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How many distinct factors does 14! have?

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How many distinct factors does 14! have?  [#permalink]

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New post 08 Dec 2019, 23:58
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A
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C
D
E

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Re: How many distinct factors does 14! have?  [#permalink]

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New post Updated on: 09 Dec 2019, 00:50
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How many distinct factors does 14! have?

A. 2592
B. 2718
C. 3142
D. 3654
E. 3660

\(14! = 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2\)
\(14! = 2*7 * 13 * 2^2*3 * 11 * 2*5 * 3^2 * 2^3 * 7 * 2*3 * 5 * 2^2 * 3 * 2\)
\(14! = 13 * 11 * 7^2 * 5^2 * 3^5 * 2^{11}\)

Taking power of all the prime numbers, Number of distinct factors = (1+1)*(1+1)*(2+1)*(2+1)*(5+1)*(11+1) = 2592
(A little while after multiplication it can be seen that unit digit of the resultant number is 2, so only options left are A and C)


Answer A.
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Originally posted by lnm87 on 09 Dec 2019, 00:47.
Last edited by lnm87 on 09 Dec 2019, 00:50, edited 2 times in total.
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 00:48
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\(14! = 14*13*12*11*10*9*8*7*6*5*4*3*2\)

To simplify, lets factorize each of these numbers
\(14 = 2*7\)
\(12 = 2^2*3\)
\(10 = 2*5\)
\(9 = 3^2\)
\(8 = 2^3\)
\(6 = 2*3\)
\(4 = 2^2\)

14! when simplified will become \(2^{11}*3^5*5^2*7^2*11*13\)

The distinct factors of any number can be determined using formula \((p+1) (q+1) (r+1)\)

if we can represent the number as a product of its prime factors
\(N = a^p * b^q * c^r\)
Here a,b,c and so on are the prime numbers and p,q, and r are their powers.


Therefore, 14! has \(12*6*3*3*2*2\) = 2592(Option A) distinct factors which will include 1 and the number itself.
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 00:59
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The answer is A.
Explaination:
14!= 2^11 x 3^5 x 5^2 x 7^2 x 11 x 13
Number of factors = (11+1)(5+1)(2+1)(2+1)(1+1)(1+1) = 2592
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 01:12
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14! ; prime factors ; 2^11*3^5*5^2*7^2*11*13
; 12*6*3*3*2*2 = 2592
IMO A

How many distinct factors does 14! have?

A. 2592
B. 2718
C. 3142
D. 3654
E. 3660
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 01:18
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14! = 14*13*12*11*10*9*8*7*6*5*4*3*2
Convert the numbers to prime factorization. In simple terms, how many primes are there in this number.
2 - 11
3 - 5
5 - 2
7 - 2
11 - 1
13 - 1
Add 1 to each of the numbers and multiply to get the distinct factors.
12*6*3*3*2*2
72*9*4
=2592
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 02:33
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14!=2^11. 3^5. 5^2. 7^2. 11^1. 13^1.

Thus, distinct factors of 14! =
(11+1).(5+1).(2+1).(2+1).(1+1).(1+1) =2592

Answer is (A)

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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 04:40
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Quote:
How many distinct factors does 14! have?

A. 2592
B. 2718
C. 3142
D. 3654
E. 3660


\(14!=1,2,3,4,5,6,7,8,9,10,11,12,13,14\)
\(14!=2,3,5,7,11,13,2^2,2*3,2^3,3^2,2*5,2^2*3,7*2\)
\(14!=13,11,7^2,5^2,3^5,2^{11}\)
\(f(14!)=product[powers.of.primes+1]\)
\(f(14!)=(1+1)(1+1)(2+1)(2+1)(5+1)(11+1)\)
\(f(14!)=12*12*18=2592\)

Ans (A)
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 10:59
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The distinct factors in 14!

14!=1*2*3*4*5*6*7*8*9*10*11*12*13*14
= 2*3*2^2*5*2*3*7*2^3*3^2*5*2*11*3*2^2*13*7*2
= 2^11 * 3^5 * 5^2 * 7^2 * 11 * 13
Number of unique factors of 14!=(12*6*3*3*2*2) = 108 * 24 = 2592.

The right answer is option A.
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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 16:18
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How many distinct factors does 14! have?

—> How many prime factors are there up to 14? —2,3,5,7,11,13

—> we’ll find the exponents of prime factors by using the formula:
—>\([\frac{14}{2}] + [\frac{14}{4}]+ [\frac{14}{8}]...= 7+3+1=11\)

—> \([\frac{14}{3}]+ [\frac{14}{9}]...= 4+1= 5\)

—> \([\frac{14}{5}]...= 2\)

—> \([\frac{14}{7}]...= 2\)

—> \([\frac{14}{11}]...= 1\)

—> \([\frac{14}{13}]...= 1\)

—> (11+1)(5+1)(2+1)(2+1)(1+1)(1+1)= 12*6*3*3*2*2= 12*6*36= 12*216 = (10+2)*216= 2160 +432= 2592

The answer is A

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Re: How many distinct factors does 14! have?  [#permalink]

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New post 09 Dec 2019, 18:28
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Writing down 14! In the form of powers of prime factors....2^11,3^5,5^(2),7^2,11,13

Total factors=(11+1)(5+1)(2+1)(1+1)(1+1)=2592

OA:A

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Re: How many distinct factors does 14! have?   [#permalink] 09 Dec 2019, 18:28
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