Increase the number so that it becomes a multiple of 16

The most banal way: we divide 102 by 16, we get the quotient (6) and the remainder (6). Check if the remainder is zero. If equal, then we found our number. If not, then we take the quotient, add a unit to it. Then multiply by 16 and get 16*7=112 . Now from the result (112), we subtract the initial number (102) and have that the initial number should be increased by 10.

 am = (a + 0x0FL) & ~0x0FL; // ближайшее сверху число, кратное 16 d = am - a; // разница 

Let's look at the bits! :)

For multiples of 16, the lower 4 bits are zeros:

  0: 0000 0000 16: 0001 0000 32: 0010 0000 

To increase any number to a multiple of 16, you need to add "1" in the positions where there are now zeroes and another 1:

 11: 0000 1011 - только третий бит = "0" добавить 0100 - "1" для третьего бита добавить 0001 - и ещё +1, чтобы все младшие биты обнулились. ------------- не хватало 0100 (это 4) и ещё 1 = итого +5 22: 0001 0110 - нули в первом и четвертом битах добавить 1001 (это 9) добавить 0001 (единицу) --------------- сумма: 1010 (это десять) 

Calculate the required increase can also be bitwise operations:

1) from the initial number only the lower 4 bits are written. The bit operation "And" will leave only those bits that are = 1 in both operands:

  исходное число: 0101 1100 маска мл. 4 бит: 0000 1111 --------------------------- результат «И»: 0000 1100 

2) the low-order bits should be inverted with the “NOT” operation

 1100 «НЕ» ---- 0011 

3) add one:

  0011 = 3 +0001 = 1 ---- 0100 = 4 

This means that you need to add 0000 0100 (4) to the original number 0101 1100 (92) to get 96, which is probably a multiple of 16.

The next multiple of 16 is the following: - Add 16-1 - this is the number whose residue from dividing by 16 is -1. - Divide integrally by 16 - we get the result of dividing the original number by 16 rounded upwards - Multiply by 16 - we get a number not less than the initial one and a multiple of 16

 Math.floor((102+15) / 16) * 16 

Well, as a function for any numbers:

 function calc(x, m) { return Math.floor((x+m-1) / m) * m; } 

To find out how much to increase, you must subtract the original number from the resulting number.

Pseudocode:

 исходное_число = 102 кратное_число = 16 частное = 6 если остаток_от_деления(исходное_число) == 0 возвращаем исходное_число иначе возвращаем (частное + 1) * кратное_число - исходное_число 

C code:

 #include <stdio.h> int main() { int input_number = 29; int multiple = 16; int quotient = input_number / multiple; int answer = 0; if (input_number % multiple == 0) { answer = input_number; } else { answer = (quotient + 1) * multiple - input_number; } printf("%u\n", answer); return 0; } 

I would do this: Divide this number by 16. I would drop the shot, add one, multiply by 16 and take away the number that was given. It looks like this: 102/16 = 6,375. +1 = 7. 7 * 16 = 112. 112-102 = 10. Answer >> 10

with integer division (positive numbers) of the делимого by the делитель , the частное and the остаток .
искомое number will be the result of subtracting the остатка from the делителя .

1. If in the language used there is a функция or an оператор for obtaining the остатка of integer division, then:

 искомое = делитель - функция ( делимое, делитель ) искомое = делитель - ( делимое оператор делитель ) 

2. if, in the language used, there is only a функция or an оператор for obtaining a частного from integer division, then:

 искомое = делитель - ( делимое - делитель * функция ( делимое, делитель ) ) искомое = делитель - ( делимое - делитель * ( делимое оператор делитель ) ) 

3. and if there is neither one nor the other, but there is оператор (ordinary) division оператор and функция rounding функция , then:

 искомое = делитель - ( делимое - делитель * функция ( делимое оператор делитель ) ) 

4. But if in the language used there is neither one nor the other, then one can get by with subtraction and comparison. algorithm in abstract language:

 если делимое > 0 и делитель > 0 число = делимое пока число >= 0 число = число - делитель конец цикла искомое = -число конец условия 

 n = 102; m = n%16 ? 16-n%16 : 0;