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16 Dec 2019, 02:40
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E
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Difficulty:
65%
(hard)
Question Stats:
48% (01:38) correct
52%
(01:32)
wrong
based on 209
sessions
History
Date
Time
Result
Not Attempted Yet
If x is a positive integer, what is the number of different positive factors of 39x ?
(1) x is a two-digit number
(2) x^2 has 3 positive factors
Are You Up For the Challenge: 700 Level Questions
M36-71
(1) x is a two-digit number
(2) x^2 has 3 positive factors
Are You Up For the Challenge: 700 Level Questions
M36-71
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09 Nov 2021, 07:44
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Bunuel
Official Solution:
If \(x\) is a positive integer, what is the number of different positive factors of \(39x\) ?
Factorize: \(39x=3*13*x\)
(1) \(x\) is a two-digit number
Clearly insufficient.
(2) \(x^2\) has 3 positive factors
The above means that \(x\) is a prime number. Only the squares of primes (\(p^2\)) have three factors: 1, \(p\), and \(p^2\)
If \(x=11\) (or any other prime except 3 and 13), then \(39x=3*11*13\) will have \((1+1)(1+1)(1+1)=8\) factors;
If \(x=13\), then \(39x=3*13^2\) will have \((1+1)(2+1)=6\) factors.
Not sufficient.
(1)+(2) Examples we used for (2) (\(x=11\) and \(x=13\)) also satisfy (1), so even taken together the statements are not sufficient to get the answer: \(39x\) can have 8 or 6 factors. Not sufficient.
Answer: E
General Discussion
firas92
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Bunuel
(1) x is a two-digit number
If x is 11 (or any 2 digit prime), the number of factors of \(39x=3^1*11^1*13^1\) is \((1+1)*(1+1)*(1+1)=8\)
If x is 12, the number of factors of \(39x=2^2*3^2*13\) is \((2+1)*(2+1)*(1+1)=18\)
1 is not sufficient
(2) x^2 has 3 positive factors
Note that only squares of prime numbers can have 3 positive factors
So x is a prime number and so the number of factors for \(39x=3^1*13^1*x^1\) is \((1+1)*(1+1)*(1+1)=8\)
But if x=13, the number of factors is (1+1)*(2+1)=6
2 is not Sufficient
(1)+(2)
x can still be 13 or any other 2 digit prime
Not sufficient
Answer is (E)
