santorasantu wrote:
Set S consists of n consecutive positive integers, each less than 25. If n > 3, what is the value of n?
(1) The number of factors of 2 contained in set S is equal to the number of factors of 3 contained in set S.
(2) n is odd
OFFICIAL SOLUTION
Did you pick (B) or (D)? If so, you fell into a trap. Here’s a hint regarding how to avoid it: look only at the second sentence of the question stem and statement (2). Ignore the first sentence and statement (1). Think that through before you keep reading.
According to the question stem, n > 3, so n is at least 4. What does n represent? It is the number of terms in set S. For example, if n = 4, then set S consists of 4 consecutive positive integers (each of which is less than 25). The set could be 6, 7, 8, 9, for example.
(2) INSUFFICIENT: Statement (2) is definitely easier, so start here. If n is odd, it could be 5, 7, or multiple other larger odd numbers (up to 23). This is not sufficient to determine the value of n.
(1) INSUFFICIENT: Test some cases to figure out whether there is a pattern.
In any list of consecutive integers, every other integer is a multiple of 2. Every fourth number is a multiple of \(2^2\). Every eighth is a multiple of \(2^3\) and so on.
A similar pattern holds for 3. In any list of consecutive integers, every third number is a multiple of 3, every ninth is a multiple of 32 and so on.
What does this mean for the problem? If n = 4, then there are two even numbers in the set. For example, the set could be 1, 2, 3, 4. In this case, the number 2 contains one factor of 2 and the number 4 contains two factors of 2, for a total of three factors of 2. In addition, the number 3 contains one factor of 3. This doesn’t fit statement (1), though, because there are more factors of 2 than factors of 3. You need to bring more factors of 3 into the mix.
Try bringing 9 into the mix: now you have two factors of 3 instead of just one. But you need one more! If the first number in the set is a multiple of 3, then the fourth number in the set will also be a multiple of 3. Try the set 9, 10, 11, 12.
9: two factors of 3
10: one factor of 2
11: nothing
12: two factors of 2 and one factor of 3
Total: three factors of 2 and three factors of 3 (bingo!)
Okay. Now could this work with a different number for n? Try n = 5. What if the set is 9, 10, 11, 12, 13?
13 doesn’t bring in any new factors of either 2 or 3, so you still have three factors of 2 and three factors of 3. There are at least two possible sets that work, so statement (1) is insufficient.
(1) AND (2) INSUFFICIENT: How do things change if n must be odd? Discard the n = 4 case, of course.
You’ve already proven n = 5. What about n = 7?
If n = 7, then you’ll have at least three consecutive even numbers, for a minimum of four factors of 2, and a minimum of two consecutive multiples of 3, for a minimum of two factors of 3. What numbers can you choose that will add in two more factors of 3 without adding more factors of 2?
First, to avoid adding extra factors of 2, don’t include any multiples of 8 in the list.
Next, make the first, fourth, and seventh numbers in the set multiples of 3. Start with 9 again (right after 8 to avoid 8!): 9, 10, 11, 12, 13, 14, 15.
9: two factors of 3
10: one factor of 2
12: two factors of 2 and one factor of 3
14: one factor of 2
15: one factor of 3
There are four factors of 2 and four factors of 3, so n = 7 is a possible solution. Since n could also be 5, the two statements together are not sufficient.
The correct answer is (E).
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