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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 15 Aug 2023, 06:00
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For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?

A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)
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D
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 15 Aug 2023, 06:38
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Bunuel
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?


A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)
Show more

Two approaches to solve this question -

1) Using Values

Let \(n = 4\)

\(n^* = 2 * 4 * 6 * 8\)

\(n! = 1 * 2 * 3 * 4\)

\(n^*÷n!\) = \(\frac{2 * 4 * 6 * 8 }{ 1 * 2 * 3 * 4} = 2 * 8 = 16\)

Let's check which option yields us a value of \(16\) for \(n = 4\)

A. (n - 1)! → 3! = 6
B. n! → 4! = 24
C. (n + 1)! → 5! = 120
D. 2^n → \(2^4 = 16\)
E. 2^(n + 1) 2^5 = 32

2) Algebraic

\(n^*÷n!\)

\(\frac{2 * 4 * 6 * 8 *...*2n}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{(2 * 1) * (2 * 2) * (2 * 3) * (2 * 4) *...* (2 * n)}{ 1 * 2 * 3 * 4 ... * n}\)

We see that the numerator has two multiplied n times, hence we can write the expression as \(2^n\)

\(\frac{2^n (1 * 2 * 3 * 4 ... n)}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{2^n (n!)}{ n!} = 2^n\)

Option D
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 28 Dec 2023, 00:38
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JeffTargetTestPrep gmatophobia

When facing this problem in the practice exam, I thought "n" could be any positive integer, not necessarily n > 3. However, that must be mistaken given how you two solve it algebraically. What threw me off is the fact that they attribute their own definition to "n*" which made me think the same was being done to "n!" instead of thinking it was a 'normal' factorial.

Do you guys perhaps have some tips about how I can avoid making such a mistake henceforth? I.e. how do I know whether they actually mean factorial or if they are attributing their own definition to a given term?

I'm not sure I understand your question. Can you rephrase it?
Also, we need not use values of n that are greater than.
The question only tells us that n is a positive integer. So, it can be 1, 2, 3, 4, 5, etc
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?

A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)


In as simple terms as possible, I do not understand what "n" represents in this problem. To the best of my knowledge, it is supposed to represent the number of terms in the numerator or denominator respectively. I.e. since there are at least three terms [(2)(4)(6) in the numerator or (1)(2)(3) in the denominator] prior to the term containing n, then n ≥ 4. That is why I thought n must be a positive integer greater than 3.

Solving the fraction algebraically gives 16 but if n is 1 in the given answer choice (D), then the answer would be 2. That's what confuses me.
Show more

The question defines a function, denoted by *, for each positive integer n as \(n^* = (2)(4)(6)...(2n) = 2^n * n!\). For example:

  • If \(n=1\), then \(n^* = 2^n*n!=2\);
  • If \(n=2\), then \(n^* = 2^n*n!=8\);
  • If \(n=3\), then \(n^* = 2^n*n!=48\);
    and so on.

So, if \(n = 1\), then \(n^* ÷ n! = 2:1 = 2\). Note that in the case of n = 1, both options C and D yield 2. However, the question seeks the option that is ALWAYS equivalent to \(n^* ÷ n!\), not just for specific values.

For the plug-in method, it might happen, as in the above case, that for some specific number(s), more than one option may appear to be the "correct" answer. In such cases, select some other numbers and recheck these "correct" options only. For instance, if you check for \(n = 2\), then \(n^* ÷ n! = 8:2 = 4\), and among options C and D, only option D is correct, making it the right answer.

Hope this helps.
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 15 Aug 2023, 08:51
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For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?

A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)

Best is to take some values, least the better.
Since any value taken would prove the to be true for only one of the choice, only one value is enough.

Let n = 3
3! = 1*2*3 = 6
and
\(3^* = (2)(4)(6) = 48\)
Now
\(3^*÷3!\) = 48/6 = 8

Only option D i.e. 2^3 = 8 satisfies.

Answer D.
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 20 Aug 2023, 17:14
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Bunuel
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?

A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)
Show more

n* = (2)(4)(6)…(2n) = (2×1)(2×2)(2×3)…(2×n) = (2^n)(n!)

n*/n! = (2^n)(n!)/n! = 2^n

Answer: D
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 27 Dec 2023, 08:11
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JeffTargetTestPrep gmatophobia

When facing this problem in the practice exam, I thought "n" could be any positive integer, not necessarily n > 3. However, that must be mistaken given how you two solve it algebraically. What threw me off is the fact that they attribute their own definition to "n*" which made me think the same was being done to "n!" instead of thinking it was a 'normal' factorial.

Do you guys perhaps have some tips about how I can avoid making such a mistake henceforth? I.e. how do I know whether they actually mean factorial or if they are attributing their own definition to a given term?
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 27 Dec 2023, 09:40
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JeffTargetTestPrep gmatophobia

When facing this problem in the practice exam, I thought "n" could be any positive integer, not necessarily n > 3. However, that must be mistaken given how you two solve it algebraically. What threw me off is the fact that they attribute their own definition to "n*" which made me think the same was being done to "n!" instead of thinking it was a 'normal' factorial.

