Question 1189399: When the digits in the number 2005 are reversed we obtain the number 5002, and 5002 = a * b * c, such that a, b and c are three distinct primes. How many other positive integers are the products of exactly three distinct primes prime1, prime2 and prime3 such that prime1 + prime2 + prime3 = a+b+c?
Found 3 solutions by greenestamps, Edwin McCravy, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
5002 = 2*41*61 = a*b*c
a+b+c=104
The question asks how many other positive integers are the products of exactly three distinct primes whose sum is also 104.
For the sum of three distinct primes to be 104, one of them has to be 2, so the sum of the other two has to be 102.
So you need to find the number of pairs of primes other than 41 and 61 that also have a sum of 102.
You will learn nothing from this question if we work the whole problem for you, so I leave it to you to finish the problem.
(my answer was 7....)
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
You need a list of all primes less than 100:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Greenestamps has told you that the other two primes
besides 2 must add up to 102, so they must end in
either 1 and 1, 3 and 9, or 5 and 7. Can you find
them and count how many pairs there are of them?
Edwin
Answer by ikleyn(52797)  (Show Source):
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