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Math Help - Prime numbers.

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    Prime numbers.

    The year 2003 is prime, but its reversal, 3002, is not. In fact, 3002 is the product of exactly three different primes. Let N be the sum of these three primes. How many other positive integers are the products of exactly three different primes with this sum N ?
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    Quote Originally Posted by rai2003 View Post
    The year 2003 is prime, but its reversal, 3002, is not. In fact, 3002 is the product of exactly three different primes. Let N be the sum of these three primes. How many other positive integers are the products of exactly three different primes with this sum N ?
    One of the three prime factors of 3002 is obviously 2. So divide 3002 by 2, getting 1501, which must be a product of two primes. If you can't factorise that number (because both its prime factors are fairly large), you could feed it into this helpful calculator, which will do the work for you.

    You should find that the sum of the three prime factors is 100. So how many ways can 100 be the sum of three primes? Answer: not many, because the sum of three odd numbers is odd, so one of the three primes will always have to be the only even prime, namely 2. To complete the question, you just have to work out how many ways there are to write 98 as a sum of two (odd) primes.
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    thanks
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