What are the final two digits of 7^177
Phi(100) = 40, so the order of 7 (i.e. the smallest power to which you have to raise 7 to get 1) is a divisor of 40, and you easily find that it is 4.
Computing Mod 100 in the following:
7^4 = 2401 = 1
176 is a multiple of 4, so we have:
7^177 = 7
Hint: take a look at the last two digits for the first few powers of 7
7^1 = 7 or 07
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = ......49
7^7 = ......43
7^8 = ......01
7^9 = ......07
7^10 = .... 49
I am sure you can do something with this pattern.
To find the final two digits of 7^177, we can use the concept of cyclicity to observe the pattern of the last two digits when powers of 7 are calculated.
First, we notice that the units digit of 7^n repeats in a cyclical pattern of 7, 9, 3, and 1 for powers that are a multiple of 4.
Similarly, the tens digit of 7^n repeats in a cyclical pattern of 0, 0, 0, and 0 for powers that are a multiple of 4.
Now, let's work out the powers of 7 and observe the pattern of the last two digits:
7^1 = 07
7^2 = 49
7^3 = 43
7^4 = 01
7^5 = 07
7^6 = 49
7^7 = 43
7^8 = 01
Notice that after every fourth power, the last two digits repeat in a cyclical pattern. This pattern will also repeat for higher powers.
Since 7^177 is a large power, we can find its pattern by using modular arithmetic. We need to calculate 177 modulo 4.
177 divided by 4 gives us a remainder of 1. Therefore, 177 is congruent to 1 modulo 4.
Now, using the pattern we observed earlier, we know that the last two digits of 7^1 is 07. Therefore, the last two digits of 7^177 will be the same as 7^1, which is 07.
Hence, the final two digits of 7^177 are 07.