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6- If f(n) is the product of n consecutive

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EgmatQuantExpert
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6- If f(n) is the product of n consecutive [#permalink] New post   Updated on: 13 Aug 2018, 02:29
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e-GMAT Question:



If \(f(n)\) is the product of \(n\) consecutive integers from \(1\) to \(N\), then what is the number of odd factors of\(f(10)\).
A) 30
B) 60
C) 120
D) 256
E) 512

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Question 11 of The e-GMAT Number Properties Marathon




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A

Originally posted by EgmatQuantExpert on 28 Feb 2018, 03:11.
Last edited by EgmatQuantExpert on 13 Aug 2018, 02:29, edited 3 times in total.
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   28 Feb 2018, 11:44
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Solution:



We are given:
\(F(n)= 1×2×3×4×5×……….×n\)
We need to find the number of odd factors of \(f(10)\).
Thus, we need to find the value of \(f(10).\) Then, we need to write \(f(10)\) in its prime factorized form. After which we can calculate the value of odd factors of\(f(10)\).
Number of Odd factors \(f(10)\)
    \(f(10)= 1×2×3×4×5×6×7×8×9×10\)
    \(f(10)= 1×2×3×(2×2)×5×(2*3)×7×(2*2*2)×(3*3)×(2*5)\)
    \(f(10)= 2^8 * 3^4 * 5^2 *7\)
Odd factors of \(f(10)\)= (power of 3+1)*(power of 5+1)* (power of 7+1)
Odd factors of \(f(10)=5*3*2\)
Odd factors of \(f(10)=30\)
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   08 May 2018, 05:01
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EgmatQuantExpert wrote:

e-GMAT Question:



If \(f(n)\) is the product of \(n\) consecutive integers from \(1\) to \(N\), then what is the number of odd factors of\(f(10)\).
A) 30
B) 60
C) 120
D) 256
E) 512

This is

Question 11 of The e-GMAT Number Properties Marathon




Go to

the next level of the Marathon



The principle this question is testing is Finding the Number of Factors of an Integer

Number of odd factors of 10!

10!=1x2x3x4x5x6x7x8x9x10.
10!=2x3x(\(2^{2}\))x5x(2x3)x7x(\(2^{3}\))x(\(3^{2}\))x(5x2)
10!=\(2^{8}\)\(3^{4}\)\(5^{2}\)7

Take the exponents of the odd factors. Add 1 to each of them and multiply them by each other.
(4+1)(2+1)(1+1)=30

Answer = A
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   28 Feb 2018, 03:46
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EgmatQuantExpert wrote:

Question:



If \(f(n)\) is the product of \(n\) consecutive integers from \(1\) to \(N\), then what is the number of odd factors of\(f(10)\).
A) 30
B) 60
C) 120
D) 256
E) 512


10! = 1 x 2 x 3 x 2^2 x 5 x 2 x 3 x 7 x 2^3 x 3^2 x 2 x 5
10! = 2^8 x 3^4 x 5^2 x 7

factors = 9 x 5 x 3 x 2


odd factors = 5x3x2

(A) 30 imo
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   28 Feb 2018, 11:55
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Solution:



We are given:
\(f(n)= n!\)
Thus we can write:
    \(f(10)= 10!\)
    \(10! = 1×2×3×4×5×6×7×8×9×10\)
    \(10!= 2^8 * 3^4 * 5^2 *7\)
Total factors of \(10!= 9*5*3*2=270\)
Total factors = Even factors + Odd factors
Odd factors= Total factors – Even factors
Thus, if we can find the even factors of \(10!\) then we can find the odd factors of \(10!\).
\(10!= 2*(2^7 * 3^4 * 5^2 *7)\)
We can say that all the factors of \(2^7 * 3^4 * 5^2 *7\) after multiplying by \(2\) becomes a factor of \(10!\).
We also know that any number, which is multiple of \(2\), is an even number.
Thus,
    Total number of factors of\(2^7 * 3^4 * 5^2 *7\) = Even number of factors of \(10!.\)
    Even factors of \(10!\)= Total factors of \(2^7 * 3^4 * 5^2 *7\)
    Even factors of \(10!\)= \(8*5*3*2\)
    Even factors of \(10!\)=\(240\)
Therefore, Odd factors= Total factors – Even factors
Odd factors= \(270-240\)
Odd factors=\(30\)
Answer: Option \(A\)
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   27 Jun 2018, 02:02
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EgmatQuantExpert wrote:

Solution:



