SOLUTION: 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.

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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