Finding last two digits of large exponent

GRE Quant Problem Solving
1

Hello, I had a quick question about one of the tips under the PrepSwift “Remainders and Exponents” section.
In the video attached to that section, when finding the remainder when dividing by 4, the video says to just use the unit digit and the remainder method like you would use for dividing by 3.

gregmat
gregmat717×522 93.3 KB

gregmat2

However, the text below the video says to calculate the last 2 digits of your number. I assume this means if the exponent was fully solved out, what the last 2 digits would be. This would be in line with the divisibility tricks from earlier in PrepSwift.

For large exponents like 7^37, for instance, was there a trick to calculating the last 2 digits that I missed in the earlier sections? I believe I only recorded my own notes about the unit digit specifically.

Would anyone have any advice about how to efficiently find the tens digit as well? Thank you so much if you can give me any advice on this!

Edited to add: the notes for remainders when dividing by 8 reference finding the last 3 digits. This also escapes me per the earlier videos in the study plans. Any advice would be appreciated, thanks!

2

It isn’t any easier to find the last two digits of said number than to directly solve for the remainder when 7^{37} is divided by 4. Also, 7^{37} is congruent to 3^{37} when looking at remainders upon division by 4, so you’d just be recycling the work u had from above.

3

Well, I am unsure how to link it to the notes. But because I’ve done some modular arithmetic one trick to simplify is that a * b = c (\mod N) \equiv (a \mod N) * (b \mod N) \equiv (c \mod N)

By this you can partially complete remainders and multiply them. In this case I noticed that 7^4 = 2401 hence 7^{37} (\mod 100)\equiv 7^4*7^4*7^4*7^4*7^4*7^4*7^4*7^4*7^4*7 \equiv 1*1*1*1*1*1*1*1*1*7 \equiv 7

Which is correct 7^{37} \equiv 7 \mod 100

Only Greg can tell you if studying a little modular arithmetic is a waste of time but I think knowing this principle and formulae can save a lot of time for such questions.

4

In this case, the “little modular arithmetic” doesn’t really help because it’s not any different from the cyclicity argument they presented above.