Correct way to add 22 to 4 to get 9999

Question

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.

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 09:58
@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

Element118 · Answer 1 · 2016-05-30T12:14:47.843

11

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 $9999$.

edited May 30 '16 at 12:14

answered May 30 '16 at 11:41

Element118

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score 3 · Answer 2 · answered May 28 '16 at 17:47

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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 $x$.

answered May 28 '16 at 17:47

ahorn

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This could literally be used for any integer 27 or greater. To add 22+4 to get $n$, simply consider operating in $\mathbb{Z}_{n-26}$. – May 29 '16 at 00:49
@JoeZ. this method can be used for any integer in $\mathbb Z_{|n-26|}$. – ahorn May 29 '16 at 07:33

Correct way to add 22 to 4 to get 9999

2 Answers2