Forum - Arithmetics

The last digit of a power

Razy
Visitor
2015-10-27 05:58:45
#1
Hi,

I have to find out which is the last digit of the expression:
n=737-736+735-734+...+73-72+71-70

 

RE: The last digit of a power

Pixy
Visitor
2015-10-27 06:13:03
#2
I don't know exactly, but i think this could be semethin'
7 ^ 1 = 7 (the latest digit), 7 ^ 2 = 9 (the latest digit), 7 ^ 3 = 3 (the last digit) 7 ^ 4 = 1 (last digit) 7 ^ 5 = 1 (last digit; again one and so on). So time is 4

RE: The last digit of a power

Steve
Visitor
2015-10-27 06:22:20
#3

71 = 7


72 = 9


73 = ...3


7= ...1


75 = ...7


....

So though, we got

n = 737 - 736 + 735 - 734 + ... + 71 - 70 =
=736(7-1) + 734(7-1) + 732(7-1) + ... + 72(7-1) + (7-1) =
=6(736 + 734 + 732 + ... +72 + 1)
But 736 last digit is 1



734 last digit is 9



732 last digit is 1



730 last digit is 9


....

We group them bt two and for every pair we have the last digit as being 0 (because 9+1 = 10)



So, we have.



0*6=0
Then, the last digit of the number n is 0.

 
 

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