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

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Joined: 02 Sep 2009
Posts: 49977

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07 Nov 2013, 05:50
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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 $$b=a-1$$).

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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Joined: 11 Aug 2011
Posts: 12
Location: India
GMAT 1: 620 Q46 V30
GPA: 3
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07 Nov 2013, 06:02
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 $$b=a-1$$).

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.
Math Expert
Joined: 02 Sep 2009
Posts: 49977

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07 Nov 2013, 06:14
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 $$b=a-1$$).

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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Joined: 21 Apr 2014
Posts: 39
Re: If x, a, and b are positive integers such that when x is  [#permalink]

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21 Oct 2014, 18:26
1
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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Eliza
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Joined: 30 May 2013
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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, 02:44
Hello Everyone,

How is option B wrong?
Math Expert
Joined: 02 Sep 2009
Posts: 49977
Re: If x, a, and b are positive integers such that when x is  [#permalink]

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22 Oct 2014, 03:50
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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Joined: 21 Apr 2014
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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, 15:20
1
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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Eliza
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bestgmatprepcourse.com

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Re: If x, a, and b are positive integers such that when x is  [#permalink]

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25 Jan 2018, 03:12
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Re: If x, a, and b are positive integers such that when x is &nbs [#permalink] 25 Jan 2018, 03:12

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