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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

any easy way to solve this?

Dear santorasantu, I'm happy to respond.

My friend, I am going to chide you on your question, "any easy way to solve this?" With all due respect, that is a very poor question. A good question would tell me: what do you consider easy? What do you understand about this problem? What do you not understand? You see, if you want to achieve excellent results, you must make a habit of excellence in your studies, and part of that is asking excellent questions. See: http://magoosh.com/gmat/2014/asking-exc ... questions/

I looked at this and thought: OK, if n > 3, that eliminates the cheap cases. When could the number of multiples of 2 and numbers of multiples of 3 be equal?

Because the multiples of three are more spaced out, it is going to be harder for them to "keep up" with the more densely packed multiples of 2. That's why I started all my example sets at 3, so that the multiples of 3 would get a "head start."

Example #1 = {3, 4, 5, 6} = 2 of each kind of multiple (n = 4)

Example #2 = {3, 4, 5, 6, 7} = 2 of each kind of multiple (n = 5)

Example #3 = {3, 4, 5, 6, 7, 8, 9} = 3 of each kind of multiple (n = 7)

All of these are consistent with Statement #1, which is insufficient. Statement #2 by itself is useless and insufficient.

Together, we can eliminate Example #1, but the other two examples both work, so n could be 5 or 7.

We can't give a definitive answer to the prompt, so everything is insufficient. OA = (E)

You see, my friend, if you are comfortable with number sense and picking numbers, this is a very easy solution, but expect to do everything with a formula, then you will be sadly disappointed by this and so many other GMAT math questions.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: Set S consists of n consecutive positive integers, each less than 25. [#permalink]

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14 Sep 2015, 04:09

mikemcgarry wrote:

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

any easy way to solve this?

Dear santorasantu, I'm happy to respond.

My friend, I am going to chide you on your question, "any easy way to solve this?" With all due respect, that is a very poor question. A good question would tell me: what do you consider easy? What do you understand about this problem? What do you not understand? You see, if you want to achieve excellent results, you must make a habit of excellence in your studies, and part of that is asking excellent questions. See: http://magoosh.com/gmat/2014/asking-exc ... questions/

I looked at this and thought: OK, if n > 3, that eliminates the cheap cases. When could the number of multiples of 2 and numbers of multiples of 3 be equal?

Because the multiples of three are more spaced out, it is going to be harder for them to "keep up" with the more densely packed multiples of 2. That's why I started all my example sets at 3, so that the multiples of 3 would get a "head start."

Example #1 = {3, 4, 5, 6} = 2 of each kind of multiple (n = 4)

Example #2 = {3, 4, 5, 6, 7} = 2 of each kind of multiple (n = 5)

Example #3 = {3, 4, 5, 6, 7, 8, 9} = 3 of each kind of multiple (n = 7)

All of these are consistent with Statement #1, which is insufficient. Statement #2 by itself is useless and insufficient.

Together, we can eliminate Example #1, but the other two examples both work, so n could be 5 or 7.

We can't give a definitive answer to the prompt, so everything is insufficient. OA = (E)

You see, my friend, if you are comfortable with number sense and picking numbers, this is a very easy solution, but expect to do everything with a formula, then you will be sadly disappointed by this and so many other GMAT math questions.

Does all this make sense? Mike

Thanks for the reply. Indeed I followed the similar approach to count the number of factors and ended up with E, but it took me almost 5 mins to arrive at the answer.

I agree with you that it is a very poor question. The source of the question is Manhattan and do you think that it could be a GMAT question?

Thanks for the reply. Indeed I followed the similar approach to count the number of factors and ended up with E, but it took me almost 5 mins to arrive at the answer.

I agree with you that it is a very poor question. The source of the question is Manhattan and do you think that it could be a GMAT question?

Dear santorasantu, I'm happy to respond. I think this MGMAT question is a brilliant question, and it is very GMAT-like. It is unconventional and out-of-the-box, designed to frustrate formulaic thinkers, and yet it can be solved with tremendous elegance very quickly. That is precisely the "sweet spot" of GMAT questions, and MGMAT has written an excellent one here.

