I have
$$ {n \choose 2} = 21 $$
and as the title mentions I have to solve for $n$, but so far all I have managed to get to is
$$n^2 -n =42 $$
and from there I'm completely lost. Any hints would be greatly appreciated.
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http://en.wikipedia.org/wiki/Completing_the_square – lab bhattacharjee Sep 12 '14 at 10:37
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Life, the universe, and everything: http://en.wikipedia.org/wiki/Phrases_from_The_Hitchhiker%27s_Guide_to_the_Galaxy#42_Puzzle – Sep 12 '14 at 10:41
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By observation, $42=(7)(6)=(-6)(-7)$ so the quadratic equation $42=(n)(n-1)$ has solutions $7$ and $-6$. – MPW Sep 12 '14 at 11:20
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Why is this tagged ([tag:probability])? – Martin Sleziak Sep 12 '14 at 12:32
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@MPW: $n$ is a positive integer – Sep 12 '14 at 12:44
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Yes, that's the next step--take the solutions to the quadratic and see which one(s) satisfy the original question. So you would eliminate one of the solutions to the quadratic. – MPW Sep 12 '14 at 13:11
4 Answers
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Note that $$n^2-n=42\iff n^2-n-42=0\iff (n-7)(n+6)=0$$ and that $n\ge 2$.
mathlove
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I think my brain has left for a holiday at this point. I can't believe i didn't see this before. Thanks for the help – Abby Sep 12 '14 at 10:48
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$$n^2-n=(n-1)n.$$ You need to find two consecutive integers with product $42$.
Check among $2,6,12,20,30,42,56,72,90...$.
Hint: the nearest perfect squares are $6^2=36$ and $7^2=49$.
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Another approach, if you know the formula for the sum of the first $n-1$ integers, which is $$ \sum_{k=1}^{n-1}k=\binom{n}{2} $$ We can check at each step until: $1+2+3+4+5+6=21$, and we get $n=7$.
robjohn
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Dear Teacher, If ı ask you, can you look my problem, if you have few minutes? Is it possible? I need a mathematician's answer. Best Regards. Thank you so much. https://math.stackexchange.com/questions/2559814/is-the-proof-i-am-using-sufficient-correct-for-the-system-of-equation – MathLover Dec 12 '17 at 20:53