Calculating Last Two Digits of a Number Raised to Power
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- Опубліковано 29 вер 2024
- A video that explains how to calculate the last two digits of a number raised to a power by JustQuant.com.
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I found this the video to be the best among all the others on UA-cam to find the last two digits.
Easy calculations, moreover explained in a superb way :-)
Such an easy way of learning an otherwise difficult concept to apply. Keep up the good work! :-)
tysm
what will we do if x is not divisible by 4 in case 2
then use the concept that we have used in remainder theorem and in the end multiply with the result
or you can see the first example of case 3, its also similar
5:02 how did you arrive at the conclusion that the last two digits of 79^2 are 41? You are actually trying to prove a theory for the numbers ending in 3,7,9 and you cannot use the same theory to prove it. Could you please explain how you ended up with 41? Thanks much!
jammerules80
Hi,
The last two digits of 79^2 is obtained by multiplying 79 by itself.
JustQuant.com Yeah, but this trick also applies to 79^2, doesn't it?
It should be like (79)^2 = (79^2)^1 and then there is no more thing to do ! So is it only working for bigger numbers ?
+Spherical Square (fsp98) last two digits of 79^2 are the same as the last two digits of (100-79)^2 i.e. 21^2. hopefully you're aware that 21^2 ends in 41.
@@Justquantpage if we do like that it would be difficult to do big numbers like 1879^2 , 1243^4 , 1547^4, .......
But this method is not working for 17^256. Can u try by this method.
how did [(79)^2]^71 become (41)^71 in case 2, example 1?
BRO, thankz alot for these EASE calculations.. :) feeling confident on this PART :) thankz agian
I am glad it helped you nanda kishore. Thanks to you too for leaving such a wonderful comment. It gives me the motivation to create more videos. :)
Dude, Case 2 is useless if the power is not divisible by 2. Waste of time..
Lmao noob
hi could you figure out how to do this
Brilliant...thnks a lot
Very helpful, thanks a lot!
Please tell how to find last 2 digits of 79^101 here 101 is not divisible by 2
it'll be: 79^101 = (79^100) * 79. then, (79^100 )*79 = ((79^2)^50)*79 = ((6241)^50)* 79. Therefore, the last two digits of (6241)^50 is equal to the unit of the base (1), the tens of base (4) times the unit of the power (0) which is equal to (4*0=0), hence 01 is the last two digits of (6241)^50. Now 01 times 79 is equal to 079 or 79. Hence the last two digits of 79^ 101is 79.
This is very helpful in my competitive math career I hope to see more videos like this. But, could you please explain last 3 digits as well?
N what about the odd powers
It is very helpful
How to solve for odd power
can someone explain the benefit of finding the unit digit of an exponent? I don't see the application of this
What are the last two digits of x, if x = (11¹+ 11²+ 11³+ ... + 11⁵⁵)? I m stuck pls help me what is the answer
Thank you
Very under appreciated. Good video! 👍🏻👍🏻
nice video,
simple language , easy to understand
this doesn't work for 21^2016
Hi,
It does work for 21^2016. Based on the technique described in the video:
Units digit of 21^2016 is 1
Tens digit will be equal to the units digit of 2x6, which is 2.
Hence last two digits of 21^2016 are 2 and 1.
Very nice
How we deriving that the 2 ND digit how the formula is coming
what if the powers of odd values are having odd values....??? only even values are explained.
bro explain each case with example
what if a num is 12^2 then is trick is failed...as last digit is 4 but 2nd last digit would come 1*2=2 but dis wrong
mranky111
Hi,
In 12^2, base ends in 2. I guess u are trying to solve this by applying case 1, which is not correct.
If the number is in the form 12^x.
12^x => (4 * 3)^x => (2^2 * 3)^x => (2^2)^x * 3^x
Now to calculate the last two digits
-> find last two digits of (2^2)^x using case 3 explained in the video
-> find last two digits of 3^x using case 3.
In your example, i.e 12 ^ 2, we have, x = 2
-> last two digits of (2^2)^2 is 16
-> last two digits of 3^2 is 09
Multiply 16 and 09 we get 44 as the last two digits.
done
how to solve for last digit in the case of 3^3^3^3........ to 2015 times
Last 2 digits of 6 to the power of 2018?
Sir can you please provide a method to get a 100 digits of a exponential number and btw awesome explanation thanks a lot
how about last three digits how to find ?
thanks justquant....really appreciate your video!
hey sir.. what if in case 2.. the exponent is not divisible by a number that when 9 is raised to would not give number 1 as the units... like 9^2015 something like that.. tnx
Skhiburdhurs
Hi,
9^2015 can be written as
(9^2014) x 9
=> (81^1007) x 9
Calculate the last two digits of 81^1007 and multiply it with 9 to get the last two digits of 9^2015
=> (..61) x 9
=> 49
JustQuant.com thank you my good sir!
17^7^7^7^7.......2008 times iska kaise niklega
sir what about 71 raised to 143?
Really helpful...thanq sir
Wow. Sir it was just easy after your video
good good good
♥️♥️♥️♥️♥️♥️♥️♥️
What will be last two digit if number ends with 0?
always will be 00
So technically each one ends in 1?
Thank you very much
Great work.
Sir but 5 raised to 3 is 125 and last two digits are 25
Hi Manoj, in 5 raised to 3 the tens place digit in the base 5 is 0, which is even (For parity of zero check this link - en.wikipedia.org/wiki/Parity_of_zero) and the exponent is an odd number. When the tens place digit of a number ending in 5 is even and exponent is odd, the number ends in 25.
Loved it.... Thank u so much!
Thanks a lot
Thank you sir @13:22
Actually helps.
This is great!!!!!!!!!!!!!!!
Thank u sir.
Great!
Great
Great explanation!
👌
Thank you