Do you guys perhaps have some tips about how I can avoid making such a mistake henceforth? I.e. how do I know whether they actually mean factorial or if they are attributing their own definition to a given term?
Show more

I'm not sure I understand your question. Can you rephrase it?
Also, we need not use values of n that are greater than.
The question only tells us that n is a positive integer. So, it can be 1, 2, 3, 4, 5, etc
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 27 Dec 2023, 16:53
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JeffTargetTestPrep gmatophobia

When facing this problem in the practice exam, I thought "n" could be any positive integer, not necessarily n > 3. However, that must be mistaken given how you two solve it algebraically. What threw me off is the fact that they attribute their own definition to "n*" which made me think the same was being done to "n!" instead of thinking it was a 'normal' factorial.

Do you guys perhaps have some tips about how I can avoid making such a mistake henceforth? I.e. how do I know whether they actually mean factorial or if they are attributing their own definition to a given term?

I'm not sure I understand your question. Can you rephrase it?
Also, we need not use values of n that are greater than.
The question only tells us that n is a positive integer. So, it can be 1, 2, 3, 4, 5, etc
Show more

In as simple terms as possible, I do not understand what "n" represents in this problem. To the best of my knowledge, it is supposed to represent the number of terms in the numerator or denominator respectively. I.e. since there are at least three terms [(2)(4)(6) in the numerator or (1)(2)(3) in the denominator] prior to the term containing n, then n ≥ 4. That is why I thought n must be a positive integer greater than 3.

Solving the fraction algebraically gives 16 but if n is 1 in the given answer choice (D), then the answer would be 2. That's what confuses me.
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 28 Dec 2023, 08:30
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JeffTargetTestPrep gmatophobia

I'm not sure I understand your question. Can you rephrase it?
Also, we need not use values of n that are greater than.
The question only tells us that n is a positive integer. So, it can be 1, 2, 3, 4, 5, etc
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?

A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)


In as simple terms as possible, I do not understand what "n" represents in this problem. To the best of my knowledge, it is supposed to represent the number of terms in the numerator or denominator respectively. I.e. since there are at least three terms [(2)(4)(6) in the numerator or (1)(2)(3) in the denominator] prior to the term containing n, then n ≥ 4. That is why I thought n must be a positive integer greater than 3.

Solving the fraction algebraically gives 16 but if n is 1 in the given answer choice (D), then the answer would be 2. That's what confuses me.

The question defines a function, denoted by *, for each positive integer n as \(n^* = (2)(4)(6)...(2n) = 2^n * n!\). For example:

  • If \(n=1\), then \(n^* = 2^n*n!=2\);
  • If \(n=2\), then \(n^* = 2^n*n!=8\);
  • If \(n=3\), then \(n^* = 2^n*n!=48\);
    and so on.

So, if \(n = 1\), then \(n^* ÷ n! = 2:1 = 2\). Note that in the case of n = 1, both options C and D yield 2. However, the question seeks the option that is ALWAYS equivalent to \(n^* ÷ n!\), not just for specific values.

For the plug-in method, it might happen, as in the above case, that for some specific number(s), more than one option may appear to be the "correct" answer. In such cases, select some other numbers and recheck these "correct" options only. For instance, if you check for \(n = 2\), then \(n^* ÷ n! = 8:2 = 4\), and among options C and D, only option D is correct, making it the right answer.

Hope this helps.
Show more

Thank you, Bunuel, that clarified things. I should take prompts at face value and solve accordingly.
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 01 Jan 2024, 13:40
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Bunuel
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?


A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)

Two approaches to solve this question -

1) Using Values

Let \(n = 4\)

\(n^* = 2 * 4 * 6 * 8\)

\(n! = 1 * 2 * 3 * 4\)

\(n^*÷n!\) = \(\frac{2 * 4 * 6 * 8 }{ 1 * 2 * 3 * 4} = 2 * 8 = 16\)

Let's check which option yields us a value of \(16\) for \(n = 4\)

A. (n - 1)! → 3! = 6
B. n! → 4! = 24
C. (n + 1)! → 5! = 120
D. 2^n → \(2^4 = 16\)
E. 2^(n + 1) 2^5 = 32

2) Algebraic

\(n^*÷n!\)

\(\frac{2 * 4 * 6 * 8 *...*2n}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{(2 * 1) * (2 * 2) * (2 * 3) * (2 * 4) *...* (2 * n)}{ 1 * 2 * 3 * 4 ... * n}\)

We see that the numerator has two multiplied n times, hence we can write the expression as \(2^n\)

\(\frac{2^n (1 * 2 * 3 * 4 ... n)}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{2^n (n!)}{ n!} = 2^n\)

Option D
Show more

Great explanation gmatophobia
To clarify in below that 2^n will be correct up to 2^2 as if 2^3 = 8. What happened with (2 * 3)=6 in here then? Therefore wondering why is it not 2n here instead? Could you kindly help clarify? Thanks

\(\frac{2^n (1 * 2 * 3 * 4 ... n)}{ 1 * 2 * 3 * 4 ... * n}\)
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 14 Apr 2024, 09:01
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gmatophobia

Bunuel
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?


