An RMO problem

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

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Gian Sanjaya - 5 years, 9 months ago

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

Dev Sharma - 5 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?

Gian Sanjaya - 5 years, 9 months ago

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

Hari Om Sharma - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$