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