I mean no offense, my friend, but what I called a "poor question" was your question, "any easy way to solve this?" You are obviously an intelligent and talented individual, and I would say that this question of yours falls far well short of your highest capabilities. You see, if you want to achieve excellence on the exam, you have to strive to make excellence a habit in everything you do, even in the way you ask questions. Every problem you solve, every question you ask, is another opportunity to bring the absolute best of yourself forward, and you have to practice this continually so that you "own" it by the time you sit for the test. Don't focus on the "easy way" in anything. Focus on deep understanding. Focus on pushing yourself to your edge. I am chiding you precisely because I believe in you and want to see you reach your highest potential.

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

We need to know the first term and the overall number of terms in the original condition. Therefore we have 2 variables and we need 2 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), there is high probability that C is the answer. (Using both 1) and 2) together from the start saves us time)

Using both 1) & 2) together, {3, 4, 5, 6, 7} --> n = 5, {3, 4, 5, 6, 7, 8, 9} --> n=7 therefore the answer is not unique. Thus the answer is E.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

Re: Set S consists of n consecutive positive integers, each less than 25. [#permalink]

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06 Oct 2016, 08:05

2

This post received KUDOS

santorasantu wrote:

mikemcgarry wrote:

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

any easy way to solve this?

Dear santorasantu, I'm happy to respond.

My friend, I am going to chide you on your question, "any easy way to solve this?" With all due respect, that is a very poor question. A good question would tell me: what do you consider easy? What do you understand about this problem? What do you not understand? You see, if you want to achieve excellent results, you must make a habit of excellence in your studies, and part of that is asking excellent questions. See: http://magoosh.com/gmat/2014/asking-exc ... questions/

I looked at this and thought: OK, if n > 3, that eliminates the cheap cases. When could the number of multiples of 2 and numbers of multiples of 3 be equal?

Because the multiples of three are more spaced out, it is going to be harder for them to "keep up" with the more densely packed multiples of 2. That's why I started all my example sets at 3, so that the multiples of 3 would get a "head start."

Example #1 = {3, 4, 5, 6} = 2 of each kind of multiple (n = 4)

Example #2 = {3, 4, 5, 6, 7} = 2 of each kind of multiple (n = 5)

Example #3 = {3, 4, 5, 6, 7, 8, 9} = 3 of each kind of multiple (n = 7)

All of these are consistent with Statement #1, which is insufficient. Statement #2 by itself is useless and insufficient.

Together, we can eliminate Example #1, but the other two examples both work, so n could be 5 or 7.

We can't give a definitive answer to the prompt, so everything is insufficient. OA = (E)

You see, my friend, if you are comfortable with number sense and picking numbers, this is a very easy solution, but expect to do everything with a formula, then you will be sadly disappointed by this and so many other GMAT math questions.

Does all this make sense? Mike

Thanks for the reply. Indeed I followed the similar approach to count the number of factors and ended up with E, but it took me almost 5 mins to arrive at the answer.

I agree with you that it is a very poor question. The source of the question is Manhattan and do you think that it could be a GMAT question?

Sorry to disagree; statement 1 mentions factors of 2 and factors of 3, not multiples.

The only factors of 2 are 1 and 2, and the only factors of 3 are 1 and 3. The only think that we can infer is that the first term of the sequence is not 3.
_________________

Re: Set S consists of n consecutive positive integers, each less than 25. [#permalink]

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06 Oct 2016, 08:13

MathRevolution wrote:

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

We need to know the first term and the overall number of terms in the original condition. Therefore we have 2 variables and we need 2 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), there is high probability that C is the answer. (Using both 1) and 2) together from the start saves us time)

Using both 1) & 2) together, {3, 4, 5, 6, 7} --> n = 5, {3, 4, 5, 6, 7, 8, 9} --> n=7 therefore the answer is not unique. Thus the answer is E.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

I thing that statement 1 has been misinterpreted.

Statement 1 mentions factors of 2 and factors of 3, not multiples.

