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Inspired by this, this, and this.


I'm not great at math, I'll admit. However, I am quite sure that there is a way to add 22 to 4 to obtain the number 613867. Now, that's a pretty big number, so my calculations might be a little off, so let's say the addition of 22 to 4 is within 1% of 613867.


As with all of the other puzzles, consider these numbers in base 10.

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  • $\begingroup$ Is that base 10 as in "the number of periods in ... ... ... ." ? $\endgroup$ – Deusovi May 28 '16 at 4:10
  • $\begingroup$ @Deusovi Yes. Or A in hexadecimal, if you so prefer. $\endgroup$ – Arcturus May 28 '16 at 4:11
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    $\begingroup$ I tried something and got 17592186044416, think I overshot the mark some :) $\endgroup$ – Jasen May 28 '16 at 6:32
  • $\begingroup$ hmm 28^4 get me close enough, but I can't mangle 22+4 into that. $\endgroup$ – Jasen May 28 '16 at 6:38
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    $\begingroup$ ERROR CANNOT STORE AS SHORT (aka NumberOverflow) $\endgroup$ – ev3commander May 28 '16 at 11:25
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The atomic number of titanium is $22$, and the atomic number of beryllium is $4$. Titanium and beryllium have a combined molar mass of $56.8792\text{ u}$. Methyl isocyanate has a molar mass of $57.051\text{ u}$, whch is within $1\%$ of the sum of titanium and beryllium.

Reasoning:

You can interpret the number $613867$ as a chemical formula, with the numbers either representing the atomic number of the element or the subscript representing the number of atoms in the molecule. The most reasonable solution along this logic is methyl isocyanate, $\text{CH}_3\text{OCN}$. Similarly, $22$ is the atomic number of titanium, and $4$ is the atomic number of beryllium. Titanium and beryllium have a combined molar mass of $56.8792\text{ u}$. Methyl isocyanate has a molar mass of $57.051\text{ u}$, whch is within $1\%$ of the sum of titanium and beryllium.

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  • $\begingroup$ could you explain how to get from 613867 to methyl isocyanate in more detail please $\endgroup$ – JMP May 29 '16 at 2:47
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    $\begingroup$ @JonMarkPerry The atomic numbers of carbon, hydrogen, oxygen, and nitrogen are (respectively) 6, 1, 8, and 7. Since there are 3 hydrogen atoms, a 3 is added after the 1. $\endgroup$ – Arcturus May 29 '16 at 13:43
  • $\begingroup$ @Eridan was this the intended solution? $\endgroup$ – ahorn May 31 '16 at 10:01
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If your math isn't that great you might have misplaced the decimal points, so $2.2+4=6.2$

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  • $\begingroup$ That is not the intended solution. 613867 is definitely within 1% of the answer. $\endgroup$ – Arcturus May 28 '16 at 2:53
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    $\begingroup$ and 6.2 is within 1% of 6.13867 $\endgroup$ – JMP May 28 '16 at 2:54
  • $\begingroup$ There is no error in the placement of a decimal place. $\endgroup$ – Arcturus May 28 '16 at 2:55
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    $\begingroup$ the way you have phrased the Q implies there might be $\endgroup$ – JMP May 28 '16 at 3:04
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In the additive cyclic group of integers modulo $613841$ $(\mathbb Z_{613841})$:

$\overline{22}+\overline{4 }=\overline{22+4}=\overline{26}= \overline{613841+26}=\overline{613867}$, where $\overline x$ denotes the equivalence class of $x$.

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  • $\begingroup$ @JonMarkPerry that overuse of notation just hurts my eyes. Equal equivalence classes can be expressed as such using "=". $\endgroup$ – ahorn May 28 '16 at 18:34
  • $\begingroup$ I think this answer meets the requirements of the question most accurately: "I'm not great at math, however," implies that the answer should be mathematical. $\endgroup$ – ahorn May 28 '16 at 20:09
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If you

Convert 61 38 67 as decimal to ASCII

You get

=&C

So, I think

You are trying to make the point that you can effectively get any number by "adding" 22 to 4 i.e, 26, 2, 84, 70, &c (which is an archaic form of etc, meaning "and so on" or "and other things")

or 22 "+" 4 = &c

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add 22% green to 4 "about 10% red and blue" to get color #613867

http://www.colorhexa.com/613867

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