2004 AMC 10B Problems/Problem 21
Contents
Problem
Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ?
Solution 1
The two sets of terms are and .
Now . We can compute . We will now find .
Consider the numbers in . We want to find out how many of them lie in . In other words, we need to find out the number of valid values of for which .
The fact "" can be rewritten as ", and ".
The first condition gives , the second one gives .
Thus the good values of are , and their count is .
Therefore , and thus .
Solution 2
We can start by finding the first non-distinct term from both sequences. We find that that number is . Now, to find every
other non-distinct terms, we can just keep adding . We know that the last terms of both sequences are and
. Clearly, is smaller and that is the last possible common term of both sequences. Now, we can
create the inequality . Using the inequality, we find that there are common terms. There are 4008
terms in total.
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See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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