A Record Breaking Prime

If you look at the powers of 2, you will notice there is a pattern with the last digit. The last digit is 2, 4, 8, 6, and then repeats.

74,207,280 is divisible by 4, so if you add one to the power, you are at the first possible last digit (2).

Subtract 1 from 2 (our current last digit) and what results is a last digit equal to 1.

Using Modular Arithmetic :

$\large {2}^{74,207,281} - 1 \equiv 2^1 - 1 \equiv 1 \, (\text{mod 10})$

Therefore the last digit of $\large {2}^{74,207,281}-1$ is $\Large \color{#20A900}{\boxed{1}}$ .

As Ethan Godden did, I just reviewed the first 40 powers of 2 and noticed the pattern 2, 4, 8,6. Hence, I found the remainder of 74207280 divided by 4; which is 1. That means it fits the first digit in the pattern above: 2.

So, even though 2 raised to that power is a tremendously large number, I know at least that it ends with a 2 due to above explanation. So, substracting 1, will result in a figure ending with 1.

For such questions always look for numbers that give the same last digit irrespective of what power we raise them to. e.g The last digit for numbers like 11,21,31,41, raised to power anything would always be 1; The last digit for numbers like 6,16,26, raised to power anything would always be 6

For this specific question we see that it has been represented as 2^n-1; we also know that 2^4=16 so the last digit for any number that is 16^n would always be 6.

=>2^74207281-1 =(2 (2^74207280))-1 =(2 (2^4)^18551820)-1 =(2 (16)^18551820)-1 The last digit of 16^18551820 would be 6; so when we multiply that number by 2; the last number would be 2 with a carry over of 1. Hence last digit of (2 (16)^18551820) is 2, when we subtract 1 from this number the last digit of the resulting number would be 1.

Notice the pattern 2^x follows for x until you have a repeating pattern. 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32

Because the two repeated in the ones digit, the pattern is 4 steps long, starting from x=1 to x=4. Therefore, you should look at 74,207,281/4

You know that 74,207,200 is divisible by 4 for sure, because any # ending with X00 for X 1-9 is divisible by 4. 81/4 = 20 with a remainder of 1. This remainder of 1 means you're on the first step of the four step cycle, which is 2. After subtracting, you get 1 as the answer.

The pattern with units digits of powers of $2$ is $2, 4, 8, 6,$ repeating. Since $74,207,280$ is a multiple of $4$ , $74,207,281$ should have a units digit of $2$ . Subtracting $1$ from this units digit of $2$ gives us $1$ as our final answer.

Good work..Ethan!