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Let b and x be positive integers. If b is the greatest divisor of x th

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Let b and x be positive integers. If b is the greatest divisor of x th  [#permalink]

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New post 20 Jan 2017, 07:40
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A
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C
D
E

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  95% (hard)

Question Stats:

24% (02:53) correct 76% (02:36) wrong based on 55 sessions

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Re: Let b and x be positive integers. If b is the greatest divisor of x th  [#permalink]

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New post 18 Dec 2017, 12:50
1
Bunuel wrote:
Let b and x be positive integers. If b is the greatest divisor of x that is less than x, is the sum of the divisors of x, which are less than x itself and greater than one, greater than 2b?

(1) b^2 = x
(2) 2b = x


Statement 1: \(b\) is the greatest divisor of \(x\) and \(x=b^2\), so \(b\) must be prime.

For eg. \(b=7\), then \(x=49\). Divisors of \(x=1,7,49\). Greatest divisor \(7=b\)

\(b\) cannot be a composite number. for eg. if \(b=4\), then \(x=16\). Divisors of \(x= 1,2,4,8,16\). Greatest divisor \(8\) which is not equal to \(b\).

So if \(b =p_1\), a prime number and as \(x\) is square of a prime number, \(x={p_1}^2\), divisors of \(x\) will be \(1\), \(p_1\) & \({p_1}^2\) only.

Now the divisors of \(x\), which are less than \(x\) itself and greater than one is \(p_1=b\) which is not greater than \(2b=2p_1\). Sufficient

Statement 2: if \(b=3\), \(x=6\). Divisors of \(x= 1,2,3,6\). sum of divisors of \(x\), which are less than \(x\) itself and greater than one is \(2+3=5<2*3\)

if \(b=6\), \(x=12\). Divisors of \(x=1,2,3,4,6,12\). sum of divisors of \(x\), which are less than \(x\) itself and greater than one is \(2+3+4+6=15>2*6\). Hence Insufficient

Option A
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Re: Let b and x be positive integers. If b is the greatest divisor of x th  [#permalink]

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New post 20 Jan 2017, 10:33
Ans is C.
Considering question stem, we have b as greatest divisor of x. For any no, it's greatest divisor will be half or less than half of itself, for it being a non prime.
Consider st1.
B^2=x

5x5 =25, sum of divisors = 5 lesser than 5.
6×6=36,sum of divisors = 2+3+4+6+9+18 greater than 6.cant answer. Not sufficient

Consider st2
2b=x
Eg
2x4=8
Sum of divisors of 8 = 2+4 less than 8
2×6=12
Sum of divisors of 12 = 2+3 less than 12.
2x8=16
Sum of divisors of 16 = 2+4+8 less than 16
2x9=18
Sum of divisors of 18 =2+3+6+9 greater than 18.
Not sufficient.

Considering both st1 n st2

b^2=2b gives b=2
For b=2 sum of divisors of 2b is 2 which is less that 2b=4. Hence sufficient.




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Re: Let b and x be positive integers. If b is the greatest divisor of x th  [#permalink]

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New post 18 Dec 2017, 11:24
I picked D: Why is it A?
St 1 - x^2 = b -->x = b * b
Thus,x > 2b > b (as asked by the question. I think I have made a mistake here in the interpretation of the question). Thus x is the largest value here.

St 2 --> 2b = x--> 2b/b = 2. So, one factor is 2. b = x/2. What I got was --> x > b and x = 2b. So I am not sure why its not suff
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Let b and x be positive integers. If b is the greatest divisor of x th  [#permalink]

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New post 18 Dec 2017, 12:54
Madhavi1990 wrote:
I picked D: Why is it A?
St 1 - x^2 = b -->x = b * b
Thus,x > 2b > b (as asked by the question. I think I have made a mistake here in the interpretation of the question). Thus x is the largest value here.

St 2 --> 2b = x--> 2b/b = 2. So, one factor is 2. b = x/2. What I got was --> x > b and x = 2b. So I am not sure why its not suff


Hi Madhavi1990

I think you interpreted the question wrongly. we need to calculate sum of divisors of x but you are focusing only on the greatest divisor of x i.e b. for statement 2 if b=6, then x=12 and it will have other factors as well apart from 2.
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Let b and x be positive integers. If b is the greatest divisor of x th   [#permalink] 18 Dec 2017, 12:54
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