A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)
Two approaches to solve this question -

1) Using Values

Let \(n = 4\)

\(n^* = 2 * 4 * 6 * 8\)

\(n! = 1 * 2 * 3 * 4\)

\(n^*÷n!\) = \(\frac{2 * 4 * 6 * 8 }{ 1 * 2 * 3 * 4} = 2 * 8 = 16\)

Let's check which option yields us a value of \(16\) for \(n = 4\)

A. (n - 1)! → 3! = 6
B. n! → 4! = 24
C. (n + 1)! → 5! = 120
D. 2^n → \(2^4 = 16\)
E. 2^(n + 1) 2^5 = 32

2) Algebraic

\(n^*÷n!\)

\(\frac{2 * 4 * 6 * 8 *...*2n}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{(2 * 1) * (2 * 2) * (2 * 3) * (2 * 4) *...* (2 * n)}{ 1 * 2 * 3 * 4 ... * n}\)

We see that the numerator has two multiplied n times, hence we can write the expression as \(2^n\)

\(\frac{2^n (1 * 2 * 3 * 4 ... n)}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{2^n (n!)}{ n!} = 2^n\)

Option D
Show more
I also solved it with the second way but a bit different, I'm missing something.­ In the nominator I factored out "2" and it became 2*(1*2*3*...*n) which is equal to 2*4*6*...*2n as it's mentioned in the stem. In this way I crossed out the parenthesis in both nominator and denominator and I ended up to bare "2". How 2^n did come up?­ JeffTargetTestPrep GMATCoachBen ScottTargetTestPrep KarishmaB Bunuel
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 17 Apr 2024, 07:34
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Bunuel
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?


A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)
Two approaches to solve this question -

1) Using Values

Let \(n = 4\)

\(n^* = 2 * 4 * 6 * 8\)

\(n! = 1 * 2 * 3 * 4\)

\(n^*÷n!\) = \(\frac{2 * 4 * 6 * 8 }{ 1 * 2 * 3 * 4} = 2 * 8 = 16\)

Let's check which option yields us a value of \(16\) for \(n = 4\)

A. (n - 1)! → 3! = 6
B. n! → 4! = 24
C. (n + 1)! → 5! = 120
D. 2^n → \(2^4 = 16\)
E. 2^(n + 1) 2^5 = 32

2) Algebraic

\(n^*÷n!\)

\(\frac{2 * 4 * 6 * 8 *...*2n}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{(2 * 1) * (2 * 2) * (2 * 3) * (2 * 4) *...* (2 * n)}{ 1 * 2 * 3 * 4 ... * n}\)

We see that the numerator has two multiplied n times, hence we can write the expression as \(2^n\)

\(\frac{2^n (1 * 2 * 3 * 4 ... n)}{ 1 * 2 * 3 * 4 ... * n}\)

\(\frac{2^n (n!)}{ n!} = 2^n\)

Option D
I also solved it with the second way but a bit different, I'm missing something.­ In the nominator I factored out "2" and it became 2*(1*2*3*...*n) which is equal to 2*4*6*...*2n as it's mentioned in the stem. In this way I crossed out the parenthesis in both nominator and denominator and I ended up to bare "2". How 2^n did come up?­ JeffTargetTestPrep GMATCoachBen ScottTargetTestPrep KarishmaB Bunuel
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Each term in the numerator has its own 2; therefore, there are n number of 2s:

\(2 * 4 * 6 * 8 *...*2n =\)

\(= (2 * 1) (2 * 2) (2 * 3)(2 * 4) ... (2 * n) = \)

\(=2^n(1 * 2 * 3 * 4 ... n)\)­
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 12 May 2024, 04:44
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­Good test of divisibility with factorials. Probably best demonstrated by choosing a value for n:

­
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 20 Jul 2024, 05:37
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Hi Bunuel,

I belive the factoring out of 2^n only applies for multiplication right. For factoring out integers in addition/subtraction we simply factor out the common integer one time.
Suppose, I have to subtract =10-6. I can factor 2 out just one time and write 2 (5-3) = 4.
However, when we take factors out in multiplication cases we need to write the factors this way instead -> 10*6 = (2*5.2*3)= 2^2(5.3)=4*15=60?­
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For each positive integer n, let n* = (2)(4)(6)...(2n) and n! = (1)(2 22 Aug 2025, 13:34
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Plug in 5 and get (2*4*6*8*10)/(2*3*4*5)

Simplify down to 2^4*10 / 5

2^5

if n=5 the answer is 2^5 therefore 2^n

**you don't have to pick 5 I chose a random number any number between 3 and 8 would work and not take too long**
Bunuel
For each positive integer n, let \(n^* = (2)(4)(6)...(2n)\) and \(n! = (1)(2)(3)...(n)\). Which of the following is equivalent to \(n^*÷n!\)?

A. (n - 1)!
B. n!
C. (n + 1)!
D. 2^n
E. 2^(n + 1)
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