Modulo 100. The Mod operator is defined as follows:

x Mod y is, by definition, the remainder you get when you divide x by y. This is the same as subtracting from x that multiple of y that makes the resulting number smaller than y.

Example

12 mod 5 = 2

If you divide 12 by 5 then the remainder is 2.

12 - 2*5 = 2

If we take Mod 100 a number, then we obtain the last two digits. Of course, in case of this problem we don't want to calculate the number first and then take Mod 100. Fortunately this is not necessary as the Mod of a product is the Mod of the product of the Mods.

To avoid awkward notations with the Mod symbol, it is customary to simply use the equals sign when doing calculations Mod some fixed number. So, let's calculate 5^347 Mod 100 and let's write a = b whenever a Mod 100 = b Mod 100.

We have:

5^3 = 125 = 25 = 5^2

This implies that 5^n = 25 for any
n > 1. If n = 2 or 3 then we already know that this is true. If n is larger, then:

5^n = 5^3 5^(n-3) = 5^2 5^(n-3) =
5^(n-1)

So, you can just subtract 1 from the exponent and eventually the exponent will become 3 in which case the number is 25.

Oh boy, I'm gonna need my clown calculator for this one! *brings out an oversized calculator* Let's see... *presses buttons dramatically* 5 to the power of 347 equals... drumroll, please... 25! So the last two digits are 2 and 5! Ta-da!

To find the last two digits of 5 raised to the power of 347, we can look for a pattern.

When we raise 5 to different powers, we can observe the pattern of the last two digits:

5^1 = 5
5^2 = 25
5^3 = 125
5^4 = 625
5^5 = 3125
...

We can see that the pattern repeats in cycles with a length of 20. So, to find the last two digits of 5^347, we simply need to find the remainder when 347 is divided by 20.

347 divided by 20 is equal to 17 with a remainder of 7. Therefore, the last two digits of 5^347 are the same as 5^7.

Calculating 5^7, we get:
5^7 = 78125

The last two digits of 78125 are 25.

Therefore, the last two digits of 5^347 are 25.

To find the last two digits of 5 to the power of 347, we can start by observing the pattern of the last two digits of powers of 5:

5^1 = 05
5^2 = 25
5^3 = 125
5^4 = 625
5^5 = 3125
...

From the pattern, we can see that the last two digits of the powers of 5 repeat in cycles of 20. In other words, 5^1 has the same last two digits as 5^21, 5^41, 5^61, and so on.

Now, let's analyze the exponent 347. We can divide it by 20 to see how many complete cycles exist within the exponent:

347 รท 20 = 17 remainder 7

This means that there are 17 complete cycles of 20, and an additional power left over, 5^7.

Using the pattern, we know that 5^7 has the same last two digits as 5^1:

5^1 = 05

Therefore, the last two digits of 5^347 are 0 and 5.