# Correct way to add 22 to 4 to get 82

Inspired by this other puzzle, tell me a correct way by which adding 22 to 4 will give 82.

As in that other puzzle, these numbers are all expressed in base 10.

• (I almost posted this little tidbit as a comment on the other puzzle, before deciding that it might be interesting enough to stand on its own.) Commented May 26, 2016 at 12:45
• Looks like you already get 3 correct answer and didnt match your desire solution. Maybe you should add more info. Commented May 26, 2016 at 17:15
• You say "As in that other puzzle, these numbers are all expressed in base 10." But that's not quite true based on the answer you selected in the other puzzle. Commented May 26, 2016 at 17:58
• @gtwebb there's a difference between base and modulo. Those are still base-10 because 10 = 9+1. If it were base-8 (for example) then 10 = 7+1. In base-12 10 = 9+3. A 24-hour clock is modulo-24, meaning 23+1 = 0, but it is still base-10. Similarly month numbers are modulo-12 but again still base-10. Commented May 26, 2016 at 18:10
• I am completely baffled about the answers. Ths site is just great. Commented May 26, 2016 at 22:43

## 11 Answers

In French:

Quatre (4) can be added to vingt deux (22) to make "Quatre-vingts deux" (82)

• Brilliant! It is also valid answer to question "Can you subtract 22 from 4 and get 82?" (note the dash) :-D Commented May 26, 2016 at 19:20
• “Vingt” only takes an s at the end of “quatre-vingts” when it is not followed by anything else. Here, you should write “quatre-vingt-deux” (and “vingt-deux”, actually, with a dash as well). Commented May 27, 2016 at 4:18
• This is not base 10. Commented May 27, 2016 at 18:26
• Taemyr, the question might have been wordplay, but the numbers are most definitely base 10. Commented May 27, 2016 at 18:52
• @JATerroba 'Quatre-vingts' is clearly base twenty. Commented May 28, 2016 at 6:26

If you superimpose a seven-segment display $4$ onto the first $2$, it becomes an $8$:

So we have the 'sum':

$$22+$$ $$4\;=$$ $$82\;\;\;$$

• this could just as easily give 28 instead of course. also you'd get the same answer adding 5, 6, 8, 9 or even 0!
– jk.
Commented Jun 1, 2016 at 12:04

My suggestion:

1. Form with matchsticks roman representation of 22 - XXII
2. Add 4 matchsticks in front forming LX
3. The result is LXXXII, which is 82

• Also not even remotely like my intended answer, but this one is awesome, too, and probably satisfies the terms of the puzzle as I phrased it! But the answer I'm really looking for involves actually adding the number 22 to the number 4, not to 4 matchsticks which are cleverly arranged to represent the number 60. :) Commented May 26, 2016 at 14:53
• @MichaelMcGriff +1 for making pun of a box of matches. Take note of the difference between "out of the box" (=standard solution) and "outside the box" (=original solution), though. Commented May 27, 2016 at 11:47

4 groats + 22 threepenny bits = 82 pence (in old money, UK).

And thank goodness we don't use those any more.

• You win one guinea's worth of internet. Commented May 29, 2016 at 3:05

Make the text Upside down / Or rotate text to 180 degree:

So the upside down text reads:

Twenty Eight equals Six plus Twenty Two

My guess:

22 weeks + 4 Semesters (15 weeks in each semester) = 82 weeks

• where I am from, a such a thing lasts 6 months. Commented May 26, 2016 at 20:44
• Semester mean six months, isnt? so at least 24 weeks? But better aproach is half year or 26 weeks. Commented May 26, 2016 at 21:16
• @JuanCarlosOropeza The name "Semester", linguistically, should be six months, but generally it means "half of an academic year" - and in the US, there's normally a "summer break" (or "summer semester" if you're taking classes). Commented May 26, 2016 at 21:33
• Similar and not so ambiguous could be present with minutes and quarters ;) Commented May 26, 2016 at 23:15
• but still, 10/2=5 months. 5 months*>4 wks/month = >20 wks =/= 15 wks
– user17008
Commented May 27, 2016 at 0:57

In the set of integers modulo 14 $(\mathbb Z_{14})$:

$\overline{22} + \overline 4 = \overline {14+8} + \overline 4 = \overline {8} + \overline 4 = \overline {12} = \overline {12} + \overline{5\times 14} = \overline {12+5\times 14}=\overline {82}$

Alternatively, use the integers modulo 56:

$\overline{22}+\overline{4 }=\overline{22+4}=\overline{26}= \overline{56+26}=\overline{82}$

where $\overline x$ denotes the equivalence class of $x$.

