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If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre

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If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 09 Feb 2015, 06:39
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If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.

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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 09 Feb 2015, 08:05
1
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


I think it's C

statement 1: x=(5a+b)(5a-b)=(25\(a^2\)-\(b^2\)) If a=3 and b=1 then x=225. hundreds digit of x is not a perfect cube. If a=3 and b=9 then x=144 and hundreds digit of x is a perfect cube.

statement 2: \(a^2\)+84=20a ---> (a-14)(a-6)=0.

1+2) if a=14 let's check what happens at the extremes of b. If b=9 then x=4819. Hundreds digit is a perfect cube. Since b is an integer greater than 0 the hundreds digit will always be 8. If a=6 the outcome is the same. We will always end up with 8 in the hundreds digit.

Answer C.

it took me more than two minutes I am not gonna lie.
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 09 Feb 2015, 09:09
2
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


Rephrased, the question is asking if the hundreds digit of x is either 1 or 8.
Hundreds digit 1 or 8?

Statement 1: \(25a^2 - b^2\)
We can manipulate a and b (under the constraints 2 < a, and 0 < b < 10) so that the hundreds digit is either 1 or 8, or neither.
Insufficient.

Statement 2: \((a-6)(a-14)\)
a=6 or 14
Nothing is mentioned about its relation to x.
Insufficient.

Combined, \(25a^2 - b^2\) and plugging in a when a = 6, we can see that the equation is \(900 - b^2\). Both the min value for b, which is 1, and the max value for b, which is 9, both make the hundreds digit 8.
when a = 14, we can see that the equation is \(4900 - b^2\). Both the min value for b, which is 1, and the max value for b, which is 9, both make the hundreds digit 8.
Sufficient

Answer: C
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 16 Feb 2015, 05:26
1
1
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Solution: C.

Start by simplifying the question itself. The only three perfect cubes which could be expressed as a single digit are 0, 1 and 8, so the question is really asking us whether x = 0, 1 or 8.

Now consider the statements. Statement (1) is a difference of squares equation that gives us 25(a^2) - b^2 = x. Trying a few values for a and b shows that x could have many different hundreds digits: if a = 3 and b = 4, for example, then x = 225 - 16 = 209, but if a = 3 and b = 9, then x = 225 - 81 = 144; INSUFFICIENT. Statement (2) tells us nothing about x at all; INSUFFICIENT.

Statement (2) does give us two possible values for a, however: if we subtract 20a from both sides we have a quadratic equation -- a^2 - 20a + 84 = 0 -- with solutions of a = 6 and a = 14, respectively. Taking the two statements together, then, we have either:

a = 6, so x = (30+b)(30-b) = 900 - b^2
a = 14, so x = (70+b)(70-b) = 4900 - b^2

Since b is a single digit integer, b^2 must be between 0 and 100. This means that 900 - b^2 must be between 800 and 900, so its hundreds digit is 8; much the same is true of 4900 - b^2, which must be between 4800 and 4900, so its hundreds digit is also 8. SUFFICIENT; C.
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 17 Feb 2015, 02:57
use statement 2 to get value of a and then put the value of a in statement 1 and check hence C.
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 19 Feb 2015, 23:24
Hi all,
Answer is Together sufficient.
Statement 1 is not sufficient:
Given x= 25a2-b2
a=8 b=1 hundred place of x is a perfect square.
a=10 b=2 hundred place of x is not a perfect square.
So insufficient.
Statement 2 is not sufficient:
Its quadratic equation where the roots are 6 and 14
This doesn’t talks about x.
Together it is sufficient.
If we take a=6, statement 1 becomes 900-b^2 where hundreds digit of x value is always has 8 which is a perfect cube irrespective of b value chosen between 0 and 10.
If we take a=14, statement 1 becomes 4900-b^2 where hundreds digit of x value is always has 1 which is a perfect cube irrespective of b value chosen between 0 and 10.
So together it is sufficient.
So answer is C
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If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 03 Mar 2015, 04:21
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


The only perfect cubes in 0-9 are 0,1,8 , so question is asking whether hundredth digit is one of them.

Lets first examine option B which seems easy one.
B) We do not know what is X in terms of 'a' and 'b' ; NS
a^2 + 84 = 20a ; A=14,6;
even if we know what X equals from option A, as we do not have information about 'b' this stmt is NS.