The only factors of 2 are 1 and 2, and the only factors of 3 are 1 and 3. The only think that we can infer is that the first term of the sequence is not 3.
_________________

Sorry to disagree; statement 1 mentions factors of 2 and factors of 3, not multiples.

The only factors of 2 are 1 and 2, and the only factors of 3 are 1 and 3. The only think that we can infer is that the first term of the sequence is not 3.

Dear cledgard,

I'm happy to respond.

My friend, I agree with you that I wasn't precise in following the specifications of the question, but I think you are misunderstanding something as well. It is not asking for the factors of 2 and 3, each of which of course has only two numbers, because each is prime.

They are asking about how many times can we locate among all the factors a factor of 2. For example, 6 has one factor of two, 12 has two factors of two, and 8 has three factors of two. In asking for the "number of factors of 2 contained in set S," the question is asking for the total number of factors of 2 in all the numbers in the set.

Suppose Set S = {2, 3, 4, 5, 6} 2 has one factor of 2 3 has one factor of 3 4 has two factors of 2 5 doesn't have a factor of either 6 has one factor of 2 and one factor of 3 Thus, this particular set S has a total of four factors of 2 and two factors of 3. That is what the question means by asking for the "number of factors of 2" or 3 contained in the set. I admit, I was too casual in interpreting it simply as factors above, rather than counting the factors in each set.

Sets that would meet the requirements of both statements: Set S1 = {17, 18, 19, 20, 21} five members, an odd number number of factors of 2 = three factors number of factors of 3 = three factors n = 5

Set S2 = {9, 10, 11, 12, 13, 14, 15} seven members, an odd number number of factors of 2 = four factors number of factors of 3 = four factors n = 7

Two different possibilities consistent with both statements give two different answer to the prompt question. Answer = (E)

Does all this make sense? Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Re: Set S consists of n consecutive positive integers, each less than 25. [#permalink]

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06 Oct 2016, 12:39

mikemcgarry wrote:

cledgard wrote:

Sorry to disagree; statement 1 mentions factors of 2 and factors of 3, not multiples.

The only factors of 2 are 1 and 2, and the only factors of 3 are 1 and 3. The only think that we can infer is that the first term of the sequence is not 3.

Dear cledgard,

I'm happy to respond.

My friend, I agree with you that I wasn't precise in following the specifications of the question, but I think you are misunderstanding something as well. It is not asking for the factors of 2 and 3, each of which of course has only two numbers, because each is prime.

They are asking about how many times can we locate among all the factors a factor of 2. For example, 6 has one factor of two, 12 has two factors of two, and 8 has three factors of two. In asking for the "number of factors of 2 contained in set S," the question is asking for the total number of factors of 2 in all the numbers in the set.

Suppose Set S = {2, 3, 4, 5, 6} 2 has one factor of 2 3 has one factor of 3 4 has two factors of 2 5 doesn't have a factor of either 6 has one factor of 2 and one factor of 3 Thus, this particular set S has a total of four factors of 2 and two factors of 3. That is what the question means by asking for the "number of factors of 2" or 3 contained in the set. I admit, I was too casual in interpreting it simply as factors above, rather than counting the factors in each set.

Sets that would meet the requirements of both statements: Set S1 = {17, 18, 19, 20, 21} five members, an odd number number of factors of 2 = three factors number of factors of 3 = three factors n = 5

Set S2 = {9, 10, 11, 12, 13, 14, 15} seven members, an odd number number of factors of 2 = four factors number of factors of 3 = four factors n = 7

Two different possibilities consistent with both statements give two different answer to the prompt question. Answer = (E)

Does all this make sense? Mike

Dear Mike: Of course what you say makes sense; however, I don't believe I misunderstood the problem. As a matter of fact, I understood the problem exactly as you did. Nevertheless, when I stopped to think about it, I believed that I should follow literally what statement 1 said, not my interpretation. I believe that that statement 1 should be worded differently in order to avoid confusion. How to word it is another question.

