Problem is below:
To solve it, I first found the unit digit of 32^32, which is 6. Then I found the unit digit of 32^6. which is 4. I divided 4/7 to get Remainder 4.
Greg Prime Factored 32 to 2^5, and then found (2^5)^32 = 2^160. From there he took (2^160)^32 to get 2^5,120. Using the UD Remainder sequence, this also comes out to Remainder 4.
Is my methodology valid? Or did I just get lucky on this one? It seems a lot more straight forward but I can’t tell if I am correct.
This may have been a factor of luck. It is true that the unit’s digit of 32^32 is 6. But the full number could be one that is divisible by 4 or one which is not divisible by 4 (the unit’s digits of 32 run in a sequence of 4). For example, hypothetically if 32^32 resulted in 4636 (unit digit is 6), then you would need to find 32^4636. Given 4636 is divisible by 4, the units digit of the final number would have been 6, not 4. But instead, if 32^32 resulted in 4626 (unit digit is still 6), then you would need to find 32^4626. Since 4626 is not divisible by 4, the units digit of the final number would have been 4, as in your case. So, prime factorisation would be the safer route to solve this question.
Got it- super helpful! Figured as much but good to see it drawn out this way.
