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If k = 10!(1 + 1/2 + 1/3 + 1/4 + 1/5....

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BrentGMATPrepNow
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If k = 10!(1 + 1/2 + 1/3 + 1/4 + 1/5.... [#permalink] New post   23 Aug 2022, 08:26
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If \(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \), then \(k\) is divisible by all of the following numbers EXCEPT:

(A) 5
(B) 6
(C) 7
(D) 8
(E) 12
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C
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Re: If k = 10!(1 + 1/2 + 1/3 + 1/4 + 1/5.... [#permalink] New post   23 Aug 2022, 09:22
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BrentGMATPrepNow wrote:
If \(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \), then \(k\) is divisible by all of the following numbers EXCEPT:

(A) 5
(B) 6
(C) 7
(D) 8
(E) 12



\(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \)
\(k=10!\frac{abcde}{LCM(2,3,4…..9,10)}=10!*\frac{abcde}{2^3*3^2*5*7}=2*3*4*6*10(abcde)\).
Now, 2*3*4*6*10 surely has 5, 6, 8 and 12 as factors.
Only 7 left.


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Re: If k = 10!(1 + 1/2 + 1/3 + 1/4 + 1/5.... [#permalink] New post   23 Aug 2022, 09:30
given
\(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \)


LCM of 1 to 10 value is 2520
sum of the number in fraction will be even but with addition of 1 the Nr value will become odd
10! * ( odd value) /2520
or say 1440 * (odd value)
2^5*3^2*5 * ( odd value)
of all given values only 7 will not be factor of k
OPTION C


BrentGMATPrepNow wrote:
If \(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \), then \(k\) is divisible by all of the following numbers EXCEPT:

(A) 5
(B) 6
(C) 7
(D) 8
(E) 12
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Re: If k = 10!(1 + 1/2 + 1/3 + 1/4 + 1/5.... [#permalink] New post   17 Jun 2023, 22:00
1st) find the prime factorization of 10!

only the primes (2 , 3, 5, 7) will be part of it

Prime 2:
10/2 = 5
——> 5/2 = 2
—————> 2/2 = 1

(2)^8

Prime 3:
10/3 = 3
——-> 3/3 = 1

(3)^4

Prime 5:
10/5 = 2

(5)^2

Prime 7:
10/7 = 1

(7)^1

So
10! = (2)^8 * (3)^4 * (5)^2 * (7)^1

When we multiply 10! through the parenthesis, the effect will be the removal of the prime factors that make up the denominator

Each fraction will not completely remove any of the prime factors EXCEPT (1/7), since there is only ONE Prime 7 that makes up 10!’s prime factorization

This, the other 9 terms will all contain a prime 7 except:

(10!) * (1/7) = NON Multiple of 7

When we add the other 9 terms (a multiple of 7) to the term (10!) * (1/7) (which is a NON multiple of 7 ——-> we get

Multiple of 7 + NON Multiple of 7 = NON multiple of 7

So the result can not be divisible by 7

(D) 7

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Re: If k = 10!(1 + 1/2 + 1/3 + 1/4 + 1/5.... [#permalink]
17 Jun 2023, 22:00
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