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Re: m16 #35
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07 Nov 2013, 04:50
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
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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
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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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Re: If x, a, and b are positive integers such that when x is
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21 Oct 2014, 17:26
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
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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
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22 Oct 2014, 02:50
siddharthmk wrote: Hello Everyone,
How is option B wrong? If x = 5, a = 3, and b = 2, then option B is not true: x + b = 7 is NOT divisible by a = 3.
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Re: If x, a, and b are positive integers such that when x is
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13 Nov 2014, 14:20
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
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02 Dec 2018, 23:16
abhi758 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"means b < a (Remainder is always less than the divisor) "when x is divided by b, the remainder is a2"means (a  2) < b (Remainder is always less than the divisor) So (a  2) < b < a b is between (a  2) and a which means it must be (a  1). Answer (D)
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Re: If x, a, and b are positive integers such that when x is
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02 Apr 2019, 09:47
Very nice question. I am in love with this question and the approach provided by Bunuelabhi758 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



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Re: If x, a, and b are positive integers such that when x is
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18 May 2020, 22:24
Great question. Got confused at the a2<b step when I was solving it so came here to check.
+1 to Bunuel, the king of Quant.




Re: If x, a, and b are positive integers such that when x is
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18 May 2020, 22:24



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