Find the last two digits!

We look at a classic number theory puzzle question.

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[altynbeknurtaza]

11:20
Michael: That's a good place to ...
Me: No it's not!
Michael: ...start
Me: Oh, I see

[thayanithirk]

Please upload your research papers and explanation as your 100 k special.whoever agrees hit a like

[AndrejPanjkov]

A little brute force but with a calculator :
3^7 = -13 mod 100
3^(7^2) = (-13)^7 = -17 mod 100
3^(7^3) = (-17)^7 = 27 mod 100
3^(7^4) = 27^7 = 3 mod 100
so the remainders repeat over a cycle of 4 terms, with -13-17+27+3 = 0

which is not to diminish Michael's proof using phi. Consider this a confirmation via a different path.

[jonathanhanon]

Since we have a 5, we can say N/5 = (3^7 + 3^9 + 3^23 + 3) (mod 20), giving us phi(20) = 8, allowing us to write this as 3^7 + 3 + 3^7 + 3 (mod 20)
Since 3(3^7) = 1, 3^7 must be 7 mod 20, making the sum 0, and that's a good place to stop.

[joaopedrobmenezes]

I miss those exercises that ask you to find the last numbers of something. Even though you can usually solve it using mod 10^x ( for some natural x), it stills funny

[keithmasumoto]

I find it amazing that these problem creators can come up with problems like this. I just don't see how they back into a problem like this where the huge sum ends up being divisible by 100 and at the same time involves a large exponent equal to the current year.

[HeyKevinYT]

(Spoiler alert)







Woww that entire sum is divisible by 100 🤯

[xevira]

When you were working on the 3^23 mod 100 section, I realized the number was a multiple of 100 and went bastard... " at the problem. XD

[NegativeAccelerate]

I like that you name the theorems that you’re using. So if I don’t know them I can google them and then attempt the question

P.s. feel free to link any videos you have done on any of the theorems you name in the description.

[micomrkaic]

Amazingly clear explanation! Thank you for this.

[videolome]

I don’t known why you don’t like to use the Chinese Remainder Theorem.

[fmakofmako]

You can skip the calculations at the end once you have that N is congruent to 5*(3^7+3^9+3^23+3) mod 100 because by inspection it is 0 mod 5 and in mod 4 it is 1*(-1+-1+-1+-1) since all the powers of 3 are odd, so -4 mod 4 = 0 mod 4.

[Archik]

last 3 digit: 500
c++ help calculate last 6 digit: 768500

[karthikkrishnaswami]

5^2 is 1 mod 8, 5^574 is 1 mod 8 so 5^575 is 5 mod 8

[cheaterman]

I just love how you keep saying "Fermat's little theorem" - I mean, there's probably a good reason (like Fermat has a "big theorem" or something) but it's still a very cute name for such a sharp and powerful theorem hahaha!

[dujas]

There is another shortcut that can be taken in here. If x=20q+r, then 5x=100q+5r. Therefore, you only need to calculate 3^7+3^9+3^23+3^1 mod 20. Here, you can use Euler's Theorem again and find that you only need the exponents mod 4. The result is then This is equivalent to 2(3^1+3^3) which is congruent to 2(3+7) which is congruent to 0 (mod 20). Multiply by 5, and the result is 0 (mod 100).

[SiuHongLi-tk]

11:33
As 20*5=100

Only need to consider mod20
And 3^4=81=4*20+1
(3^7+3^9+3^23+3)
=3(3^6+3^8+3^22+1)
Mod20 obtain
3(3^2+1+3^2+1)=20*3=60
Mod 20 obtain 0

[joaopedrobmenezes]

And also, congratulations, a few more subscribers and you will reach 100k!

[chamelius]

So, what I have found out is:

I really need to work on my Number Theory proficiency.

[shailendragupta]

find rational number a, b, c such that