Visitor

2015-10-27 05:58:45

#1

Hi,

I have to find out which is the last digit of the expression:

n=7^{37}-7^{36}+7^{35}-7^{34}+...+7^{3}-7^{2}+7^{1}-7^{0}

I have to find out which is the last digit of the expression:

n=7

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

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

Visitor

2015-10-27 06:22:20

#3

7^{1} = 7

7^{2} = 9

7^{3} = ...3

7^{4 }= ...1

7^{5} = ...7

....

So though, we gotn = 7^{37} - 7^{36} + 7^{35} - 7^{34} + ... + 7^{1} - 7^{0} =

=7^{36}(7-1) + 7^{34}(7-1) + 7^{32}(7-1) + ... + 7^{2}(7-1) + (7-1) =

=6(7^{36} + 7^{34} + 7^{32} + ... +7^{2} + 1)

But 7^{36} last digit is 1

7^{34} last digit is 9

7^{32} last digit is 1

7^{30} 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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