Question

If x^y denotes x raised to the power y, find last two digits of (1941^3843) + (1961^4181)

Answer 1

Answer:

Last two digits = 42

Explanation:

1941^3843

first take last two digit 41

first last digit is 1

then take the 4 (1941) and multiply with 3 (3843)

4*3=12,take last digit from this is 2

now (1)(2)

so last two digit is 21

now same for 1961^4181

first take last two digit 61

first last digit is 1 ..............(1)

then take the 6 (1961) and multiply with 1 (4181)

6*1=6,take last digit from this is 2 ...........(2)

now (1)(2)

so last two digit is 61

last two digits of (1941^3843)+(1961^4181) is 61+21 = 82

That's the final answer.