A) x = (5a+b)(5a-b) ; x=25a^2 - b^2;
a=3, b=5 --> x=200; is hundred th digit 0/1/8 ? NO
a=6, b=4---> x=884; is hundred th digit 0/1/8 ? YES
so option A is also NS

Lets combine A&B
a=14,6 in both
x= 4900-b^2 when a=14
x= 900-b^2 when a=6
in both cases we know the hundred th digit is 8;

C is the answer
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 03 Mar 2015, 04:25
Bunuel wrote:
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Solution: C.

Start by simplifying the question itself. The only three perfect cubes which could be expressed as a single digit are 0, 1 and 8, so the question is really asking us whether x = 0, 1 or 8.

Now consider the statements. Statement (1) is a difference of squares equation that gives us 25(a^2) - b^2 = x. Trying a few values for a and b shows that x could have many different hundreds digits: if a = 3 and b = 4, for example, then x = 225 - 16 = 209, but if a = 3 and b = 9, then x = 225 - 81 = 144; INSUFFICIENT. Statement (2) tells us nothing about x at all; INSUFFICIENT.

Statement (2) does give us two possible values for a, however: if we subtract 20a from both sides we have a quadratic equation -- a^2 - 20a + 84 = 0 -- with solutions of a = 6 and a = 14, respectively. Taking the two statements together, then, we have either:

a = 6, so x = (30+b)(30-b) = 900 - b^2
a = 14, so x = (70+b)(70-b) = 4900 - b^2

Since b is a single digit integer, b^2 must be between 0 and 100. This means that 900 - b^2 must be between 800 and 900, so its hundreds digit is 8; much the same is true of 4900 - b^2, which must be between 4800 and 4900, so its hundreds digit is also 8. SUFFICIENT; C.


Hi Bunuel,

for the explanation provided for STMT 2 , we do not what is 'x' in terms of 'a' and 'b' so this option is straight forward out.
Please let me know if i am missing something.

thanks
Lucky.
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 03 Mar 2015, 06:04
Lucky2783 wrote:
Bunuel wrote:
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION

Solution: C.

Start by simplifying the question itself. The only three perfect cubes which could be expressed as a single digit are 0, 1 and 8, so the question is really asking us whether x = 0, 1 or 8.

Now consider the statements. Statement (1) is a difference of squares equation that gives us 25(a^2) - b^2 = x. Trying a few values for a and b shows that x could have many different hundreds digits: if a = 3 and b = 4, for example, then x = 225 - 16 = 209, but if a = 3 and b = 9, then x = 225 - 81 = 144; INSUFFICIENT. Statement (2) tells us nothing about x at all; INSUFFICIENT.

Statement (2) does give us two possible values for a, however: if we subtract 20a from both sides we have a quadratic equation -- a^2 - 20a + 84 = 0 -- with solutions of a = 6 and a = 14, respectively. Taking the two statements together, then, we have either:

a = 6, so x = (30+b)(30-b) = 900 - b^2
a = 14, so x = (70+b)(70-b) = 4900 - b^2

Since b is a single digit integer, b^2 must be between 0 and 100. This means that 900 - b^2 must be between 800 and 900, so its hundreds digit is 8; much the same is true of 4900 - b^2, which must be between 4800 and 4900, so its hundreds digit is also 8. SUFFICIENT; C.


Hi Bunuel,

for the explanation provided for STMT 2 , we do not what is 'x' in terms of 'a' and 'b' so this option is straight forward out.
Please let me know if i am missing something.

thanks
Lucky.


Yes, from (2) we know nothing about x, hence it's not sufficient.
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre [#permalink]

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New post 12 Sep 2017, 00:52
Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?

(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a

Kudos for a correct solution.


solution: C

1) x= 25a^2 -b^2
a=3 => x= 225-b^2
if b=1 H(hundreds digit = 2)
if b=9 H=1 perfect cube of 1
A,D eliminated

2) a^2+84=20a
=> a = 6,14
but we dont know what is x => insufficient B is eliminated

Combine

a=6 => x= 900-b^2 , but 10<b<0 so H =8 for any b => perfect cube of 2
a=14 => x= 4900-b^2 , but 10<b<0 so H =8 for any b => perfect cube of 2

Definite YES

C is the answer
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre   [#permalink] 12 Sep 2017, 00:52
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