Re: -4 to 4 puzzle combos

fzpowerman47 wrote:

cool, thank you clm, it's clear in my head, now !

your table is perfect for me, I'll succeed to finish my 4 book puzzles, I know it now, it's nice, thank again

You are welcome, I am happy if the table can help. It's not difficult to extend it to - 7 to 7, for instance, so covering the - 7 to 1 or the -1 to 7 book puzzles.

However, do not forget that you could find situations (depending on the software or, in our site, until Patrick clarifies about the sign of the remainder, ... ), when dealing with mod operation and negative numbers, where two different results can be obtained, both correct, and both complying with |r| < |d| (as exposed in the Wikipedia), so there is, let's say, a "duplicity" or "duality" (in the remainder and in the quotient).

Those two solutions are part of the same class of "equivalence". For instance, you can see in the table that -4 mod -5 = - 4 (choosing the "natural" "integer" 0 as the quotient) but if you choose a quotient of 1 then 1 x (- 5) = - 5 and the remainder is: - 4 - (- 5) = - 4 + 5 = 1. In the class of equivalence created by the divisor - 5, the

**- 4** and the

**1** are both representative members, exactly as it is the - 14, for instance, quotient 2 >>> 2 x (- 5) = - 10 >>> remainder is - 14 - (- 10) = - 14 + 10 =

**- 4**. Or the - 9, another representative of the "class", quotient 1 >>> 1 x (- 5) = - 5 >>> remainder: - 9 - (- 5) = - 9 + 5 =

**- 4**, etc. Using different quotients is like using the subtraction process to create representatives of the "class".

Consequently in the table provided each result has another "colleague", "in the shadow", as a possible solution of the mod operation (both absolute values below the absolute value of the divisor). The modulo operation is easily understood with positive numbers while with negative numbers the "field of possibilities" expand to all numbers "to the left" of zero

.

Best, clm.