JOMO 6, Long 3

A teacher writes down three numbers, 1, 2 and 3, on the whiteboard. Now, every student take turns to the whiteboard and erase one number, and then replace it by the sum of the two numbers left. After some turns, is it possible to have the numbers: $6^{2012}, 7^{2013}, 8^{2014}$ on the whiteboard at the same time? Give proof.

#Combinatorics #JOMO #Jomo6

Easy Math Editor

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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

Notice that a curious characteristic (even/uneven) of the initial sum of numbers never changes. We have that $1+2+3$ is even, but $6^{2012} + 7^{2013} + 8^{2014}$ is uneven. By the Invariance Principle, it is not possible to have these numbers on the whiteboard. $\boxed{\mathbb{QED}.}$

Guilherme Dela Corte - 6 years, 11 months ago

After every turn, the sum of the numbers on the whiteboard will be even (from a,b,c on the whiteboard we will gen 2*(a+b) or the analogs) Since 6^2012+7^2013+8^2014 is odd the answer is NO

Tudor Darius Cardas - 5 years, 4 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}`	$\boxed{123}$