Report divisibility with arithmetic integer operations only

Question

Given two numbers, the first is greater than the second, check whether the first number is divisible by the second without a remainder. If divisible, answer "yes"(or any other string), otherwise "no"(or any other string you like).

Conditions: You cannot use conditional operators, loops, arrays, bitwise operations, functions and methods. You can use variables, all arithmetic integer operations (that is, integer division), even the operation of taking the remainder of a division.

Numbers can range from 1 to 1,000,000. Maximum running time is 1 second.

You can of course write in any language, but here is my non-golfed solution in C++:

int main() {
  int a,b; cin >> a >> b;
  int c = a%b;
  int n = (c+1-(c-1)((2(c-1)+1)%2))/2;
  int y = 121, e = 101, s = 115;

  cout << (char)(y-11*n) << (char)(e+10*n) << (char)(s-83*n) << endl;
}

Answer 1

05AB1E, 1 byte

Ö

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Outputs 1 if divsible 0 if not

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Old version with string must be Yes or No

05AB1E, 13 12 10 bytes

Ö'…Ü„nor.D

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Ö              // Test to see if they are divisible
 '…Ü           // The string yes
     „no      // the string no
         r     // reverse the stack 
          .D   // push whether they are divisible copies of "yes"
               // Implicitly output the result 

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Answer 2

JavaScript (ES6), 17 bytes

Takes input as (a)(b). Returns "Y" for yes or "undefined" (as a string) for no.

a=>b=>"Y"[a%b]+''

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JavaScript (ES6), 50 bytes

Takes input as (a)(b). Returns "yes" or "no", as per the original rules.

a=>b=>(22724-21872*((2*(a%b-1)+1)%2)).toString(36)

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Answer 3

APL (Dyalog Unicode), 17 3 bytes^SBCS

Anonymous tacit infix function. Takes lesser and greater numbers as left and right arguments. Prints the string "0" if divisible and "1" if not.

1⌊|

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| division remainder when dividing right argument by left argument (gives 0 for divisible or strictly positive integer for non-divisible)

1⌊ the minimum of that and 1 (gives 0 for divisible or 1 for non-divisible)

Answer 4

Runic Enchantments, 6 bytes

0ii%p@

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Outputs 1 if divisible 0 if not.

Determines 0/1-ness by raising 0 to the power of the remainder, which returns 0 for all x except 0 (due to the indeterminate nature of 0⁰ having been decided to be 1).

Version with string output

Runic Enchantments, 20 bytes

10ii%p-4´36*6X+d-E@

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Utilizes the word dictionary to convert from a 0/1 value to the strings "yes" and "no" by multiplying the result by 436 and adding 47 (ok, it adds 60 and subtracts 13; the byte cost is ironically the same either way: 4X7++ vs. 6X+d- vs. 4´7+).

May be invalid due to using an array? There is certainly no array in the submission code but rather in the the interpreter due to how E is executed when the top of the stack is not a string.

Answer 5

Befunge-98 (FBBI), 6 bytes

%1\k0.

Arguments should be pushed onto the stack, and requires the current IP delta to be (1,0). Prints 1 if divisible and 0 if not.

Try it online! Link includes I/O boilerplate.