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If the answer is greater than a single digit, simply carry over the extra digit (which will be a 1 or 2) to the next operation. The remaining digit is one digit of the final result. Example: Determine neighbors in the multiplicand 0316: digit 6 has no right neighbor; digit 1 has neighbor 6; digit 3 has neighbor 1
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 ...
If the n + 1 digit is greater than 5 or is 5 followed by other non-zero digits, add 1 to the n digit. For example, if we want to round 1.2459 to 3 significant figures, then this step results in 1.25. If the n + 1 digit is 5 not followed by other digits or followed by only zeros, then rounding requires a tie-breaking rule. For example, to round ...
(For example, the sixth row is read as: 0 ⁄ 6 1 ⁄ 2 3 ⁄ 6 → 756). Like in multiplication shown before, the numbers are read from right to left and add the diagonal numbers from top-right to left-bottom (6 + 0 = 6; 3 + 2 = 5; 1 + 6 = 7). The largest number less than the current remainder, 1078 (from the eighth row), is found.
Evaluating the multiplication starts from the units column. The tip of each triangle points to the next digit of the result. The digits visited, by following the triangles from base to tip are those of the desired product: 210996.
Any two consecutive integers have opposite parity. A number (i.e., integer) expressed in the decimal numeral system is even or odd according to whether its last digit is even or odd. That is, if the last digit is 1, 3, 5, 7, or 9, then it is odd; otherwise it is even—as the last digit of any even number is 0, 2, 4, 6, or 8.
A proper prime is a prime p which ends in the digit 1 in base 10 and whose reciprocal in base 10 has a repetend with length p − 1. In such primes, each digit 0, 1,..., 9 appears in the repeating sequence the same number of times as does each other digit (namely, p − 1 / 10 times). They are: [10]: 166
In particular, if n is 2 k, for some integer k, and the recursion stops only when n is 1, then the number of single-digit multiplications is 3 k, which is n c where c = log 2 3. Since one can extend any inputs with zero digits until their length is a power of two, it follows that the number of elementary multiplications, for any n, is at most
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