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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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09 Feb 2015, 08:05

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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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09 Feb 2015, 09:09

2

This post received KUDOS

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

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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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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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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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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Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

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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