All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Get the last digit II (Posted on 2011-02-23)
Let A=1^n+2^n+3^n+...9^n

For a positive integer n what is
the last digit of A?

Comments: ( Back to comment list | You must be logged in to post comments.)
 Pissible solution | Comment 1 of 4

The last digits of the powers of n are periodic with period 4.

Proof:Let n=5 for x{1...10}; the last digit of x^5=x (see below).So the effect of multiplying the last digit of x^1 by x is exactly the same as the effect of multiplying the last digit of x^5 by x, to produce x^6; and the effect of multiplying the last digit of x^2 by x is exactly the same as the effect of multiplying the last digit of x^6 by x, to produce x^7; and so on.

x^1 x^2 x^3 x^4 x^5
1       1     1     1    1    1
2       2     4     8    6    2
3       3     9     7    1    3
4       4     6     4    6    4
5       5     5     5    5    5
6       6     6     6    6    6
7       7     9     3    1    7
8       8     4     2    6    8
9       9     1     9    1    9
10     0     0     0    0    0
sum:(4)5(4)5(4)5(3)3(4)5

Accordingly, we just need to do the sums for n=1 to 4 to get all the results for all n: {5,5,5,3} and the sum of powers terminates with a 5 unless 4 divides n, in which case the sum of powers terminates with a 3.

On review, that probably ought to be 'possible solution'.

Edited on February 23, 2011, 3:01 pm
 Posted by broll on 2011-02-23 14:59:33

 Search: Search body:
Forums (0)