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Prove that the tenths digit of powers of 3 are even....
Hi Satyaram,
Firstly note that powes of 3, can end in 3,9,7,1
Now lets consider some 3^x. The last digit can be 3,9,7, 0r 1.
Now 3^(x) = 10a + b.
Where b = 3/7/9/1.
Hence 3^(x+1) = 30a+3b.
Now what ever b is (ie 3/7/9/1), note that 3b = 09/21/27/03.... ie the 10s digit is always even. -----------(1)
Now 3^1 = 03
3^2 = 09
3^3 = 27 = 20+7.....
So from (1) we can conclude that the higer powers of 3^x [for x>3], will always have 10s digit even, because 3b always has 10s digit even, and a is alwas even.
Hence the 10s digit of 3^x -----[for any positive x] is always even.
Hope it helps.
Best Regards,
Ashwin (IIT Madras).
Thanks
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