If a =5, b =-3, try to determine the last digit of the 2011 power of a + the 2010 power of b

If a =5, b =-3, try to determine the last digit of the 2011 power of a + the 2010 power of b

1. When the base number is 5, no matter the index is a number of times (of course, it is an integer, a child has not learned), the single digit of the result is always 5, the 2011 power of the single digit of 5 is 52, the base number is -3. The index is 2010, first we have to determine that the result of this number is a positive number (negative even times is positive) bottom number is -3.

1. When the base number is 5, no matter the index is a number of times (of course, it is an integer, a child has never learned), the single digit of the result is always 5, the 2011 power of the single digit of 5 is 52, the base number is -3. The index is 2010, first we have to determine that the result of this number is a positive number (negative even times is positive) bottom number is -3.

A=25, b=-3, the last digit of the 2010 power of a + the last digit of the 2011 power of b is 8? The end of several powers of a is 5, and the end of b is -3,-27,81,-243, regularized, four in each group, end 3,9,7,1 Yong 2011 divided by 3 of 4, so 2011 of negative 3 must be how much negative 7, so the final mantissa is 8 I understand the process, but do not understand why the last 8 ah!

A =25, the bits of the positive integer power of a are all 5.
By calculating the 1,2,3,... powers of 3, we find that their individual bits have the following laws:
3,9,7,1,3,9,7,1,3…
2011=4×502+3,(-3)*2011 Bits are 7.
A positive number plus a negative number equals 15-7=8.

A =25, the individual bits of the positive integer power of a are all 5.
By calculating the 1,2,3,... powers of 3, we find that their individual bits have the following laws:
3,9,7,1,3,9,7,1,3…
Four cycles.2011=4×502+3,(-3)*2011 bits are 7.
A positive number plus a negative number equals 15-7=8.