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If x, a, and b are positive integers such that when x is

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Re: m16 #35  [#permalink]

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New post 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 \(a-2\), then which of the following must be true?

A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\)
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 \(a-2\) --> \(x=bp+(a-2)\) --> \(remainder=(a-2)<b=divisor\).

So we have that: \(a-2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a-1\) (there is only one integer between \(a-2\) and \(a\), which is \(a-1\) and we are told that this integer is \(b\), hence \(b=a-1\)).

Answer: D.


Hi Bunuel,

I solved this problem with a bit different approach
x = p*a + b..........eqn(1)
and x = q*b + (a-2)..............eqn(2)

now, equating eqn(1) and eqn(2)
p*a + b = q*b + (a-2)

a(p-1) = b(q-1) - 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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New post 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 \(a-2\), then which of the following must be true?

A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\)
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 \(a-2\) --> \(x=bp+(a-2)\) --> \(remainder=(a-2)<b=divisor\).

So we have that: \(a-2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a-1\) (there is only one integer between \(a-2\) and \(a\), which is \(a-1\) and we are told that this integer is \(b\), hence \(b=a-1\)).

Answer: D.


Hi Bunuel,

I solved this problem with a bit different approach
x = p*a + b..........eqn(1)
and x = q*b + (a-2)..............eqn(2)

now, equating eqn(1) and eqn(2)
p*a + b = q*b + (a-2)

a(p-1) = b(q-1) - 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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Joined: 02 Sep 2009
Posts: 52292
Re: m16 #35  [#permalink]

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New post 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 \(a-2\), then which of the following must be true?

A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\)
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 \(a-2\) --> \(x=bp+(a-2)\) --> \(remainder=(a-2)<b=divisor\).

So we have that: \(a-2<b<a\), as \(a\) and \(b\) are integers, then it must be true that \(b=a-1\) (there is only one integer between \(a-2\) and \(a\), which is \(a-1\) and we are told that this integer is \(b\), hence \(b=a-1\)).

Answer: D.


Hi Bunuel,

I solved this problem with a bit different approach
x = p*a + b..........eqn(1)
and x = q*b + (a-2)..............eqn(2)

now, equating eqn(1) and eqn(2)
p*a + b = q*b + (a-2)

a(p-1) = b(q-1) - 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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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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New post 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>a-2, put these two together and we get: a>b>a-2 and since these are all positive integers then there b must equal a-2
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Re: If x, a, and b are positive integers such that when x is  [#permalink]

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New post 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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New post 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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New post 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 a-2 is smaller than b, because a-2 is the remainder when x is divided by b. If we put those two together we know that:
a-2<b<a

Because we know that x, a and b are all positive integers we also know that b must be a-1, because it is an integer that is bigger than a-2 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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New post 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 a-2, then which of the following must be true?

A. a is even
B. x+b is divisible by a
C. x-1 is divisible by a
D. b=a-1
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 a-2"

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 &nbs [#permalink] 02 Dec 2018, 23:16

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