Division

[cinelli]

.
          x x x x
    -------------
x x ) x x x x 0 x
        x x
      -----
          x x x
          x x 1
          -----
              x x
              2 x
              ---

This puzzle is to find all the x's in this long division which has no remainder.

Cinelli

[GoSeigen]

.
          x x x x
    -------------
x x ) x x x x 0 x
        x x
      -----
          x x x
          x x 1
          -----
              x x
              2 x
              ---

This puzzle is to find all the x's in this long division which has no remainder.

Cinelli

I can't make the penultimate column (containing a 0, 1 and 2) work.

Is it formatted correctly, or is it just me? Perhaps I'm not familiar with the long-division method being used...

GS

[UncleEbenezer]

Perhaps I'm not familiar with the long-division method being used...

GS

I'm so unfamiliar with it it doesn't even ring a bell from anything in my schooldays.

Perhaps it's something they teach to today's kids, that has mutated since our time?

[GoSeigen]

Perhaps I'm not familiar with the long-division method being used...

GS

I'm so unfamiliar with it it doesn't even ring a bell from anything in my schooldays.

Perhaps it's something they teach to today's kids, that has mutated since our time?

Hmm, the layout looks exactly like the way long division was taught in my school c40 years ago, but those particular numbers don't fit into that method, so perhaps there's a difference.

Hopefully Cinelli can add some information without spoiling the puzzle...


GS

[jfgw]

Is it base 3?

Julian F. G. W.

[jfgw]

Spoiler

.
          1 0 2 1
    -------------
2 2 ) 1 0 1 0 0 2
        2 2
      -----
          2 0 0
          1 2 1
          -----
              2 2
              2 2
              ---

Julian F. G. W.

[Gengulphus]

.
          x x x x
    -------------
x x ) x x x x 0 x
        x x
      -----
          x x x
          x x 1
          -----
              x x
              2 x
              ---

This puzzle is to find all the x's in this long division which has no remainder.

A spoiler with reasoning...

First, fill in details of digits (in the looser sense - as the solution turns out, "trits" would be the more precise term) that are known or must be the same as each other due to the way long divisions are laid out and/or it ending with no remainder, using capital letters to indicate ones that must be identical (though unlike in normal alphametics, different capital letters need not be different):

.
          x 0 x x
    -------------
x x ) x x x B 0 A
        x x
      -----
          x B 0
          x x 1
          -----
              2 A
              2 A
              ---

Subtracting 1 from 0 in any base produces a result digit equal to the base minus 1, with a borrow from the column to its left. Since such a subtraction produces a result digit of 2 in this case, the base must be 3.

Every value subtracted during the long division must therefore be either the divisor or twice the divisor. So the 2-trit values being subtracted must both be equal to the divisor, which therefore is 2A, and the 3-trit value xx1 twice the divisor. The divisor can therefore only be 20, 21 or 22, making twice the divisor 110, 112 or 121 respectively. Only 121 fits the pattern, so the divisor must be 22 and the three subtracted values from top to bottom 22, 121 and 22 respectively:

.
          1 0 2 1
    -------------
2 2 ) x x x B 0 2
        2 2
      -----
          x B 0
          1 2 1
          -----
              2 2
              2 2
              ---

It's then a matter just a matter of reversing the two incompletely-known subtractions: xB0-121=2 makes xB0 be 121+2=200, then xxx-22=2 makes xxx be 22+2=101:

.
          1 0 2 1
    -------------
2 2 ) 1 0 1 0 0 2
        2 2
      -----
          2 0 0
          1 2 1
          -----
              2 2
              2 2
              ---

Gengulphus

[cinelli]

Well solved Julian and Gengulphus. I am glad the art of long division has not been lost. As well as with numerical examples, you can also do long division algebraically. For instance you can show that

f(x) = (2*x^3 - x^2 + 5*x - 3) / (x+2) = 2*x^2 - 5*x + 15 - 33 / (x+2)

and this would be the way to proceed if you wanted to integrate f(x).

Cinelli