457 % 10 = 7 *
457 / 10 = 45
45 % 10 = 5 *
45 / 10 = 4
4 % 10 = 4 *
4 / 10 = 0 done
Get it?
Here's a C implementation of the algorithm that my answer implies. It will find any digit in any integer. It is essentially the exact same as Shakti Singh's answer except that it works for negative integers and stops as soon as the digit is found...
const int NUMBER = 457; // This can be any integer
const int DIGIT_TO_FIND = 5; // This can be any digit
int thisNumber = NUMBER >= 0 ? NUMBER : -NUMBER; // ?: => Conditional Operator
int thisDigit;
while (thisNumber != 0)
{
thisDigit = thisNumber % 10; // Always equal to the last digit of thisNumber
thisNumber = thisNumber / 10; // Always equal to thisNumber with the last digit
// chopped off, or 0 if thisNumber is less than 10
if (thisDigit == DIGIT_TO_FIND)
{
printf("%d contains digit %d", NUMBER, DIGIT_TO_FIND);
break;
}
}
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