weird math problem involving representation

Question

We can write the number $384$ as $4\overline{2}4$ where the bar above the 2 denotes a negative digit, so that $4\overline{2}4$ means $4\times 100−2\times 10+4$. How could we write 1988 in this way?

what type of math is this? does it have a specific name? if not, does anyone get the question?

Answer 1

It's like a different base system.

We have $2\overline{1}9\overline{2}$ as the solution.

Alternatively we have $1\overline{8}0\overline{2}8$, which is much harder to find.

Answer 2

This is the negadecimal number system, also known as base $-10$ ("negative ten"), assuming the bars are on every other digit and you never place them so there is a bar over the last digit.

Conversion from base ten can be done much like conversion into any other base: divide by the base, leaving a non-negative remainder, and the remainder gives one digit of the result. Repeat until done.

So for $1988,$ dividing by $-10$ with positive remainder: $$1988=-198\times-10+8,$$ so the last digit is $8.$ Next we continue with the quotient, $-198$: $$-198=20\times-10+2,$$ so the digit in the "$-10$s" place is $2.$ Next, $$20=-2\times-10+0,$$ then $$-2=1\times-10+8,$$ and finally $$1=0\times-10+1.$$ Reading the remainders from last to first we have $1,8,0,2,8,$ which gives us the negadecimal number $1\bar80\bar28.$

Answer 3

$384 = 3\times 100 + 8\times 10 + 4\times 1$

$=(3\times 100 + 8\times 10 + 4\times 1)+0$

$=(3\times 100 + 8\times 10 + 4\times 1)+100-100$

$=(3\times 100+100) +(8\times 10-100) + 4\times 1$

$=4\times 100 -2\times 10 + 4\times 1 = 4\overline{2}4$

In essence, what happened is that we take the tens-complement of the digit we wish to modify and increase the digit to its left by one. If we wish to do this for multiple digits, its best to approach from right-to-left.

$1988=1\times 1000 + 9\times 100+8\times 10+8\times 1$

$ = 1\times 1000 + 9\times 100 + 8\times 10+8\times 1 + 100-100$

$= 1 \times 1000 + (9\times 100+100)+(8\times 10 - 100)+8\times 1$

$=2\times 1000+0\times 100-2\times 10+8\times 1$

$=2\times 1000+0\times 100-2\times 10+8\times 1+10000-10000$

$=1\times 10000 +(2\times 1000 - 10000)+0\times 100 - 2\times 10+8\times 1$

$=1\times 10000 - 8\times 1000 + 0\times 100 - 2\times 10 +8\times 1$

$=1\overline{8}0\overline{2}8$

With practice, one can do this without going through each step like I did above and arrive quickly at an answer. E.g. rewriting $3334$ would be $1\overline{7}4\overline{7}4$. Take the tens-complement of the digit you wish to change and increase the digit to its left by one.