Ten words (Just for fun)

How well can we summarise proofs?

Come up with proofs for the following statements in ten words or less. Who can make the most rigorous, accurate proof in just ten words?

$$1$$. Among $$51$$ integers from $$1$$ to $$100$$ inclusive, there exists two summing to $$101$$.

$2$. $8^{n}-5^{n}$ is a multiple of $3$ where $n$ is a positive integer.

$3$. The point $O$ in equilateral triangle $ABC$ such that $OA+OB+OC$ is minimised is the center of $ABC$.

All operations, except for brackets, are words.

Note by Joel Tan
6 years ago

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1) There are $50$ pairs with sum $101$. PHP.

- 6 years ago

2) $8^n - 5^n \equiv 2^n - 2^n = 0 \pmod{3}$

- 6 years ago

Isn't it supposed to be $8^{n}-5^{n}$ in the very first step...??

- 6 years ago

Yes, typoed it.

- 6 years ago

Proofs:

1) Gauss' pairing trick paired with pigeonhole principle.

2) $=$ GP with ends $5^{n-1}, 8^{n-1}$, ratio $1.6$, summed thrice.

3) By construction of Fermat point, $O = X(13)$.

1) Alliteration 'p's yay! No I really did not want to refer to any language stuff.

2) It is trivial to see that all terms of the GP are integers, and ends refer to initial and final terms. I would have preferred $3\sum^{n-1}_{k=0}8^{n-1-k}5^k$ but I am not sure how many words would that be.

3) Sorry for the overkill :P But anyway I could have mentioned an equilateral triangle is acute for clarity.

By the way, nice choice of problems!

- 6 years ago

$8^{n}-5^{n}=8^{n}-(8-3)^{n}=3A$

- 6 years ago

2.8^n-5^=(8-5)k

- 6 years ago

Q.2 It has one factor(8-5) = 3 means it is divisible by 3 Q.3 O is crcumcentre

- 6 years ago

Maximum is at the fermat point, not at the circumcentre, although it is true in this case since it is equilateral.

- 6 years ago

There can be 49 pairs summing to 101

- 6 years ago

2) If N>=0, 8^N - 5^N = 0 (mod 3)

- 5 years, 11 months ago

1. Prove that sum of two integers within 100 is 101.

- 6 years ago

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