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The problem I'm trying to solve is:

Let a sequence contain the members:
1040, 1050, 1060,..., 10(n+103),..., 9990 for all integral n such that 1<(or equal to) n <(or equal to) 896. For how many distinct members of the sequence will the members be divisible by 33?

Is there a formula for this problem? I cant't think of a way to solve this without going through all the numbers in the sequence and seeing if they are divsable by 33. Thanks.

Sincerely,
Isaac

Answer
Questioner:Isaac
Country:Illinois, United States
Category:Advanced Math
Private:Yes  <<<<<<<<<<<<<<<<, changed; no private questions.
Subject:Number sequence problem
Question:

Let a sequence contain the members:
1040, 1050, ..., 10(n+103),..., 9990 for all 1 <= n <= 896. For how many distinct members of the sequence will the members be divisible by 33?

Is there a formula for this problem? I cant't think of a way to solve this without going through all the numbers in the sequence and seeing if they are divisible by 33. Thanks.

Sincerely,
Isaac

.........................................

If S is your sequence, each member of S is divisible by 10.

10 and 33 are relatively prime.

So any number divisible by both 10 and 33 is divisible by their LCM, which is 330.

Your last is 9990.  9990 / 330 = 30 (plus)

Your first is 1040.  1040 / 330 = 3 (plus)

So you have elements in S from 330(4) to 330(30).  That is 27 elements.

How does that look?

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