The first 25 positive integers are 1 through 25. A number with exactly four positive integer divisors must be the product of two (equal or unequal primes). If the primes are different, then the product's divisors will be the two primes, 1, and the product itself. If the primes are equal, then the product's divisors will be the prime, the prime squared, 1, and the product itself. The first five primes are 2, 3, 5, 7, and 11. The next prime, 13, and all higher primes are irrelevent, because 13*2 = 26, so all of its multiples exceed the parameters.
So, how many number can you form by multiplying any two of the first five primes but without exceeding 25? We take the numbers in pairs starting with 2 and going up until we exceed 25, then start over from 3 and multiply only by numbers equal to or larger than it. That gives us 2*2 = 4, 2*3 = 6, 2*5 = 10, 2*7 = 14, 2*11 = 22, 3*3 = 9, 3*5 = 15, 3*7 = 21, (not 3*11 = 33 or anything greater), 5*5 = 25 (not 5*7 = 35 or anything greater, and not 7*7 = 49 or anything greater). So we have 4, 6, 10, 14, 22, 15, 21, and 25. We can put them in order if we want, so that's 4, 6, 10, 14, 15, 21, and 22, a total of 7 numbers. 7/25 = 28%. 28% of the first 25 positive integers are cute.
2006-10-02 09:38:26
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answer #1
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answered by DavidK93 7
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There are 7 cute numbers. Those are; 6, 8, 10, 14, 15, 21, 22
That makes 28% of the first 25 integers
2006-10-02 16:43:32
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answer #2
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answered by daniel_cohadier 3
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If you include the number itself and 1, then it's 7/25 = 28% (6,8,10,14,15,21,22). Ifyou don't, then it's 3/25 = 12% (12,18,20)
2006-10-02 16:46:46
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answer #3
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answered by Anonymous
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your avatar must have exactly four positive integer divisors, it's cute.
2006-10-02 16:43:14
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answer #4
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answered by echiasso 3
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