x&7

x-(8 * (x/8))

int a=0;
int remainder=0;
a=x/8;
remainder=x-(a*8);

will give the same result as of x%8

int remainder=0;
int i=0;
i=x/8;
remainder=x-(i*8);

this will give you the same output as x%8

Alternative expression:
x&0x7(x bitwise and with 0x7)

x % 8 = x & 7

the equivalent expression for x%8 is x-(x/8)*8

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