What is the last digit of 2^2019 ?

This video shows a simple math skill to help you find the last digit of large exponents.
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[@Harshavardhantej9000]

Love and Respect from India

[@ultramegabarneyplayz3318]

WOW, I have searched so many youtube videos, and in this video I finally learned the steps you take to solve it. I thought this question was impossible before I watched this video. You are such a good math teacher!!!

[@PrimeNewtons]

I'm glad it helped. Good luck.

[@rabekasultana7097]

kzbin.info/www/bejne/mYLQkJJvptOLf9k This should help more my bro. Easiest method.

[@devonspeaks93]

All I can say is, "WOW!" Thank you so much. I just watched so many other videos and you explained it so well.

[@korboon6320]

Thank you so much without you I wouldn’t pass my exponent exam

[@maricarsereno7623]

Thank God I found your video sir.🙂🥰. I find it hard to understand this lesson, not until I saw this.🙂

[@bayondip12345]

Love and respect from India.

[@sackwqwokz208]

Thank you for precise and concise explanation, sir. This video is very helpful.

[@idkdikdidkd]

pleasure learning from Newton himself !

[@kamehamehaaaaa]

Simple explaination to the point. Thanks

[@Sacred_ray]

I watched the other videos of my country, but your video helped me a lot, thank you sir

[@YeanYaneth-hl2eo]

I watch some vdo on my country that how to do but I don’t understand anything at all 😂 but when I saw ur vdo I feel better so thanks u sm ❤

[@PumpkinHorn]

This makes more sense than what I put in my notes

[@bibimehrabi9377]

You explained it SO WELL! Thank you!!

[@Pugboy35Man]

This was very clear and helpful, you really deserves more subs :)

[@PrimeNewtons]

I'm glad it helped

[@endyeyvlogs475]

SAme here sir.

[@flashfocusbruh5144]

The beginning of a legend

[@LexandroxX]

Great Video Mr Ok!

[@gini302]

BROWN! But also I agree.

[@edlkmm]

Yea nice job

[@DEVILOP-f2d]

Sir you helped me a lto today coz tommorow is my exam and i was bit confused with this topic you deserve it ❤️

[@ouchypouchy4402]

Wow,What a nice video!

[@navyaakkageorge7635]

That's a very beautiful explanation sir. Thanks a lot :)

[@goooz3369]

Well done

[@Monestfiks]

beginning of a legend

[@tinascott7496]

Awesome explanation!! Thank you!

[@aklyrics746]

Again, it's another good video and I congratulate you on doing it in the purpose of helping students around the world improve their math skills. You have already made two videos on how to find last digits numbers when we have any kind of number raised to any kind of power. I think that it will be also helpful to make videos about the same subjects but explaining how to find last digits for any factorial numbers. Thanks !

[@soulseekerbs5301]

For factorials, if n>4, the last digit of n! will always be 0, do you see why?

[@aklyrics746]

@@soulseekerbs5301 Not really !

[@soulseekerbs5301]

@@aklyrics746 4!= 24. But when we get 5!, We have 120. Notice that n!=n(n-1)! Since 5! is 120, and includes a 10, all n! with n>=5 will have a 10, and having 10 as a factor will always make the last digit 0.

[@aaryagupta7748]

THANK YOU You helped me on my homework so much

[@PrimeNewtons]

You're welcome

[@indadulhoque9979]

Love from india on assam

[@rikshikakashyap7929]

No video explained it so well. Thank you so much.

[@AdminTubeZ]

Hi, can we use Euler's totient theorem to find the last 2 digits of 2^2019 and 2^2016?

[@PrimeNewtons]

Yes. Search the newer video. I did this a long time ago when I avoided number theory because I didn't want non mathematics students to get intimidated.

[@AdminTubeZ]

Can provide the link to that video, please?

[@chicakenz9616]

Mr okay it's me nasira :)

[@rabekasultana7097]

Noice vid and explanation

[@neilclay5835]

Great teaching

[@oliviagow-smith1767]

Thank you so much for this video! So helpful! I was wondering what I do if there's no remainder after dividing by the number of numbers in the sequence? Studying for my final! :)

[@PrimeNewtons]

I'm glad it helped. Good luck on your final. Please share this channel with friends. Thank you.

[@saucybaka1585]

I think we just use the last digit in the cycle

[@imawesome.]

what if we don't get a remainder? like for 9^2020 9^1 =9 9^2=81 9^3= 729 9^4= 6561 cycle= 9, 1, 9, 1..... 2020 divided by 4 = 505 but no remainder....... what should i do? btw your teaching is so fun! edit: got it!!

[@PrimeNewtons]

Notice that 9 has a cycle of size 2. You should divide 2020 by 2. If there's no remainder, then your answer is a complete cycle. Answer is 1.

[@imawesome.]

@@PrimeNewtons thank you!

[@indadulhoque9979]

Thans sir i am under stand I am from india

[@Jagruk2098]

make moreeeee!!!!!!!!!!!!!!!!!!!!!!!!!!!! great video

[@FunnyGuyexe4]

I hope this will help me in the informatica Olympiad

[@Problematica.]