• Soon someone will complain that you are using another base than 10... Commented May 28, 2016 at 19:17
• @Olba12 Let them display their ignorance. Commented May 28, 2016 at 19:20
• Of course, in $\mathbb{Z}$ the difference between $82$ and $22+4$ is $56$, so if you pick any divisor $d$ of $56$, then this will work modulo $d$ (i.e. in the ring $\mathbb{Z}/d\mathbb{Z}$), and it will work with no other modulus. The positive divisors of $56$ are: $1, 2, 4, 7, 8, 14, 28, 56$. Commented May 31, 2016 at 14:50

by ear:

"tell me a correct way by which adding two two to four will give 82."
is the same as
"tell me a correct way by which adding 2224 will give 82."

And so the answer is

-2142

Or written longly

The way by which adding 2224 will give 82, is by adding it to -2142.

Since there's an accepted answer, I'm not going to bother hiding...

$(2_{10} 2_{10})_{38} + (4_{10})_{38} = 26_{38} = 76_{10} + 6_{10} = 82_{10}$

where subscripts indicate the base in which the number is written.

EDIT: Comments indicate this is not a new idea. (Every number in base $b$ is in base $10_b$.)

• "these numbers are all expressed in base 10."
– f''
Commented May 27, 2016 at 7:22
• @f'' : Apparently my fingers are smarter about the idea I was trying to capture than my perverse brain is, and filtered out the intended content. Bad fingers. Bad. Commented May 27, 2016 at 7:29
• @f'' : Base 10 is not a limitation. Commented May 27, 2016 at 12:20
• @aluriak That one is lateral-thinking, this one isn't.
– f''
Commented May 27, 2016 at 18:25
• @f'' : Not sure that a base-implicit number belongs to lateral-thinking. Commented May 28, 2016 at 9:47

Since several answers are making use of modulo arithmetic (and I know they were not the selected answers), I'm just going to throw this stupid solution out there.

$[22]_1$ + $[4]_1$ = $[82]_1$

• Haha! Ofc the most obvious one. I choosed $[.]_4$ becuase then I could relate to winter, spring, summer autumn... heh Commented Jun 4, 2016 at 23:23

We have summer, autumn, winter and spring. If we say that summer is $0$ then autumn is $1$, winter $2$, and spring $3$. Then $4 + 22 = 26$ which is $6$ cycles of summer->autumn->winter->spring then followed by autumn->winter. Which of course is equal to $20$ cycles of summer->autumn->winter->spring followed by autumn->winter. That is why $22+4=82,$ because winter is coming.

or explained with math

$[22]_4 + [4]_4 = [26]_4 = [2]_4 = [80+2]_4 = [82]_4$ using that $[a]_4 = \{ b \ | \ b \equiv a \pmod{4} \}$ since we have that $b \equiv a \pmod{n}$ iff $[a]_n = [b]_n$ it follows directly that since $82 \equiv 26 \pmod{4}$ then $[82]_4 = [26]_4$ and by using the arithmetic rules of congruence classes modulo $n$. We get have that $[82]_4 = [26]_4 = [22 + 4]_4 = [22]_4 + [4]_4$

• Is this a joke?  It doesn't make sense to me; please expand on the explanation. And add spoiler markdown. Commented May 27, 2016 at 0:55
• No, im using base 10, with modulo arithmetic. @Prem Commented May 27, 2016 at 7:10
• I have added a spoiler and updated my explanationen @PeregrineRook Commented May 27, 2016 at 7:30
• THIS IS NOT BASE 4?! Commented May 27, 2016 at 19:41
• An 8 and a 6 are digits that do not exist in base 4. Thus, this is not base 4. It is the integers modulo 4. Commented May 28, 2016 at 16:31