We are given:
\(F(n)= 1×2×3×4×5×……….×n\)
We need to find the number of odd factors of \(f(10)\).
Thus, we need to find the value of \(f(10).\) Then, we need to write \(f(10)\) in its prime factorized form. After which we can calculate the value of odd factors of\(f(10)\).
Number of Odd factors \(f(10)\)
    \(f(10)= 1×2×3×4×5×6×7×8×9×10\)
    \(f(10)= 1×2×3×(2×2)×5×(2*3)×7×(2*2*2)×(3*3)×(2*5)\)
    \(f(10)= 2^8 * 3^4 * 5^2 *7\)
Odd factors of \(f(10)\)= (power of 3+1)*(power of 5+1)* (power of 7+1)
Odd factors of \(f(10)=5*3*2\)
Odd factors of \(f(10)=30\)


Hello!

I have a doubt here.. Why is 1 not included in the odd factors ?
Sorry for such a silly question but isn't that supposed to be counted too ?

It would be really helpful if you could provide your insights on the same.

Thanks in advance
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   27 Jun 2018, 02:41
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kabirchaudhry92 wrote:
EgmatQuantExpert wrote:

Solution:



We are given:
\(F(n)= 1×2×3×4×5×……….×n\)
We need to find the number of odd factors of \(f(10)\).
Thus, we need to find the value of \(f(10).\) Then, we need to write \(f(10)\) in its prime factorized form. After which we can calculate the value of odd factors of\(f(10)\).
Number of Odd factors \(f(10)\)
    \(f(10)= 1×2×3×4×5×6×7×8×9×10\)
    \(f(10)= 1×2×3×(2×2)×5×(2*3)×7×(2*2*2)×(3*3)×(2*5)\)
    \(f(10)= 2^8 * 3^4 * 5^2 *7\)
Odd factors of \(f(10)\)= (power of 3+1)*(power of 5+1)* (power of 7+1)
Odd factors of \(f(10)=5*3*2\)
Odd factors of \(f(10)=30\)


Hello!

I have a doubt here.. Why is 1 not included in the odd factors ?
Sorry for such a silly question but isn't that supposed to be counted too ?

It would be really helpful if you could provide your insights on the same.

Thanks in advance


Hi kabirchaudhry92,

This is a good question. Let me explain with some small examples.

Q1. Find the number of factors of 25.

Step 1: Do the prime factorization. 25 = 5*5 = \(5^2\)

Step 2: Number of factors = (2+1) = 3.

Why have we added 1? By adding "1" we ensure that we count 1 as one of the factors.

Factors of 25: \(5^0 = 1, 5^1 = 5,\) and \(5^2 = 25.\) Hence, the answer is 3.

Q2. Find the number of factors of 75.

Step 1: Prime factorization = 3*5*5 = \(3^1 * 5^2\)

Step 2: Number of factors = (1+1) * (2+1) = 6.

Factors of 75:
  • \(5^0 * 3^0 = 1\)
  • \(5^0 * 3^1 = 3\)
  • \(5^1 * 3^0 = 5\)
  • \(5^1 * 3^1 = 15\)
  • \(5^2 * 3^0 = 25\)
  • \(5^2 * 3^1 = 75\)
.

Hence, 1 is already included in the given solution.

I hope this helps.

Thank you.
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   13 May 2018, 08:43
Expert Reply
HI
I feel question would have been more interesting if there is an option 270 in the answer choices.
If someone fail to read odd factors (as I did ;) ), one would mark 270 as answer.
Fortunately, answer choices saved me.

Regards

EgmatQuantExpert wrote:

Solution:



We are given:
\(F(n)= 1×2×3×4×5×……….×n\)
We need to find the number of odd factors of \(f(10)\).
Thus, we need to find the value of \(f(10).\) Then, we need to write \(f(10)\) in its prime factorized form. After which we can calculate the value of odd factors of\(f(10)\).
Number of Odd factors \(f(10)\)
    \(f(10)= 1×2×3×4×5×6×7×8×9×10\)
    \(f(10)= 1×2×3×(2×2)×5×(2*3)×7×(2*2*2)×(3*3)×(2*5)\)
    \(f(10)= 2^8 * 3^4 * 5^2 *7\)
Odd factors of \(f(10)\)= (power of 3+1)*(power of 5+1)* (power of 7+1)
Odd factors of \(f(10)=5*3*2\)
Odd factors of \(f(10)=30\)
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Re: 6- If f(n) is the product of n consecutive [#permalink] New post   22 Jun 2023, 11:27
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