The current answers all focus on decimal digits, when applying the “add all digits and see if that divides by 3”. That trick actually works in hex as well; e.g. 0x12 can be divided by 3 because 0x1 + 0x2 = 0x3. And “converting” to hex is a lot easier than converting to decimal. Pseudo-code: … Read more
If you can’t use a big-integer library, and you don’t have a native uint128_t (or similar), you’ll need to do this manually. One option is to express a as the sum of two 32-bit quantities, i.e. a = 232b + c, where b contains the 32 msbs, and c contains the 32 lsbs. Squaring is … Read more
You can use the division version of Russian Peasant Multiplication. To find the remainder, execute (in pseudo-code): X = B; while (X <= A/2) { X <<= 1; } while (A >= B) { if (A >= X) A -= X; X >>= 1; } The modulus is left in A. You’ll need to implement … Read more
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds: h = s / 3600; m = (s / 60) % 60; s = s % 60;
There’s a fairly well-known trick for determining whether a number is a multiple of 11, by alternately adding and subtracting its decimal digits. If the number you get at the end is a multiple of 11, then the number you started out with is also a multiple of 11: 47278 4 – 7 + 2 … Read more
Edit Apr 19, 2013: Modulo function updated to handle boundary cases as noted by aka.nice and arr_sea: static const double _PI= 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348; static const double _TWO_PI= 6.2831853071795864769252867665590057683943387987502116419498891846156328125724179972560696; // Floating-point modulo // The result (the remainder) has same sign as the divisor. // Similar to matlab’s mod(); Not similar to fmod() – Mod(-3,4)= 1 fmod(-3,4)= -3 … Read more
actually, this is not a bug and it is pretty common. its called a floating point “error” and in a nutshell, it has to do things with how decimal numbers are stored within a google sheets (even excel or any other app) more details can be found here: https://en.wikipedia.org/wiki/IEEE_754 to counter it you will need … Read more
Exponentiation by squaring still “works” for modulo exponentiation. Your problem isn’t that 2 ^ 168277 is an exceptionally large number, it’s that one of your intermediate results is a fairly large number (bigger than 2^32), because 673109 is bigger than 2^16. So I think the following will do. It’s possible I’ve missed a detail, but … Read more
Here’s an answer from the MSDN documentation. When you divide two integers, the result is always an integer. For example, the result of 7 / 3 is 2. To determine the remainder of 7 / 3, use the remainder operator (%). int a = 5; int b = 3; int div = a / b; … Read more
Unlike C or C++, Python’s modulo operator (%) always return a number having the same sign as the denominator (divisor). Your expression yields 3 because (-5) / 4 = -1.25 –> floor(-1.25) = -2 (-5) % 4 = (-2 × 4 + 3) % 4 = 3. It is chosen over the C behavior because … Read more