Does the integer g have a factor f such that 1<f<g ?
1) g > 3!
2) 11!+2 <= g <= 11!+11
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ranjithreddy.k9
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Frankenstein
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Hi,
From(1):
if g=7, there is no f which satisfies the conditions
if g=8, there is f (can be 2,4 )
Not sufficient
From(2):
11!+2 = 2(1*3*4...*11+1), one possible value of f is 2
11!+3 = 3(1*2*4*...*11+1), one possible value of f is 3
.
.
11!+11 = 11(10! + 1), one possible value of f is 11
So for all integer values of g, g has a factor f satisfying the conditions.
Sufficient
Hence, B
From(1):
if g=7, there is no f which satisfies the conditions
if g=8, there is f (can be 2,4 )
Not sufficient
From(2):
11!+2 = 2(1*3*4...*11+1), one possible value of f is 2
11!+3 = 3(1*2*4*...*11+1), one possible value of f is 3
.
.
11!+11 = 11(10! + 1), one possible value of f is 11
So for all integer values of g, g has a factor f satisfying the conditions.
Sufficient
Hence, B
Cheers!
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GMATGuruNY
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If the answer is NO -- if g does not have a factor between 1 and itself -- then g is prime.ranjithreddy.k9 wrote:Does the integer g have a factor f such that 1<f<g ?
1) g > 3!
2) 11!+2 <= g <= 11!+11
Question rephrased: Is g prime?
Statement 1:
3! = 6.
If g=7, then g is prime.
If g=8, then g is not prime.
Insufficient.
Statement 2: 11!+2 <= g <= 11!+11
2 is a factor of 11! and of 2, so 2 is a factor of 11!+2.
3 is a factor of 11! and of 3, so 3 is a factor of 11!+3.
Continuing with this logic, every value of g between 11!+2 <= g <= 11!+11 will have a factor bigger than 1.
Thus, g is not prime.
Sufficient.
The correct answer is B
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GMATGuruNY
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If the answer is NO -- if g does not have a factor between 1 and itself -- then g is prime.ranjithreddy.k9 wrote:Does the integer g have a factor f such that 1<f<g ?
1) g > 3!
2) 11!+2 <= g <= 11!+11
Question rephrased: Is g prime?
Statement 1:
3! = 6.
If g=7, then g is prime.
If g=8, then g is not prime.
Insufficient.
Statement 2: 11!+2 <= g <= 11!+11
2 is a factor of 11! and of 2, so 2 is a factor of 11!+2.
3 is a factor of 11! and of 3, so 3 is a factor of 11!+3.
Continuing with this logic, every value of g between 11!+2 <= g <= 11!+11 will have a factor bigger than 1.
Thus, g is not prime.
Sufficient.
The correct answer is B
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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amit2k9
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a its can be 7,8 etc. not sufficient.
b 2(11*10*9...*1 + 1) <=g <= 11(10*9*....*1 + 1)
thus 3,4,5..10 are all values of f. hence sufficient.
Thus b it is.
b 2(11*10*9...*1 + 1) <=g <= 11(10*9*....*1 + 1)
thus 3,4,5..10 are all values of f. hence sufficient.
Thus b it is.
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