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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 22 Oct 2019, 21:27
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

31% (02:42) correct 69% (02:21) wrong based on 35 sessions

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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 Updated on: 23 Oct 2019, 23:45
key line in question b is the greatest divisor of x that is less than x

#1
b^2 = x
let b=2 and x=4
is the sum of the divisors of x, which are less than x itself and greater than one, greater than 2b
divisors of x;4 , as per given condition will be 2 only and no is answer
sufficient
#2
2b = x
2*6=12
factor of 12 ; as per condition (2,3,4,6) ; sum ; 15 >12 yes
x=2 and b=1
we have 0 value as per given condition so ; 0>2 ; no
insufficient
IMO A


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

Originally posted by Archit3110 on 22 Oct 2019, 22:49.
Last edited by Archit3110 on 23 Oct 2019, 23:45, edited 1 time in total.
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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 22 Oct 2019, 23:44
1
(1) b^2 = x

If b = 5 then x = 25

Factors are 1,5,25....Sum factors below 25 is 6<25.


If b=6,x=36
Factors=1,2,3,4,6,9,12,18,36
Sum is 55 >36

So A is insufficient


(2) 2b = x

If b=5 then x=10
Factors=1,2,5,10
Sum=8<10

If b=18,x=36
Factors=
1,2,3,4,6,9,12,18,36
Sum is 55 >36

So B is also insufficient

Combing both I.e x=b^2=2b

Then b must be 2....So x=4

SO OA:C

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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 23 Oct 2019, 01:52
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?
the question stem itself give me headache lol jk

b and x >0
B greatest divisor of x but B is not X
is the sum of the divisors of X that isn't X and isn't 1 > 2B?


(1) b^2 = x
if b=2 x=4 (past the criteria given)
sum of divisors of X(excl. x and 1) is 2
2>2(2) no
if b=3 x=9 (past the criteria given)
sum of divisors of X(excl. x and 1) is 3
3>2(3) no
if b=4 x=16 (then b isn't the greatest divisor of x that isn't x; 8 is) therefore can't use this
therefore
sufficient

(2) 2b = x
x=2 b=1
sum of divisors of X(excl. x and 1) is 0
0>2(1) no
x=4 b=2
sum of divisors of X(excl. x and 1) is 2
2>2(2) no
sufficient

Therefore, D
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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 23 Oct 2019, 05:49
Given that b is the greatest divisor of x, and 1<b<x, we are to determine if the sum of all the divisors (factors) of b greater than 1 and less than b is greater than 2b.

Statement 1: b^2=x
This implies that b=√x hence x must be a perfect square. From the given condition, possible values of x are:16, 36, 64, 81, 100, 144, 196, 225, 256, etc
Lets test with the smallest possible value of x=16
b=4.
factors of b that satisfy the condition =2. but 2b=8, since 2 < 8, the answer No.
Let x=20,712, the b=144
Factors of b satisfying given condition = 2,3,4,6,8,9,12,16,18,24,36,48,72 and their sum is 258 < 2b=288. The answer is therefore No to the question asked.
Statement 1 on its own is sufficient.

Statement 2: 2b=x, hence b=x/2
values of x that satisfy the given condition are: 8,12,18,20,24,...
when x=8, b=4, sum of factors of 4 satisfying the condition is 2, and 2<8, hence No.
x=24, b=12, factors of 12 satisfying the condition = 2,3,4,6 sum of the factors = 15. Since 15<24, the answer is still No.
Statement 2 is also sufficient on its own.

The answer is D.
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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 23 Oct 2019, 08:18
1
Quote:
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


(1) b^2 = x sufic.

prime is a number that is divisible only by one and itself; ie. f(2)=2,1
prime squared is a number that is divisible only by one, itself and its prime factor; ie. f(9)=9,3,1

\(b^2=x…b=prime…x=prime^2…x=9:f(x)=[9,3,1]…sumf(x)=3…2b=6…answer=always.no\)

(2) 2b = x insufic.

\(x=2: f(x)=[2,1]…sum=0 > 2b=2(1)=2… no\)
\(x=12: f(x)=[12,6,4,3,2,1]…sum=[6,4,3,2]=15 > 2b=2(6)=12… yes\)

Answer (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 23 Oct 2019, 15:14
1
b and x — positive integers
—> b is the greatest divisor of x that is less than x —> b= x/2
————————
Is sum of the divisors of x, which are less than x itself and greater than one, greater than 2b?

(Statement1): b^{2}= x
—> In order ‘b’ be the greatest divisor of x, b should be the prime number.
If b= 5, 5^{2}=25
—> the divisors of 25 are 1,5,25
5> 2*5 (No)

If b= 11, then 11^{2}=121
—> the divisors of 121 are 1,11,121
11> 2*11(No)
Sufficient

(Statement2): 2b= x
If b= 6, then 2*6=12
—> the divisors of 12 are 1,2,3,4,6,12
2+3+4+6> 6*2
15> 12 (yes)

If b=5, then 2*5=10
—> the divisors of 10 are 1,2,5,10
2+5> 2*5
7> 10 (No)
Insufficient

The answer is 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 23 Oct 2019, 15:57
From Stmnt 1, b=2,x= 4, then, sum of divisors which are less than x itself and greater than =2<2b. For x=9, it is 3<2b.
For x= 16, sum of such divisors is 14, which is greater than 2b or 8. For x=25, it is 5, which is less than 2b. No single answer. NOT SUFFICIENT.

From Stmnt 2, b=3, x=6. here, sum of divisors (2+3)=5,less than 2b.
b=4, x= 8. sum of divisors (2+4)= 6, less than 2b.
b=6, x= 12, (2+3+4+6)= 15>2b. NOT SUFFICIENT.

Combining both statement, it is possible only when b = 2, x= 4. SUFFICIENT.
C is the answer.
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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 24 Oct 2019, 12:00
madgmat2019 wrote:
(1) b^2 = x

If b = 5 then x = 25

Factors are 1,5,25....Sum factors below 25 is 6<25.


HI madgmat2019 ,
Above you made a small blunder ,
So if you are taking b as 5 then x=25
Now all factors of x above 1 and less than 25 is 5 only.So sum of all such factors is 5. Also 2b =10
so is 5>10 , NO ( From where did you get 6 )

Quote:
If b=6,x=36
Factors=1,2,3,4,6,9,12,18,36
Sum is 55 >36


if b = 6 then x=36
NOTE: b has to be the greatest factor of x below x, the greatest factor of 36 just below 36 is 18 NOT 6, hence b=6 and x=36 case is invalid. We have to choose b in such a way that b is the greatest factor of x just below x.

In other words such a scenario is only possible if b is prime.

Take another prime and test it. Let b =7 then x=49 here b is indeed the greatest factor of x just below x.
Now sum of all such factors of x that are above 1 and below 49 is 7 only .
Also 2b is 14 hence 2b>7 . Hence 2b will always be greater and answer will always be NO.Hope this is clear .
Hence A is SUFF.

Similarly you can work out why statement 2 is INSUFF.
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Let b and x be positive integers. If b is the greatest divisor of x th   [#permalink] 24 Oct 2019, 12:00
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