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For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...
For the multiplicative inverse of a real number, divide 1 by the number. For example, the reciprocal of 5 is one fifth (1/5 or 0.2), and the reciprocal of 0.25 is 1 divided by 0.25, or 4. The reciprocal function, the function f(x) that maps x to 1/x, is one of the simplest examples of a function which is its own inverse (an involution).
The method for general multiplication is a method to achieve multiplications ... 6+6=12 (carry the 1), 2+2=4 4+4=8; 8624 × 2 = 17248. Example: 76892 × 2
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 ...
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 ...
With x=2 and y=3 for example, by positioning the top scale to start at the bottom scale's 2, the result of the multiplication 3×2=6 can then be read on the bottom scale under the top scale's 3: While the above example lies within one decade, users must mentally account for additional zeroes when dealing with multiple decades.
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.
x 3 has been divided leaving no remainder, and can therefore be marked as used by crossing it out. The result x 2 is then multiplied by the second term in the divisor −3 = −3x 2. Determine the partial remainder by subtracting −2x 2 − (−3x 2) = x 2. Mark −2x 2 as used and place the new remainder x 2 above it.