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cricket20 [7]
4 years ago
12

What times what equals 12 but when you add those numbers together it equals 3

Mathematics
1 answer:
wel4 years ago
4 0
So, we can write it like this:

x+y=3
x*y=12
from the first we know that
x+y=3
y=3-x

so:

x*y=12 can be written as:
x*(x-3)=12

so x^2-3x=12

Actually this equation does not have a solution within the real numbers

but we can calculate:

x^2-3x-12=0

and if we use the quadratic formula, we will have:

(remember the quadratic formula is the following, where for us a=1,b=-3 and c=-12)

\frac{-b- \sqrt{b*b-4a*c} }{2a}
\frac{3- \sqrt{3*3-4*(-12)} }{2} = \frac{3- \sqrt{9+48} }{2} = \frac{3- \sqrt{57} }{2}


So this would be the solution, where the other number would be
3-\frac{3- \sqrt{57} }{2}= \frac{6}{2}-\frac{3- \sqrt{57} }{2}=\frac{3- \sqrt{57} }{2}

The other solution would be the following one;

x=\frac{3+\sqrt{57} }{2}, y=1\frac{1+\sqrt{57} }{2}






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Of the total, there are 18 consisting of consecutive triplets (\{1,2,3\},\{2,3,4\},\ldots,\{18,19,20\}).

Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs (\{1,2\},\{2,3\},\ldots,\{19,20\}), and for each pair 18 possible choices for the third integer (for instance, \{1,2\} can be taken with 3, 4, ..., 20), to a total of 19\times18=342. To avoid double-counting (e.g. \{1,2\} can't go with 3; \{2,3\} can't go with 1 or 4), we subtract 1 from the extreme pairs \{1,2\} and \{19,20\} (twice), and 2 from the rest (17 times).

So, the number of triplets that don't consist of pairwise consecutive integers is

1140-(18+342-(2\times1+17\times2))=816

I don't know how useful this would be to you, but I've verified the count in Mathematica:

In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816
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