Least Residue of a big power of 7 mod 50 using congruences

How to find the nonnegative residue modulo 50 or remainder when dividing by 50

[@godsparticle7657]

I remember the scene from F.R.I.E.N.D.S when Monica screams "Seven! Seven! Seven! Seven! Seven!!!...." . 😂

[@MathsWithJay]

...but not 329 times I guess!

[@prodyungbvx7800]

Thank you so much, I finally understand, all the other videos of other people teaching this has just said "it is congruent to one" without actually explaining why... so thank you very much for doing so :)

[@MathsWithJay]

Glad it was helpful! Thank you so much for this feedback.

[@alfonsomarquez1026]

Your videos have helped me keep up during the quarantine, thank you very much. Greetings from Spain

[@MathsWithJay]

@Alfonso Márquez: Glad to hear it! London's in lockdown too! Stay safe

[@logannilson2089]

Thank you so much for this! Absolutely saved me for a school assignment.

[@MathsWithJay]

You're very welcome!

[@niranjankumarcoimbatore5842]

Hi , I am an Mathematics teacher from India....I would like to take classes like you to my students also....please tell me what tools or softwares you are using to take this math class....writing numbers etc....so this will be helpful for me to educate my students...I am grateful to you...Thank you....

[@MathsWithJay]

Hi Niranjan, I use a microsoft surface pro so I can write on the tablet directly. Good luck with your online teaching!

[@geshbenrewand1778]

God bless you

[@MathsWithJay]

@Gaş Bîn: Thank you!

[@Mathematician-uf9yy]

How can this be done using Fermat's Last theorem?

[@MathsWithJay]

@Mathematician 1010: Mmmm...do you mean "Last"? This might help: kzbin.info/www/bejne/pn6kXpd6YrB2ja8

[@StarFlatinum]

Thanks so much! I'm guessing this can be applied to any power? Even ridiculously large ones.

[@MathsWithJay]

@Luke: Such as?

[@StarFlatinum]

@@MathsWithJay say for instance 43^25483mod 37

[@StarFlatinum]

I just made this example up but these are the kind of examples we have to work with in our exam

[@MathsWithJay]

@Luke: Sometimes Fermat's Little Theorem can be used: kzbin.info/www/bejne/pn6kXpd6YrB2ja8

[@Someone____99-r9l]

at first thank you... but i wanna ask you about the "4" why did u choose exactly 4 and how can i guess that the mod is 1 ??

[@MathsWithJay]

At what time in the video?

[@chamindanawula3810]

Madam great job how i contact you

[@tunailker8]

108957006957880384854781400332988655104195780924346448594905101807142948743533497375319263496297652070491969771073582487659356699431175449777087388597611757900070061059734878450350888020156369740407014773814755940670361796273579035410270670889643488187096146596661961779984097607 (thanks to python and computers to calculate:D)

[@MathsWithJay]

@tunailker8: Thank you! Just 279 digits...

[@abenryuVoldigoad]

Whwre did you get 43 in 7³ on the introduction?

[@MathsWithJay]

343 = 50 x 6 + 43

[@GraciellaBundalian]

How about 120^50???

[@MathsWithJay]

What is the mod?

[@IODell]

Thank you for posting. 3^45 (mod 45) is 18 according to calculators. I have yet to find the iteration giving this solution. I've tried Euler's theorem and modular exponentiation. Any ideas?

[@MathsWithJay]

Yes...if I wasn't using a calculator, I'd start with 3^4....does that help?

[@IODell]

@@MathsWithJay Yes, thank you. I didn't take my calculation far enough. (3^4)^10 = 3,486,784,401 (mod 45) which reduced to 36 (mod 45). From there it was easy. I appreciate the motivation.

[@MathsWithJay]

Interesting! I was thinking of writing 81 as equivalent to -9 in mod 45, then it can be done without a calculator: (-9)^2 is also 81, which is congruent to -9, etc...

[@IODell]

@@MathsWithJay Yes, I actually calculated -9^10 which equals 3,486,784,401. I didn't see a way to solve it using Euler's theorem.

[@amartyasengupta2999]

Pls help what will be the answer of this question 3^258 mod 17

[@MathsWithJay]

kzbin.info/www/bejne/pn6kXpd6YrB2ja8 can be used for prime modulus

[@madhujadelgoda4228]

But this question, How to find the answer can please someone help. 5^10(mod7)

[@MathsWithJay]

For small numbers you could use a calculator....remember that you are just looking for a remainder from a division

[@madhujadelgoda4228]

Oh okay, I used the fermat's theorem but it won't work on this kinda one. 5^(7-1) = 1(mod 7) 5^6 = 1(mod 7) And then cannot solve. I think I did wrong or something Can you help me for this.

[@MathsWithJay]

It does help that 5^6=1 because you have to work out 5^10=5^(6+4), so now you only need to work out...

[@madhujadelgoda4228]

@@MathsWithJay Okay, Thank you miss. 👍

[@chandana.k3074]

What is the value of 11^20 mod 33 ?

[@MathsWithJay]

What have you tried so far?

[@MathsWithJay]

What have you tried so far?

[@dipeshadhikari9191]

why we add 50 to -7.

[@MathsWithJay]

@dipesh adhikari: At what time in the video?

[@aryagupta9231]

tyyyyy

[@MathsWithJay]

@Arya Gupta: ty

[@tkcoolz18]

343÷50=6.86

[@MathsWithJay]

Yes!

[@mrakshay4950]

Hi

[@MathsWithJay]

Yes - that's the one

[@nabhitha]

Y by -1

[@MathsWithJay]

Hello Nabi Chandar...is that a question?