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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 ♦
    Commented May 28, 2016 at 4:10
  • $\begingroup$ @Deusovi Yes. Or A in hexadecimal, if you so prefer. $\endgroup$
    – Arcturus
    Commented May 28, 2016 at 4:11
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    $\begingroup$ I tried something and got 17592186044416, think I overshot the mark some :) $\endgroup$
    – Jasen
    Commented May 28, 2016 at 6:32
  • $\begingroup$ hmm 28^4 get me close enough, but I can't mangle 22+4 into that. $\endgroup$
    – Jasen
    Commented May 28, 2016 at 6:38
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    $\begingroup$ ERROR CANNOT STORE AS SHORT (aka NumberOverflow) $\endgroup$
    – user17008
    Commented May 28, 2016 at 11:25

5 Answers 5

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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
    Commented May 29, 2016 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
    Commented May 29, 2016 at 13:43
  • $\begingroup$ @Eridan was this the intended solution? $\endgroup$
    – ahorn
    Commented May 31, 2016 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
    Commented May 28, 2016 at 2:53
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    $\begingroup$ and 6.2 is within 1% of 6.13867 $\endgroup$
    – JMP
    Commented May 28, 2016 at 2:54
  • $\begingroup$ There is no error in the placement of a decimal place. $\endgroup$
    – Arcturus
    Commented May 28, 2016 at 2:55
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    $\begingroup$ the way you have phrased the Q implies there might be $\endgroup$
    – JMP
    Commented May 28, 2016 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
    Commented May 28, 2016 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
    Commented May 28, 2016 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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