Jumping in Puddles

Mon Jan 4

Answer: Christmas Teasers

1. This is a tricky puzzle and admittedly the answer involves a somewhat unlikely scenario. The only way I can celebrate my birthday two days before Nicolette is if she was born east of the International Date Line on March 1st of a non-leap year. Assuming we’re travelling (for example, on a flight or a boat), I could then be born nine minutes later west of the International Date Line where it would be February 28th. If this was the case, on any subsequent leap year, I would celebrate my birthday two days before Nicolette.

2. We’re told that:

The limit as n -> ∞ of x^x^x^x^x… (n terms) is equal to 2.

Therefore, the limit as n -> ∞ of x^ [x^x^x^x… (n terms)] is equal to 2. So x^2 = 2. And hence:

x = √ 2

Here’s a crude numerical approximation courtesy of the brilliant Wolfram Alpha that illustrates the answer.

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