How to Find The Last 2 Digits of A Large Number Written in Exponent Form

  Рет қаралды 10,949

Prime Newtons

Prime Newtons

Күн бұрын

Пікірлер: 32
@l1mmg0t
@l1mmg0t Жыл бұрын
i like your video. explained in a very simple and clear way
@atomic_salt
@atomic_salt 7 ай бұрын
last two digits of 247^223 should be 23, no? 47^3 mod 100 is congruent to 23.
@osenisaheedakinsola7428
@osenisaheedakinsola7428 3 жыл бұрын
Well done... I have really learnt alot from your videos. However, in this video you got 63 as the last two digits of 47³ whereas it should be 23
@PrimeNewtons
@PrimeNewtons 3 жыл бұрын
You are correct. Thank you for the observation. I hope others can help out like this in other videos. I will leave a note in the video description. I appreciate your keen attention.
@linaelmousslih835
@linaelmousslih835 4 ай бұрын
what about 1234^5678 using exponent rules
@Dice-w7u
@Dice-w7u 2 ай бұрын
Hi sir, why do we multiply by the last digit? for example in the video you only multiplied it by the 2 at the 2nd of the 132
@ItsVarunContractor
@ItsVarunContractor 2 ай бұрын
thank you sir for this explanation. Very helpful.
@robertveith6383
@robertveith6383 3 ай бұрын
*@ Prime Newtons* -- As with another comment in another video of yours, please *stop* using the implication symbol. The equals symbols needs to be used instead.
@PrimeNewtons
@PrimeNewtons 3 ай бұрын
Noted
@Eraqon
@Eraqon 10 ай бұрын
Great video. Thank you!
@me56743
@me56743 5 күн бұрын
Find last twp dgt pf 22^33
@infinitereviver
@infinitereviver 10 ай бұрын
Thank you sooo much!
@dakshpatel7076
@dakshpatel7076 Ай бұрын
Bro u sigma
@nabeeladriansyah4817
@nabeeladriansyah4817 6 ай бұрын
This really helps. Thanks
@Alexander-oz9rz
@Alexander-oz9rz 2 жыл бұрын
How about last 3 or 4 digits?
@PrimeNewtons
@PrimeNewtons 2 жыл бұрын
I don't know how to do that
@Lovedrug96
@Lovedrug96 Жыл бұрын
please do last 3 digit
@PrimeNewtons
@PrimeNewtons Жыл бұрын
That would require a lot of number theory. Let's see.
@Aminraghibfans774
@Aminraghibfans774 2 жыл бұрын
Well done, but the rule of number 5 doesn't always work successful, because if you calucl 5^26 you'll get 96 as they are last digits and not 25 as you said... Can you explain why?
@PrimeNewtons
@PrimeNewtons 2 жыл бұрын
5^26=1,490,116,119,384,765,625
@Aminraghibfans774
@Aminraghibfans774 2 жыл бұрын
@@PrimeNewtons in fact i used my c++ compiler and unsigned long long as type of variable to store the input, and the result of calcul of 5^26 gave me : 1490116119384765696 I don't know how, perhaps there was something wrong in limit of variable or in my compiler.. Thnks any way ✌
@PrimeNewtons
@PrimeNewtons 2 жыл бұрын
Okay, I can't help with that. I'm sure you know that your answer is not a multiple of five. You should ask some c++ savvy person. I'm just a math teacher.
@Aminraghibfans774
@Aminraghibfans774 2 жыл бұрын
@@PrimeNewtons yes sir you're right, i respect you for your interest, and thank you for everything 👍👍
@johnernestsantos3625
@johnernestsantos3625 2 жыл бұрын
This really helps! For 1234ˆ432, shouldn't the answer be 36? ((17ˆ4)ˆ108) should have been simplified to (21)ˆ27 since 108 should be divided by 4? It will then give you 41. So it should be 24x4x41 = (....36).
@PrimeNewtons
@PrimeNewtons 2 жыл бұрын
That was no error. I just ran it on my computer, and the last two digits are 56.
@michalkrsik2702
@michalkrsik2702 Жыл бұрын
@@PrimeNewtons But when I run it on wolfram, the last two digis should in fact be 56.
@mohamedrafeeque4463
@mohamedrafeeque4463 Жыл бұрын
​​@@PrimeNewtons is that an error really? Because 17^4=21 and if it should be raised to power of 4. You'll get (21^4)^27 . It will give the answer as again as 81^27=61 so 56 is the answer. Please explain sir.
@PrimeNewtons
@PrimeNewtons Жыл бұрын
You're correct. There was no error. Thank you
@PrimeNewtons
@PrimeNewtons Жыл бұрын
Correct
@minute-ai
@minute-ai Жыл бұрын
I don't know Will Smith makes math tutorials, hahaha
@PrimeNewtons
@PrimeNewtons Жыл бұрын
He doesn't.
@robertveith6383
@robertveith6383 3 ай бұрын
You have a stupid comment. I reported it as spam.
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