Nice solution. Can you find the last three digits of the number of 19^97

[@artandata]

you demonstrated that for base 2 & 3, the cycle if 4 but my question: Is four the cycle for any base number ?

[@PrimeNewtons]

I think 4 works for all. Although it's excessive for some.

[@kayelabrador5729]

Sir, does units digit and last digit the same?

[@PrimeNewtons]

Yes

[@kayelabrador5729]

@@PrimeNewtons Thankyou so much Sir

[@FindingHeaven173]

Love from Bangladesh

[@GIOVELAUGUSTINE]

Thanks alot this helped so much.

[@keenantheho]

Damn this would have helped in spirit of math 3 years ago

[@Msworld230]

Ur justttttttt wowwwwwwwwww thank you so so much 🙏🏻

[@Escviitash]

Every integer has a last-digit-cycle length of 1 (1,5,6,0), 2 (4,9) or 4 (2,3,7,8). You can usually say that N*X has they same last digit as N^(X mod cycle.length). But you run into a small problem if the cycle length is 2 or 4 and X is a multiple of the cycle length. Then you remainder will be 0, but there is no zeroth digit in the cycle. And you can't just can't say that the last digit is the same as N^0, because N^0 is always 1. So you have to rememeber that if the remainder is 0 then the last digit of the result is the last digit in the cycle. I mention this because I have seen so many being bewildered in that case.

[@Nikioko]

The same as the last digit of 2¹⁹, 2¹⁵, 2¹¹, 2⁷ or 2³: 8

[@SyabongaGumede-d8d]

It's 36 not 34

[@ferciasatina740]

Hello! How about 3^2018. It has a remainder of 5. How should I do it?

[@PrimeNewtons]

Watch how to do it here: kzbin.info/www/bejne/anWtpWmId919gas

[@Harshavardhantej9000]

The Answer is 1

[@alexisreeves8799]

How do you get a reminder of 3?

[@Krystelzndr]

Hii, what if the rem is 5? Hope you notice my comment☺️

[@PrimeNewtons]

Hi, I noticed your comment. Please tell me the full question 😊

[@holyshit922]

2^2019 = 0 (mod 2) 2^2019 = 2^(4*504+3)(mod 5) 2^2019 = 0 (mod 2) 2^2019 = 3(mod 5) We have system of congruences x = 0 (mod 2) x = 3(mod 5) We can solve it with Chinese Remainder theorem 1 = 2*(3) + 5*(-1) x = (2*(3)*3 + 5*(-1)*0)mod 10 x = 18 mod 10 x = 8 mod 10

[@loading...5221]

But what about those numbers who don't have rem, for example 3¹⁰⁰?

[@PrimeNewtons]

Then your answer iis the last digit in the cycle

[@Msworld230]

Can you please solve 3^10000 this is my humble request and i have exam in 2 days so please help me 🙏🏻

[@PrimeNewtons]

What do you mean by 'solve'?

[@Msworld230]

@@PrimeNewtons means final digit of 3^10000

[@ThenSaidHeUntoThem]

@@Msworld230 This video will help you kzbin.info/www/bejne/anWtpWmId919gas

[@Harshavardhantej9000]

8:58 The answer is 7 sir

[@perseverance3925]

What about something like this 18^5677

[@davidthomas6043]

147 divided by 4 is 38 plus remainder of 3, not 34 a plus 3 as shown !!!

[@humester]

147/4 = 36R3 not 34R3. 36 x 4 = 144 (+ 3 ==> 147).

[@assamyoutubechannel7390]

❤

[@ariamahmed3434]

How would you do 2^2020 ???

[@PrimeNewtons]

2 has a cycle of 4 (2,4,8,6). When you divide 2020 by 2, the remainder is 0. That means the cycle is completed and the last digit is 6.

[@ajadali340]

7

[@tobiasbundgar7063]

now what if u want the last 5 digits or so

[@PrimeNewtons]

That's beyond me for now 😂

[@tobiasbundgar7063]

@@PrimeNewtons i guess youwill have to do this whole mod thing

[@Harshavardhantej9000]

7:07 Sir, the answer is 2!

[@SushmagowdaSusu]

147/4 =36

[@Sandeepan]

"cyclic groups"

[@paleok12]

The answer is 7

[@Straight_Talk]

I've got two better questions: 1. How many digits are there in 2^2019? 2. What are the first three digits in the number? Let me know if you want the solution.

[@PrimeNewtons]

Those are interesting questions

[@Straight_Talk]

@@PrimeNewtons The solution is simple. Using logs, 2^2019 = 10^(log2 x 2019) = 10^607.7795613183 From this, we know the number has 608 digits, and the first three digits are given by 10^0.779); i.e. the first 3 digits are 601 (using a calculator). Note: 1. Of the expression 10^607.7795613183, the numbers BEFORE the decimal point determine how many digits (zeroes) are in the number, while the numbers AFTER the decimal point determine what the digits are. 2. Only the first few decimal places determine the first few digits in the number; the digits further back after the decimal point determine the digits further back in the number. Although a bit of calculator assistance is required, this method allows one to determine the number of digits and first few digits of very large numbers that are too big to evaluate using a calculator.

[@seroujghazarian6343]

608 digits

[@rabekasultana7097]