Re: Set S consists of n consecutive positive integers, each less than 25. [#permalink]

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15 Mar 2017, 18:05

2

This post was BOOKMARKED

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).
_________________

"Be challenged at EVERY MOMENT."

“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

Re: Set S consists of n consecutive positive integers, each less than 25. [#permalink]

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15 Mar 2017, 21:58

mikemcgarry wrote:

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

any easy way to solve this?

Dear santorasantu, I'm happy to respond.

My friend, I am going to chide you on your question, "any easy way to solve this?" With all due respect, that is a very poor question. A good question would tell me: what do you consider easy? What do you understand about this problem? What do you not understand? You see, if you want to achieve excellent results, you must make a habit of excellence in your studies, and part of that is asking excellent questions. See: http://magoosh.com/gmat/2014/asking-exc ... questions/

I looked at this and thought: OK, if n > 3, that eliminates the cheap cases. When could the number of multiples of 2 and numbers of multiples of 3 be equal?

Because the multiples of three are more spaced out, it is going to be harder for them to "keep up" with the more densely packed multiples of 2. That's why I started all my example sets at 3, so that the multiples of 3 would get a "head start."

Example #1 = {3, 4, 5, 6} = 2 of each kind of multiple (n = 4)

Example #2 = {3, 4, 5, 6, 7} = 2 of each kind of multiple (n = 5)

Example #3 = {3, 4, 5, 6, 7, 8, 9} = 3 of each kind of multiple (n = 7)

All of these are consistent with Statement #1, which is insufficient. Statement #2 by itself is useless and insufficient.

Together, we can eliminate Example #1, but the other two examples both work, so n could be 5 or 7.

We can't give a definitive answer to the prompt, so everything is insufficient. OA = (E)

You see, my friend, if you are comfortable with number sense and picking numbers, this is a very easy solution, but expect to do everything with a formula, then you will be sadly disappointed by this and so many other GMAT math questions.

Does all this make sense? Mike

Hi Mike,

Please correct me if I'm wrong but I'm not sure the factors are equal here: Example #3 = {3, 4, 5, 6, 7, 8, 9} = 3 of each kind of multiple (n = 7)

3: 3 4: 2, 2 5: - 6: 2, 3 7: - 8: 2, 2, 2 9: 3

By my count we don't have an equal number of factors of 2 and 3 here...

Good point, my friend. I was thinking in terms of "multiples," which is one way to think about factors, but you are thinking about factors in another way that is perfectly consistent with the problem. I think the question needs to be tighten a bit: exactly what is meant by the "number of factors of X in a set"?

Let's count factors your way. Consistent with both statement would be the sets Example #4 = {9, 10, 11, 12} = 3 factors of each, n = 4 Example #5 = {9, 10, 11, 12, 13} = 3 factors of each, n = 5 Example #6 = {9, 10, 11, 12, 13, 14, 15} = 4 factors of each, n = 7 Example #7 = {18, 19, 20, 21} = 3 factors of each, n = 4 Example #8 = {17, 18, 19, 20, 21} = 3 factors of each, n = 5

Either way, OA = (E)

Mike
_________________

Mike McGarry Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Good point, my friend. I was thinking in terms of "multiples," which is one way to think about factors, but you are thinking about factors in another way that is perfectly consistent with the problem. I think the question needs to be tighten a bit: exactly what is meant by the "number of factors of X in a set"?

Let's count factors your way. Consistent with both statement would be the sets Example #4 = {9, 10, 11, 12} = 3 factors of each, n = 4 Example #5 = {9, 10, 11, 12, 13} = 3 factors of each, n = 5 Example #6 = {9, 10, 11, 12, 13, 14, 15} = 4 factors of each, n = 7 Example #7 = {18, 19, 20, 21} = 3 factors of each, n = 4 Example #8 = {17, 18, 19, 20, 21} = 3 factors of each, n = 5

Either way, OA = (E)

Mike

Hi Mike! Hope you're doing great. Needless to say, I fell into the trap of B In case you have further such questions, could you post them? Really liked you approach and explanation.