power problem

[raunekk]

93. What is the remainder when 43^43+ 33^33 is divided by 10?


what is the simplest method to solve this kinda sums...??

thank you!!

[Motherjane]

Hey Raunekk,
The simplest way to do such a sum where the remainder has to be found out for numbers divided by 10 is to find out the units digit.

A number is completely divisible by 10 when the units digit is 0. For eg 20,100,2367890. All numbers having the units digit as 0 is completely divided by 10.

Now in 43, 3 has a cyclicity of 4.
So 3^1 = 3
3^2 = 9
3^3 = 7 (by taking only the units digit of 27)
3^4 = 1 (by taking only the units digit of 81)

and 3^5 = 3 (by taking only the units digit of 243)

So the cycle repeats after 4.
Hence to find the units digit of 43 ^43, divide 43/4 to get 3 as remainder.
so the units digit will be 7.

similarly units digit of 33^33 is 3 (Remainder of 33/4 is 1, hence the unit digit is 3).

So the units digit of 43^43 + 33^33 is 7 + 3 = 0 (taking only the units digit of 10)

Now when you divide such a number by 10, the remainder will be 0.

The answer is 0.

Would request others to verify the answer.

[raunekk]

Thx motherjane...0 is the correct answer and the explanation is immaculate..didnt have to re-read any of the lines..
thx a ton..