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Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
[1]: 7 For example, if 20 apples are divided evenly between 4 people, everyone receives 5 apples (see picture). However, this number of times or the number contained (divisor) need not be integers. The division with remainder or Euclidean division of two natural numbers provides an integer quotient, which is the number of times the second ...
The black numbers are the addends, the green number is the carry, and the blue number is the sum. In the rightmost digit, the addition of 9 and 7 is 16, carrying 1 into the next pair of the digit to the left, making its addition 1 + 5 + 2 = 8.
If this number is truncated to 4 decimal places, the result is 3.141. Rounding is a similar process in which the last preserved digit is increased by one if the next digit is 5 or greater but remains the same if the next digit is less than 5, so that the rounded number is the best approximation of a given precision for the original number.
A natural number is divisible by three if the sum of its digits in base 10 is divisible by 3. For example, the number 21 is divisible by three (3 times 7) and the sum of its digits is 2 + 1 = 3. Because of this, the reverse of any number that is divisible by three (or indeed, any permutation of its digits) is also divisible by three. For ...
A divisor of any number of digits can be used. In this example, 1260257 is to be divided by 37. First the problem is set up as follows: 37)1260257 Digits of the number 1260257 are taken until a number greater than or equal to 37 occurs. So 1 and 12 are less than 37, but 126 is greater.
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a highly composite number has a number of positive divisors that is greater than any lesser number; that is, d(n) > d(m) for every positive integer m < n. Counterintuitively, the first two highly composite numbers are not composite numbers.