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Recently I came upon this puzzle:

1= 2= 55= 42= 102= 26262

The clue was "the power of zero". Anyone has any ideas on how to solve this? No other information was given.

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    $\begingroup$ Welcome to Puzzles SE! Please take a look at our Help Center and feel free to take the Tour. Since this looks like a outside source, please add a link to the document. $\endgroup$ – Prince North Læraðr Feb 20 '18 at 20:47
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    $\begingroup$ Is this supposed to be a number sequence, or is it an equation that you're supposed to add symbols to? For the former, the equals signs seem weird. If it's the latter though, (given the hint) seems like it could be pretty trivial, just multiply by (or raise to the power of) zero and it wouldn't matter what the numbers were. Was there any more context around the puzzle to suggest what sort of solution it was expecting (e.g. was it part of a group of similar puzzles)? $\endgroup$ – Blckknght Feb 20 '18 at 20:58
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio Mar 3 '18 at 8:45
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I thing the answer is in the title:

$1^0=2^0=55^0=42^0=102^0=26262^0 (=1)$
Basically, any +ve integer to the power of 0 is equal to any other positive integer to the power of zero.

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Is it:

$0$ to the power of all these numbers is $0$?

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Is it because

all numbers to the power of 0 is 1?

(can't comment yet, but dang, just late by a couple of minutes)

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There are infinite answers(n is the number):

n-n+C, n^0+c, 0*n+C And so on. All of them use the power of 0 (not as a mathematical power but the special properties of 0. C is any number (real, complex, doesn't matter) There are of course many more ridiculous answers such as a^(0*n)+C

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