Correct way to add 22 to 4 to get 82

Question

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.

Answer 1

In French:

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

Answer 2

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

So we have the 'sum':

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

Answer 3

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

Answer 4

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

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

Answer 5

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

So the upside down text reads:

Twenty Eight equals Six plus Twenty Two

Answer 6

My guess:

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

Answer 7

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 $x$.

Answer 8

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.

Answer 9

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 $10_b$.)

Answer 10

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.

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

Answer 11

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 $[82]_4 = [26]_4 = [22 + 4]_4 = [22]_4 + [4]_4$