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Here is a question and some answer for it. Now my question is a bit outside the box. Well, put operators between the numbers show to get the end result:

10   10   10 = 6
11   11   11 = 6

As you can see some examples in the above link, also here is an example:

2 + 2 + 2 = 6

So, what operators should I have between them for 10 and 11 ?

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    $\begingroup$ downvoted it because even though the puzzle is nice I think the explanation is wrong and incomplete. For example, you say you can only put operators between the numbers. The accepted answer doesn't have it only between the numbers. Also I think you should have said what operators are allowed. And also no mention of the use of parentheses because I think that technically isn't an operator. $\endgroup$ – Ivo Beckers Nov 19 '15 at 14:15
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    $\begingroup$ @IvoBeckers sorry, I don't know english very well. $\endgroup$ – Shafizadeh Nov 19 '15 at 14:16
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10.

$(\sqrt{10-10/10})! = 6$

11.

$\lfloor\log 11\rfloor+\lfloor\log 11\rfloor+\lfloor\log 11\rfloor=6$

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  • $\begingroup$ Oh wait, log11 is less than 1.5, So, its round will be 1, not 2. Right? $\endgroup$ – Shafizadeh Nov 19 '15 at 16:23
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    $\begingroup$ Well, in this case $\log$ refers to the natural logarithm, i.e., $\log_e$. So $\log_e(11) = 2.39...$. $\endgroup$ – Carl Löndahl Nov 19 '15 at 16:26
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    $\begingroup$ It seems to me that Carl used the natural logarithm (the one with e), ln(11), which is approximately equal to 2.4 $\endgroup$ – Akiiino Nov 19 '15 at 16:27
  • $\begingroup$ If you are not fine with the natural logarithm, consider this instead $(\lfloor\sqrt{11-11/11}\rfloor)! = 6$ :-) $\endgroup$ – Carl Löndahl Nov 19 '15 at 16:30
  • $\begingroup$ WTF !!! Your brain is not normal :P !! Bravo ... good for you $\endgroup$ – Shafizadeh Nov 19 '15 at 16:31
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For all integers $n > 1$, we can write

$$\frac{\log \left( \log(n) / \log \left( √√√√√\sqrt{n} \right) \right)}{\log(\lceil √√\dots\sqrt{n} \rceil )}=6$$

First, we use the fact that $$\log(n) / \log(√√√√√\sqrt{n}) =64.$$

To get a "free" $2$, we repeatedly take the square root of our number $n$: we know that for all $n > 1$, $1 < \sqrt{n} < n$, so eventually we approach $1$ from above, but never reach it. Then, we take the ceiling of this value, and get $\lceil 1+\varepsilon \rceil = 2$.

Then, we simply calculate $\log(64) / \log(2) = 6$.

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