...another math question..help! can you show me how you got it?

Question

Question: explain how you can tell, without evaluating the powers, what the last digit of the following sum would be. Use examples where necessary to illustrate your explanation.

11^5 + 12^5 + 13^5 + 14^5

Answer 1

11^5 = calculate last digit (i.e 1^5)
12^5 = 2^5
13^5= 3^5
14^4 = 4^5

1^5+2^5+ 3^5+4^5 = 1300

Answer: 0

Answer 2

Note that the units digit of a product of two numbers depends only on the units digits of those two numbers. To see why this is so, consider how multiplication is done by hand. Except for the first row, every row has extra zeros at the right as placeholders. When the rows are added, only the first row's rightmost digit contributes to the units digit of the final product. Where did this digit come from? It's from the product of the units digits of the given numbers.

An example:
Multiply 59 by 17. (Ignore the periods; they're there to make the numbers line up.)
.....59
x...17
...413
+.59
.1003

The 3 came from 9x7=63.

So you only need to worry about the units digits of the given exponents. 1^5=1, 2^5=32, 3^5=243, 4^5=1024. In fact, you could have dropped all but the rightmost digit after subsequent multiplications. (3x3=9, 9x3=27=7, 7x3=21=1, 1x3=3)

So the units digit is 1+2+3+4=10=0, after the carry.

Answer 3

with the power of 5, any number raised to the fifth power will end with the same last number as the beginning number. ie 24^5 will end with 4, 33^5 will end with 3, 126472^5 will end with 2.

so the answer would be 1+2+3+4 = 10 leaving the last number of 0.

Answer 4

1^5=1
2^5=32
3^5=243
4^5=1024


Therefore the last digit of the sum would be 0.

Answer 5

(11*11*11*11*11)+(12*12*12*12*12)+(13*13*13*13*13)+(14*14*14*14*14)