-5
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Create numbers from 50-100 only using 1,2,3 and 5. No repeats and you have to use each number. Also, you can use any operation! thanks :)?

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closed as too broad by Phylyp, athin, NL628, JonMark Perry, Deusovi Mar 8 '18 at 4:35

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Welcome to PSE. Do take a look at the tour page to familiarize yourself with how this site works. Also, please note that questions of this form often tend to be closed as being too broad (link). For example, Solutus Immensus's answer shows how such a broad question can result in an equally broad answer. $\endgroup$ – Phylyp Mar 8 '18 at 4:33
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    $\begingroup$ Try to avoid these kinds of questions unless you are 100% sure that this is a clever riddle that is not too broad. Almost all kinds of these questions are too broad. Welcome to Puzzling SE by the way! Hope you stay :D $\endgroup$ – NL628 Mar 8 '18 at 4:33
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    $\begingroup$ Also, try to avoid placing your entire riddle/puzzle in the title, because the point of a title is to give other users an idea about what the puzzle is about or to give an interesting hook to draw them in. $\endgroup$ – NL628 Mar 8 '18 at 6:21
  • $\begingroup$ With "any" you mean +,-,*,/,! and taking the square root and one number to the power of the other? $\endgroup$ – Thern Mar 8 '18 at 8:55
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If you can use any operation:

Let $ f_n(a,b,c,d)=n$ be a 4-ary constant operation.
Then $ f_n(1,2,3,5) = n $ for $ n = 50, 51, ..., 100 $

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    $\begingroup$ Or even better, let $\boxed{???}$ be the magical mathematical function that maps all integers to all integers and hence it will be trivial to create any integer. @zar do you see why this is too broad? $\endgroup$ – NL628 Mar 8 '18 at 6:22
  • $\begingroup$ What you described is not a function $\endgroup$ – user45835 Mar 8 '18 at 6:24
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    $\begingroup$ that's why it's magical... $\endgroup$ – NL628 Mar 8 '18 at 6:51

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