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23 Aug 2022, 08:26
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
54% (02:10) correct
46%
(01:47)
wrong
based on 177
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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
(A) 5
(B) 6
(C) 7
(D) 8
(E) 12
23 Aug 2022, 09:22
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BrentGMATPrepNow
\(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \)
\(k=10!\frac{abcde}{LCM(2,3,4…..9,10)}=10!*\frac{abcde}{2^3*3^2*5*7}=2*3*4*6*10(abcde)\).
Now, 2*3*4*6*10 surely has 5, 6, 8 and 12 as factors.
Only 7 left.
C
23 Aug 2022, 09:30
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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
\(k = 10!(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10}) \)
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
