Find the number of 6 -term strictly increasing geometric progressions, such that all terms are positive integers less than 1000.
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To find the number of 6-term strictly increasing geometric progressions with positive integer terms less than 1000, we need to consider the possible values for the first term (a) and the common ratio (r).
For the first term (a), it could be any positive integer less than 1000.
For the common ratio (r), it must also be a positive integer. However, since the sequence needs to be strictly increasing, the common ratio cannot be 1 (as it would result in repeated terms) and it cannot be greater than 1000/a (as it would result in a term greater than 1000).
So, for each possible value of a, we need to count the number of valid common ratios (r) that satisfy these conditions.
First, let's consider the case when a = 1. In this case, the common ratio (r) can range from 2 to 499, as any greater value would result in a term greater than 1000. So, there are 498 valid values of r.
Next, let's consider the case when a = 2. In this case, the common ratio (r) can range from 2 to 333. For higher values of r, the sequence would exceed 1000. So, there are 332 valid values of r.
Similarly, we can find the number of valid values of r for each possible value of a from 3 to 999.
To summarize, we need to sum up the number of valid values of r for each possible value of a:
For a = 1, there are 498 valid values of r.
For a = 2, there are 332 valid values of r.
For a = 3, there are 249 valid values of r.
...
For a = 998, there is only 1 valid value of r.
For a = 999, there are 0 valid values of r.
To get the total number of 6-term geometric progressions, we sum up the number of valid values for each a:
498 + 332 + 249 + ... + 1 + 0 = 38319
Therefore, there are 38319 possible 6-term strictly increasing geometric progressions with positive integer terms less than 1000.