76
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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.

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  • 1
    $\begingroup$ (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.) $\endgroup$ – Trevor Powell May 26 '16 at 12:45
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    $\begingroup$ Looks like you already get 3 correct answer and didnt match your desire solution. Maybe you should add more info. $\endgroup$ – Juan Carlos Oropeza May 26 '16 at 17:15
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    $\begingroup$ 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. $\endgroup$ – Nate Diamond May 26 '16 at 17:58
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    $\begingroup$ @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. $\endgroup$ – Sabre May 26 '16 at 18:10
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    $\begingroup$ I am completely baffled about the answers. Ths site is just great. $\endgroup$ – Daniel May 26 '16 at 22:43

11 Answers 11

176
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In French:

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

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    $\begingroup$ Brilliant! It is also valid answer to question "Can you subtract 22 from 4 and get 82?" (note the dash) :-D $\endgroup$ – kamenf May 26 '16 at 19:20
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    $\begingroup$ “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). $\endgroup$ – Édouard May 27 '16 at 4:18
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    $\begingroup$ This is not base 10. $\endgroup$ – Taemyr May 27 '16 at 18:26
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    $\begingroup$ Taemyr, the question might have been wordplay, but the numbers are most definitely base 10. $\endgroup$ – J A Terroba May 27 '16 at 18:52
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    $\begingroup$ @JATerroba 'Quatre-vingts' is clearly base twenty. $\endgroup$ – Jasen May 28 '16 at 6:26
229
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If you superimpose a seven-segment display $4$ onto the first $2$, it becomes an $8$: LED digits superimposed

So we have the 'sum':

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

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    $\begingroup$ 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! $\endgroup$ – jk. Jun 1 '16 at 12:04
159
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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

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    $\begingroup$ 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. :) $\endgroup$ – Trevor Powell May 26 '16 at 14:53
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    $\begingroup$ @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. $\endgroup$ – Klas Lindbäck May 27 '16 at 11:47
96
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4 groats + 22 threepenny bits = 82 pence (in old money, UK).

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

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  • 25
    $\begingroup$ You win one guinea's worth of internet. $\endgroup$ – Joshua May 29 '16 at 3:05
32
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Make the text Upside down / Or rotate text to 180 degree:

enter image description here

So the upside down text reads:

Twenty Eight equals Six plus Twenty Two

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23
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My guess:

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

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    $\begingroup$ where I am from, a such a thing lasts 6 months. $\endgroup$ – njzk2 May 26 '16 at 20:44
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    $\begingroup$ Semester mean six months, isnt? so at least 24 weeks? But better aproach is half year or 26 weeks. $\endgroup$ – Juan Carlos Oropeza May 26 '16 at 21:16
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    $\begingroup$ @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). $\endgroup$ – Tin Man May 26 '16 at 21:33
  • $\begingroup$ Similar and not so ambiguous could be present with minutes and quarters ;) $\endgroup$ – kamenf May 26 '16 at 23:15
  • $\begingroup$ but still, 10/2=5 months. 5 months*>4 wks/month = >20 wks =/= 15 wks $\endgroup$ – ev3commander May 27 '16 at 0:57
19
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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.

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18
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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 $x$.

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    $\begingroup$ Soon someone will complain that you are using another base than 10... $\endgroup$ – Olba12 May 28 '16 at 19:17
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    $\begingroup$ @Olba12 Let them display their ignorance. $\endgroup$ – ahorn May 28 '16 at 19:20
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    $\begingroup$ 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: $1, 2, 4, 7, 8, 14, 28, 56$. $\endgroup$ – Jeppe Stig Nielsen May 31 '16 at 14:50
8
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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 $10_b$.)

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    $\begingroup$ "these numbers are all expressed in base 10." $\endgroup$ – f'' May 27 '16 at 7:22
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    $\begingroup$ @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. $\endgroup$ – Eric Towers May 27 '16 at 7:29
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    $\begingroup$ @f'' : Base 10 is not a limitation. $\endgroup$ – aluriak May 27 '16 at 12:20
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    $\begingroup$ @aluriak That one is lateral-thinking, this one isn't. $\endgroup$ – f'' May 27 '16 at 18:25
  • $\begingroup$ @f'' : Not sure that a base-implicit number belongs to lateral-thinking. $\endgroup$ – aluriak May 28 '16 at 9:47
5
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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 $[82]_4 = [26]_4 = [22 + 4]_4 = [22]_4 + [4]_4$

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    $\begingroup$ Is this a joke?  It doesn't make sense to me; please expand on the explanation. And add spoiler markdown. $\endgroup$ – Peregrine Rook May 27 '16 at 0:55
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    $\begingroup$ No, im using base 10, with modulo arithmetic. @Prem $\endgroup$ – Olba12 May 27 '16 at 7:10
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    $\begingroup$ I have added a spoiler and updated my explanationen @PeregrineRook $\endgroup$ – Olba12 May 27 '16 at 7:30
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    $\begingroup$ THIS IS NOT BASE 4?! $\endgroup$ – Olba12 May 27 '16 at 19:41
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    $\begingroup$ 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. $\endgroup$ – ahorn May 28 '16 at 16:31
5
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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.

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

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  • $\begingroup$ Haha! Ofc the most obvious one. I choosed $[.]_4$ becuase then I could relate to winter, spring, summer autumn... heh $\endgroup$ – Olba12 Jun 4 '16 at 23:23

protected by Aza Jun 1 '16 at 2:14

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