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Since 9 = 10 − 1, to multiply a number by nine, multiply it by 10 and then subtract the original number from the result. For example, 9 × 27 = 270 − 27 = 243. This method can be adjusted to multiply by eight instead of nine, by doubling the number being subtracted; 8 × 27 = 270 − (2×27) = 270 − 54 = 216.
A scale factor of 1 ⁄ 10 cannot be used here, because scaling 160 by 1 ⁄ 10 gives 16, which is greater than the greatest value that can be stored in this fixed-point format. However, 1 ⁄ 11 will work as a scale factor, because the maximum scaled value, 160 ⁄ 11 = 14. 54, fits within this range. Given this set:
Since 2 represents 20, all numbers in that scale are multiplied by 10. Thus, any answer in the second set of numbers is multiplied by 100. Since 8.8 in the top scale represents 88, the answer must additionally be multiplied by 10. The answer directly reads 1.76. Multiply by 100 and then by 10 to get the actual answer: 1,760.
For 8-bit integers the table of quarter squares will have 2 9 −1=511 entries (one entry for the full range 0..510 of possible sums, the differences using only the first 256 entries in range 0..255) or 2 9 −1=511 entries (using for negative differences the technique of 2-complements and 9-bit masking, which avoids testing the sign of ...
In general, if an increase of x percent is followed by a decrease of x percent, and the initial amount was p, the final amount is p (1 + 0.01 x)(1 − 0.01 x) = p (1 − (0.01 x) 2); hence the net change is an overall decrease by x percent of x percent (the square of the original percent change when expressed as a decimal number).
In algebraic notation, widely used in mathematics, a multiplication symbol is usually omitted wherever it would not cause confusion: "a multiplied by b" can be written as ab or a b. [1] Other symbols can also be used to denote multiplication, often to reduce confusion between the multiplication sign × and the common variable x.
A linear congruential generator with base b = 2 32 is implemented as + = (+) , where c is a constant. If a ≡ 1 (mod 4) and c is odd, the resulting base-2 32 congruential sequence will have period 2 32.
For example, to multiply 7 and 15 modulo 17 in Montgomery form, again with R = 100, compute the product of 3 and 4 to get 12 as above. The extended Euclidean algorithm implies that 8⋅100 − 47⋅17 = 1, so R′ = 8. Multiply 12 by 8 to get 96 and reduce modulo 17 to get 11. This is the Montgomery form of 3, as expected.