Proof-based math puzzles

Reply #5

db101 — Thu, 10 Aug 2023 00:57:18 GMT

Clever solution Joe, but now you must prove the AM-GM inequality. (just kidding

Reply #4

Wavepack — Wed, 09 Aug 2023 16:30:38 GMT

Very nice, except for one thing. You've so far shown that the minimum = 10. Next, you need to choose a=b=c=d=e, which in the expression, evaluates to 10.

Reply #3

cottoneyedjoe — Wed, 09 Aug 2023 16:05:47 GMT

For 2) split the sum (a+d)/sqrt(bc) + (a+c)/sqrt(de) + (b+e)/sqrt(cd) + (b+d)/sqrt(ae) + (c+e)/sqrt(ab) into two sums

(a/sqrt(bc) + b/sqrt(cd) + c/sqrt(de) + d/sqrt(ae) + e/sqrt(ab)) +

(a/sqrt(de) + b/sqrt(ae) + c/sqrt(ab) + d/sqrt(bc) + e/sqrt(cd))

Now invoke the Arithmetic Mean Geometric Mean inequality, which states that for any set of n non-negative real numbers {x1, x2, ..., xn}, their sum is greater than or equal to n times the nth root of their product, with equality only when

Reply #2

db101 — Tue, 08 Aug 2023 14:18:32 GMT

Very nice solution.

Reply #1

Wavepack — Tue, 08 Aug 2023 11:48:26 GMT

1) Since the density of primes decreases as the integers increase, our best chance of proving this existence for all N is to make sure the starting odd number is large enough. We want a starting number such that the next N numbers are factorable/composite. The best way to do this is to make the starting number minus three a composite of factors of offsets from the starting number. Let S = starting number = (2N+1)! + 3 .

Candidate set of consecutive odd numbers = {(2N+1)! + 3, (2N+1)! + 5, .... [ More ]

Proof-based math puzzles

db101 — Tue, 08 Aug 2023 05:42:19 GMT

Someone showed me these problems with clever solutions, so I thought why not share them here. These are not meant to be solved by calculator or computer but by logic and math principles.

1) For every positive integer N, is it always possible to find a set of N consecutive odd numbers such that none of them are prime? Why or why not?

2) Let a, b, c, d, and e be positive real numbers. What is the minimum value of (a+d)/sqrt(bc) + (a+c)/sqrt(de) + (b+e)/sqrt(cd) + (b+d)/sqrt(ae) + (c+e)/sqr... [ More ]