# Correct way to add 22 to 4 to get 9999

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.

• Base 10 as in base $1010_2$? – ev3commander May 28 '16 at 20:41
• Yes, and as in the number of periods ending that sentence if you prefer.......... – zʏᴀʙiɴ101 May 28 '16 at 20:48
• I'm downvoting this because it is a low-quality question, probably with no definite answer. See the meta discussion – ahorn May 31 '16 at 9:58
• @ev3commander why don't you just use the roman numeral X, and not make the representation of numbers so complicated? – ahorn May 31 '16 at 10:09

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 $9999$.
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 $x$.
• This could literally be used for any integer 27 or greater. To add 22+4 to get $n$, simply consider operating in $\mathbb{Z}_{n-26}$. – Joe Z. May 29 '16 at 0:49
• @JoeZ. this method can be used for any integer in $\mathbb Z_{|n-26|}$. – ahorn May 29 '16 at 7:33