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For the numbers in (7.7), wc begin by marking every number in the list that has a factor of 2 (other than 2 itself). This is easily accomplished by simply marking every other number with - . beginning from 4, t o obtain 2 3 4 - 5 + 7 + 9 w 11 X2 13 4 15 46 17 4 19 2 4 8 % 21 23 4 25 6 27 8 29 30. The next unmarked number after 2, which is 3 , must be prime. To remove all nunibers with a factor of 3 (other than 3 itself), we mark every third number, beginning from 6, with / t o obtain 2 3 4 5 5 $ + 7 i & g w 11 @ 13 M 45 46 17 @ 19 28 2/1 23 q 25 6 8 29 $.
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The next unmarked number, 5, must be prime, so we mark every fifth number beginning frorn 10 with \ , giving 2 3 4 5 + 7 4 g I , \ g 11 @ 13 f4 @ 3 % 17 19 2 23 % 2fi 6 8 29 @. Continuing, we inark numbers having a factor of 7 with 2 11 $
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and numbers having a factor of 11 arc marked with
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and. finally. numbers with a factor of 13 are marked with n .
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(7.8)
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Since the next unmarked number is 17, which is greater than 30/2, we are finished. The primes less than 31, namely, 2 , 3 , 5 , 7 , 1 1 ,13,17,19,23,29 have passed through the sieve.
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7.2 FACTORING ALGORITHMS
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While the sieve of Eratosthenes gives us the primes, it also provides considerable information on the non-primes in the list, since the marks through any given number tell us the prime factors of the number. However, the marks do not tell us the power t o which a given prime factor occurs. For example, in (7.8), the number 24 is marked with "-" and "/", so we know that it is divisible by 2 (due t o the "--") and 3 (due t o the "/"), but from this information we do not know that 24 has a factor of 23. Now suppose that instead of crossing out the numbers, we divide out the factors (here, we also divide the prime by itself). Then, beginning again from
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
after dividing every other number by 2, beginning from 2, we have
11 21
3 13 23
1 12
5 15 25
8 17 9 19 lo
where the numbers that were divided by 2 at this step are underlined. Next, we divide every third number by 3, beginning with 3 (again, underlining the numbers that were divided a t this step), to obtain
1 2 5 1 7 4 3 5 11 2 13 7 5 8 17 3 19 10 - 11 23 4 25 13 9 14 29 5. 7
Then we divide every fifth number by 5, beginning with 5,
1 1 2 1 1 7 4 3 1 11 2 1 3 7 1 8 17 3 1 9 2 . 7 11 23 4 5 13 9 14 29 1
Dividing every seventh number by 7 yields
1 1 2 1 1 1 4 3 1 11 2 1 3 1 1 8 1 7 3 1 9 2 - 1 1 2 3 4 5 1 3 1 9 2 2 9 1
and suppose that we stop a t this point. Then the numbers that correspond t o the positions now occupied by the 1s in this array are 7-smooth, that is, they have no prime factors greater than 7. However, some of the numbers that correspond t o non-1s are also 7-smooth, such as 28, which is represented by the number 2 in the last row. The problem here is that we need to divide