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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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09 Feb 2015, 06:39
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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)(5ab) (2) a^2 + 84 = 20a
Kudos for a correct solution. I think it's C statement 1: x=(5a+b)(5ab)=(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 > (a14)(a6)=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
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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)(5ab) (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: \((a6)(a14)\) 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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16 Feb 2015, 05:26
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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)(5ab) (2) a^2 + 84 = 20a
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTIONSolution: 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)(30b) = 900  b^2 a = 14, so x = (70+b)(70b) = 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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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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19 Feb 2015, 23:24
Hi all, Answer is Together sufficient. Statement 1 is not sufficient: Given x= 25a2b2 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 900b^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 4900b^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)(5ab) (2) a^2 + 84 = 20a
Kudos for a correct solution. The only perfect cubes in 09 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)(5ab) ; 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= 4900b^2 when a=14 x= 900b^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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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)(5ab) (2) a^2 + 84 = 20a
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTIONSolution: 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)(30b) = 900  b^2 a = 14, so x = (70+b)(70b) = 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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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)(5ab) (2) a^2 + 84 = 20a
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTIONSolution: 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)(30b) = 900  b^2 a = 14, so x = (70+b)(70b) = 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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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)(5ab) (2) a^2 + 84 = 20a
Kudos for a correct solution. solution: C 1) x= 25a^2 b^2 a=3 => x= 225b^2 if b=1 H(hundreds digit = 2) if b=9 H=1 perfect cube of 1 A,D eliminated2) a^2+84=20a => a = 6,14 but we dont know what is x => insufficient B is eliminatedCombine a=6 => x= 900b^2 , but 10<b<0 so H =8 for any b => perfect cube of 2 a=14 => x= 4900b^2 , but 10<b<0 so H =8 for any b => perfect cube of 2 Definite YESC is the answer
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Re: If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundre
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