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Re: m16 #35 [#permalink]
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07 Nov 2013, 04:50
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ankit41 wrote: Bunuel wrote: If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) E. \(a+2=b+1\)
When \(x\) is divided by \(a\), the remainder is \(b\) > \(x=aq+b\) > \(remainder=b<a=divisor\) (remainder must be less than divisor); When \(x\) is divided by \(b\), the remainder is \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
Answer: D. Hi Bunuel, I solved this problem with a bit different approach x = p*a + b..........eqn(1) and x = q*b + (a2)..............eqn(2) now, equating eqn(1) and eqn(2) p*a + b = q*b + (a2) a(p1) = b(q1)  2 if we put p = q = 3 we get, 2a = 2b  2 or a = b  1 or a + 2 = b + 1 which is option E would pl tell me where am i wrong with my approach?? Thanks. You cannot assign arbitrary values to p and q and say that p = q = 3.
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Re: m16 #35 [#permalink]
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07 Nov 2013, 05:02
Bunuel wrote: ankit41 wrote: Bunuel wrote: If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) E. \(a+2=b+1\)
When \(x\) is divided by \(a\), the remainder is \(b\) > \(x=aq+b\) > \(remainder=b<a=divisor\) (remainder must be less than divisor); When \(x\) is divided by \(b\), the remainder is \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
Answer: D. Hi Bunuel, I solved this problem with a bit different approach x = p*a + b..........eqn(1) and x = q*b + (a2)..............eqn(2) now, equating eqn(1) and eqn(2) p*a + b = q*b + (a2) a(p1) = b(q1)  2 if we put p = q = 3 we get, 2a = 2b  2 or a = b  1 or a + 2 = b + 1 which is option E would pl tell me where am i wrong with my approach?? Thanks. You cannot assign arbitrary values to p and q and say that p = q = 3. But p and q are integers. Would you elaborate your point? Thanks.



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Re: m16 #35 [#permalink]
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07 Nov 2013, 05:14
ankit41 wrote: Bunuel wrote: ankit41 wrote: If \(x\), \(a\), and \(b\) are positive integers such that when \(x\) is divided by \(a\), the remainder is \(b\) and when \(x\) is divided by \(b\), the remainder is \(a2\), then which of the following must be true?
A. \(a\) is even B. \(x+b\) is divisible by \(a\) C. \(x1\) is divisible by \(a\) D. \(b=a1\) E. \(a+2=b+1\)
When \(x\) is divided by \(a\), the remainder is \(b\) > \(x=aq+b\) > \(remainder=b<a=divisor\) (remainder must be less than divisor); When \(x\) is divided by \(b\), the remainder is \(a2\) > \(x=bp+(a2)\) > \(remainder=(a2)<b=divisor\).
So we have that: \(a2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a1\) (there is only one integer between \(a2\) and \(a\), which is \(a1\) and we are told that this integer is \(b\), hence \(b=a1\)).
Answer: D. Hi Bunuel, I solved this problem with a bit different approach x = p*a + b..........eqn(1) and x = q*b + (a2)..............eqn(2) now, equating eqn(1) and eqn(2) p*a + b = q*b + (a2) a(p1) = b(q1)  2 if we put p = q = 3 we get, 2a = 2b  2 or a = b  1 or a + 2 = b + 1 which is option E would pl tell me where am i wrong with my approach?? Thanks. But p and q are integers. Would you elaborate your point? Thanks. What does p and q being integers has to do with it? The question asks which of the following MUST be true. Again, you cannot assign arbitrary values to p and q. For example, what you get if you assume that p=q=1 to p=q=10?
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New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If x, a, and b are positive integers such that when x is [#permalink]
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21 Oct 2014, 17:26
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So the thing to keep in mind with remainder problems like this is that it usually involves finding a value from knowing that the remainder must be smaller than the quotient. From this, we know that a>b and that b>a2, put these two together and we get: a>b>a2 and since these are all positive integers then there b must equal a2
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Re: If x, a, and b are positive integers such that when x is [#permalink]
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22 Oct 2014, 01:44
Hello Everyone,
How is option B wrong?



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Re: If x, a, and b are positive integers such that when x is [#permalink]
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22 Oct 2014, 02:50



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Re: If x, a, and b are positive integers such that when x is [#permalink]
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13 Nov 2014, 14:20
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Since we know that the remainder is b when x is divided by a we know that b is smaller than a. We also know that a2 is smaller than b, because a2 is the remainder when x is divided by b. If we put those two together we know that: a2<b<a Because we know that x, a and b are all positive integers we also know that b must be a1, because it is an integer that is bigger than a2 and smaller than a
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Re: If x, a, and b are positive integers such that when x is [#permalink]
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