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Inspired by four other puzzles, how could it be possible that adding 22 to 4 gives 9999? What is the correct way to do it?

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

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    $\begingroup$ Base 10 as in base $1010_2$? $\endgroup$
    – user17008
    May 28 '16 at 20:41
  • $\begingroup$ Yes, and as in the number of periods ending that sentence if you prefer.......... $\endgroup$ May 28 '16 at 20:48
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    $\begingroup$ I'm downvoting this because it is a low-quality question, probably with no definite answer. See the meta discussion $\endgroup$
    – ahorn
    May 31 '16 at 9:58
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    $\begingroup$ @ev3commander why don't you just use the roman numeral X, and not make the representation of numbers so complicated? $\endgroup$
    – ahorn
    May 31 '16 at 10:09
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Just a stab at this. Probably not the answer by a long shot, but might be interesting.

Read "22 to 4" as "Two two to four", which can be also $2$ to $24$. Adding up the numbers from $2$ to $24$ give $299$. Two $99$s concatenated give $9999$.

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In the additive cyclic group of integers modulo $9973$ $(\mathbb Z_{9973})$:

$\overline{22}+\overline{4 }=\overline{22+4}=\overline{26}= \overline{9973+26}=\overline{9999}$, where $\overline x$ denotes the equivalence class of $x$.

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  • $\begingroup$ This could literally be used for any integer 27 or greater. To add 22+4 to get $n$, simply consider operating in $\mathbb{Z}_{n-26}$. $\endgroup$
    – user88
    May 29 '16 at 0:49
  • $\begingroup$ @JoeZ. this method can be used for any integer in $\mathbb Z_{|n-26|}$. $\endgroup$
    – ahorn
    May 29 '16 at 7:33

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