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Dividing large numbers |
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| Mar21-09, 04:08 PM | #1 |
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Dividing large numbers
How would you Divide very large numbers without using a calculator?
EX. [tex]\frac{125000}{299000000}[/tex] |
| Mar21-09, 04:23 PM | #2 |
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Long ago, before calculators, logarithms were used and invented for this purpose. You'd divide by subtracting logarithms and antilog the result to get the answer.
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| Mar21-09, 05:40 PM | #3 |
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One should usually first take out the obvious powers of ten, then factorize.
e.g. [tex]\frac{125000}{299000000} = \frac{125}{299000}=\frac{5^3}{299\cdot 10^3} = \frac{5^3}{299\cdot (2\cdot 5)^3} = \frac{1}{299\cdot 2^3}[/tex] And [tex]299\cdot 8 = 3 \cdot 10^2 \cdot 8 - 8 = 24 \cdot 10^2 - 8 = 2400 - 8 = 2392[/tex], so that [tex]\frac{125000}{299000000} = \frac{1}{2392}[/tex] Which by hand is good enough for me. (This might be wrong tho, it is kinda late here) |
| Mar22-09, 12:24 AM | #4 |
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Dividing large numbers
"How would you Divide very large numbers without using a calculator? "
Long division is a correct algorithm. Are you asking whether or not there exists a faster way? |
| Mar22-09, 12:35 PM | #5 |
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| Mar22-09, 01:02 PM | #6 |
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| Mar22-09, 01:49 PM | #7 |
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You could use Newton-Raphson. Computing x = 1/y for given y amounts to solving the equation:
1/x - y = 0 Then, Newton-Raphson yields the following recursion for the nth approximation x_{n+1} = x_n - (1/x_n - y)/(-1/x_n^2) = x_n +x_n -y x_n^2 = 2 x_n - y x_n^2 The iteration doesn't involve any divisions, so it is a true division algorithm. The number of correct digits doubles after each iteration, while with long division you only get one decimal at a time, so it is much faster than long division. |
| Mar22-09, 02:01 PM | #8 |
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