Aptitude - Numbers - Discussion

Just devide the power with 4, we get the remainder 2 and if we observe the last digits in the first four consecutive powers of 7, we get 7,9,3,1, and after that last digits will repeat again, that is every period of 4, the last digits will be repeated. So, as we got the remainder 2 when we divide the power by 4 the last digit would be 9. There is no need to expand 4173^2.

No need to this question please explain again?

754 => 4*n + 2, Here n will be 188.

So, unit digit will be the unit digit of 7^2, ie 9.

Is it always that the powers are divided by 4? If yes, kindly explain the concept.

No dude not always by 4.

4 ka first power = 4 (unit digit 4).
4 ka square = 16 ( unit digit 6).
4 ka cube = 64. ( unit digit 4).
4 ka fourth power = 2401 ( unit digit 1).

4 ka fifth power = now check unit digits later on going to be repeat again. That is 4,6,4,1.

That's why. Unit digit restriction. Is. Upto 4th power in this case

That's why.. Divide the no by 754 by 4.. you will get 2 as a remainder.

By rule ,
Dividend = divisor *quotient +remainder.

Assume quotient = n.

754=4n+2.
n=188.

4137 power 754 = 7 ^ 4 ^188 + 7^2.
=9.

Power = 754.
Divide by 4(7,9,3,1).
754/4 = 2(remainder).

Given number 4137 = 7(last digit).
Last digit^remainder.
7^2 = 49.

Last digit of 49 is 9(answer).

2^4344 what will be unit digit here?

Please explain how to find unit digit in any problem?

Why it can't be solve like this (4137)^754?

As 7^4 as equal to 1.

@Manish Methani. How 4 To the power 4 is 2401? Just explain.