Question 1197927: The numbers 1998, 2997, 3996, 4995, ..., 8991 all have two distinct primes in common in their prime factorization. Find the sum of these two primes.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Those numbers are all multiples of 999, which is 27*37 = 3*3*3*37.
The two distinct primes in the factorization of each of the numbers are 3 and 37.
ANSWER: 3+37 = 40
Answer by ikleyn(52795)  (Show Source):
You can put this solution on YOUR website! .
In order to facilitate your finding of the two distinct primes, notice that the given numbers
represent an arithmetic progression with the common difference of
2997 - 1998 = 999 = 3*333 = 3*3*111 = 3*3*3*37.
Next, since the number 1998 is divisible by 3 and by 37 (check it directly !),
it implies that the given numbers ALL have 3 and 37 as their prime common divisors,
and DO NOT HAVE ANY OTHER prime common divisors.
Hence, the problem wants you find the sum of 3 and 37, which is 3+37 = 40.
ANSWER. The sum of these two primes is 40.
Solved.
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