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Is the reason why we know that p=4t is enough because there are only 4 possible remainders when 2^p is divided by 10 (2, 4, 8, 6)? Therefore 4 times a number will always end up in the 4th spot? In other words, if (B) said p=3t then it would be insufficient because every multiple of 3t would result in a different remainder (until you cycled through all 4 possibilities).

Plugging in is good here too, because it only takes a couple options to realize the remainder is always the same.

the last digit of 2^p should be 2, 4, 8,or 6, so the remainder should be one of these.

are their any other remainder generalizations such as this? I would've had to of done n amt. of probs like this to finally realize that.

Not necessarily for remainders, but it is good to know that a pattern occurs for every units digit that is raised to the nth power.

There is a pattern created. It is good to learn some and know that there is a pattern for all.
pattern of units digit when a number is raised to ^n
2^n 2,4,8,6
3^n 3,9,7,1
4^n 4,6,4,6
5^n 5,5,5,5
...and so on....

knowing the patterns exist helps with remainder questions

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