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The number 600 has 12 pairs of factors: 1 X 600; 2 x 300; 3 x 200; 4 X 150; 5 X 120; 6 x 100; 8 X 75; 10 X 60; 12 X 50; 15 X 40; 20 X 30; 24 X 25. It is very helpful to know, in advance, that there will be 12 pairs, before seeking to identify them. Following, is shown an easy way to know, without enumerating them, that there are exactly twelve pairs. First, the steps: Factorise the number 600 into its smallest elements. For example, 600 = 6 X 100 = 2 X 3 X 5 X 20 = 2 X 3 X 5 X 5 X 4 = 2 X 3 X 5 X 5 X 2 X 2 = 3 X 5² X 2³. Note that 3, 5, and 2 are prime numbers and can be factorised no further. Note, also, that their three exponents are 1, 2, and 3; increase those last three numbers, and you get 2, 3, and 4; their product is 24. This means that there are 24 integers that divide into 600; and this gives 12 pairs. For reasons of space, I won't here go into the number theory that shows why the above method works; but I daresay that, with a little playing, you can satisfy yourself that it does indeed work. If 600 had an odd number of divisors, it would mean that 600 is a perfect square, which it is not. Let's try 676, just to illustrate a small point: 676 = 26 X 26 = 2 X 2 X 13 X 13 = 2² X 13². Exponents: 2 and 2; increment exponents, by 1: 3 and 3; number of divisors of 676: 9; number of pairs of factors: 5. Why? It is because 26 appears twice in one pair. Number of divisors less than 26: 4. They are: 1, 2, 4, and 13. The five pairs: 1 X 676; 2 X 338; 4 X 169; 13 X 52; and 26 X 26. Thus, returning to the original problem of 600: It is helpful also to know that, in each pair of factors of 600, the smaller of them must be 24 or less, and the larger of them must be 25 or greater. That is because the square-root of 600 lies between 24 and 25; put otherwise, 24² = 576, and 25² = 625. Clearly, a pair of factors, both of which were at least as large as 25 would have a product of 625 or larger; both of them less than 25 would give a product of 576 or less. Then, there must be exactly twelve numbers, less than 25, that are divisors of 600; and exactly twelve that are 25 or greater. What's more, no divisor of 600 can be divisible by any number that is not, itself, a divisor of 600: thus, you can exclude the primes smaller than 25, but larger than 5, and all of their multiples. Those primes are 7, 11, 13, 17, 19, and 23. The relevant multiples are 14, 21, and 22. That gives nine of the twelve numbers to be excluded from the first 24 integers; the other three are 9, 16, and 18. Nine and its multiples are excluded, because 9 contains two threes, which is too many: you are allowed only 1, because 3 appears among the prime factors of 600, only once. 16 is similarly excluded, because it contains a factor of 2, four times, which is too many. Thus we easily found the twelve integers less than 24 that are to be excluded from the set of divisors we seek, without any tedious or fruitless division. Because 600 is the sort of number that has relatively few small prime divisors, repeated several times in a prime factorisation, divisors of 600 are relatively dense in their occurance among the integers that are less than the square-root of 600; hence, the approach of excluding divisors becomes economical. With 676, on the other hand, with only nine divisors, the exclusion method might be more trouble than it is worth. This information makes the first 12 divisors of 600, easy to find, with the assurance of missing none; the larger 12 divisors are easily found from the first 12, by easy division.

Q: What are the factor pairs for 6 hundred?

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