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# Prove that the tenths digit of powers of 3 are even....

290 Points
9 years ago

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,

Sathya
35 Points
9 years ago

Thanks