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I cooked this one up on the fly based on one of yesterdays problems so it may have more than one answer.
What you got: the numbers 2 3 4 5 7 9
The problem: you need to get 13 using only addition, multiplication, division and subtraction. everything is in base 10 and no decimal's where used and you need to use ALL 6 numbers only once.
Why is this is Cosmos? (<edit>nm - I'm not sleeping this week. Apparently I'm delusional again - for some reason I thought there was another "new" forum like the humor forum for this sort of stuff.</edit>)
57-49+2+3=13
...though I'd venture there are a few other solutions.
I dont think if you are given the numbers 5 and 7 then you can turn this into 57, Is that right Zepherin?
P
I never said you couldn't
but I'm not really about right or wrong, that wasn't the solution I had in mind if that's what your asking prank.
Tee hee hee. This one made me think for a minute or so before I came up with the actual
solution, but yet again I have not slept in like 48 hours.
Given 2, 3, 4, 5 ,7, 9 and using logical mathematical operands (+,-,*,/) and only using
each diget once in any order to reach 13. Using these systems you have to include
the quantity operand (grouping in parenthesis) because well that IS the only way you
can solve the problem.
(((7+9)-5)/3)+(4+2)
((26-5)/3) +(6)
((21)/3) +(6)
7 + 6
-----------------------
13
7+9 = 25
25-5 = 21
21/3 = 7
7+4 = 11
11+2 = 13
13 = 13
I am assuming that is what you meant.
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Lumina
another solution is ((2+5)(5)(4)-9)/7=13