# An RMO problem

Prove that the ten's digit in $$3^x$$ is always even, where $$x$$ is a natural number.

Note by Dev Sharma
2 years, 9 months ago

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We can use binomial theorem to prove this. 3^x = ( 1 + 2 ) ^ x

- 2 years, 9 months ago

See that 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, and 3^4=81 conclude us with:

$$3^{4n+k}= 81^n 3^k$$

For any natural number a, b, and a digit number y, notice that:

$$(20a+1)(20b+y) = 400ab+20(ay+b)+y)$$

400ab only determine hundreds and upper and y is a digit, so both has nothing to do with the tens. The term 20(ay+b) is the one that fix the tens as an even number. By using all of this, we are done.

- 2 years, 9 months ago

why did you take 20a +1 . 20b +y

- 2 years, 9 months ago

Applying that with a=4 and the fact from 3^0, 3^1, 3^2, 3^3, and 3^4 (all of them has even tens). Hasn't it prove the original statement?

- 2 years, 9 months ago