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# How many distinct factors does 14! have?

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Joined: 02 Sep 2009
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How many distinct factors does 14! have?  [#permalink]

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08 Dec 2019, 23:58
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55% (hard)

Question Stats:

61% (02:07) correct 39% (02:38) wrong based on 36 sessions

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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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Updated on: 09 Dec 2019, 00:50
1
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)

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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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09 Dec 2019, 00:48
1
$$14! = 14*13*12*11*10*9*8*7*6*5*4*3*2$$

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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09 Dec 2019, 00:59
1
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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09 Dec 2019, 01:12
1
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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09 Dec 2019, 01:18
1
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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09 Dec 2019, 02:33
1
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

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

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09 Dec 2019, 04:40
1
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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09 Dec 2019, 10:59
1
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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09 Dec 2019, 16:18
1
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

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

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09 Dec 2019, 18:28
1
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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