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Manager  Joined: 16 Jul 2009
Posts: 161
If x, a, and b are positive integers such that when x is  [#permalink]

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111 00:00

Difficulty:   95% (hard)

Question Stats: 51% (02:33) correct 49% (02:43) wrong based on 725 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 64216

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74
52
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 $$b=a-1$$).

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Senior Manager  Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 311
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB

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1
Awesome Bunuel. Was trying hard to figure out the answer. All the information was just readily available but was looking all around.

+1 to you.
Manager  Joined: 16 Jul 2009
Posts: 161

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Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?
Retired Moderator Joined: 03 Aug 2010
Posts: 156

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Simply amazing bunuel...... you are a genius....
Intern  Joined: 25 Mar 2009
Posts: 35

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cool, BUenel, i had go arround with the formulations but can not find that A-2<B tks much
Math Expert V
Joined: 02 Sep 2009
Posts: 64216

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4
12
abhi758 wrote:
Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?

What do you mean by "these problems"? Remainder problems or must be true problems?

Anyway...

Remainders:
Theory: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

Practice PS:
remainder-101074.html
remainder-problem-92629.html
number-properties-question-from-qr-2nd-edition-ps-96030.html
remainder-when-k-96127.html
ps-0-to-50-inclusive-remainder-76984.html
good-problem-90442.html
remainder-of-89470.html
number-system-60282.html
remainder-problem-88102.html

Practice DS:
remainder-problem-101740.html
remainder-101663.html
ds-gcd-of-numbers-101360.html
data-sufficiency-with-remainder-98529.html
sum-of-remainders-99943.html
ds8-93971.html
need-solution-98567.html
gmat-prep-ds-remainder-96366.html
gmat-prep-ds-93364.html
ds-from-gmatprep-96712.html
remainder-problem-divisible-by-86839.html
gmat-prep-2-remainder-86155.html
remainder-94472.html
remainder-problem-84967.html

COULD or MUST be true questions:
ds-number-theory-101025.html?hilit=must%20true
number-properties-question-101150.html?hilit=must%20true
must-be-true-101575.html?hilit=must%20true
gmat-prep-question-101282.html?hilit=must%20true
ab-2-c-is-even-101751.html?hilit=must%20true
mgmat-inequalities-101732.html?hilit=must%20true#p788920
division-and-inequalities-87707.html?hilit=could%20true%20following#p666131

Hope it helps.
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Senior Manager  Joined: 18 Feb 2008
Posts: 352
Location: Kolkata

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2
Bunuel wrote:
abhi758 wrote:
Thanks Bunnel!! Do you suggest using a particular strategy for these problems or using different strategy for every problem and whichever fits the bill for the given question..?

What do you mean by "these problems"? Remainder problems or must be true problems?

Anyway...

Remainders:
Theory: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

Practice PS:
remainder-101074.html
remainder-problem-92629.html
number-properties-question-from-qr-2nd-edition-ps-96030.html
remainder-when-k-96127.html
ps-0-to-50-inclusive-remainder-76984.html
good-problem-90442.html
remainder-of-89470.html
number-system-60282.html
remainder-problem-88102.html

Practice DS:
remainder-problem-101740.html
remainder-101663.html
ds-gcd-of-numbers-101360.html
data-sufficiency-with-remainder-98529.html
sum-of-remainders-99943.html
ds8-93971.html
need-solution-98567.html
gmat-prep-ds-remainder-96366.html
gmat-prep-ds-93364.html
ds-from-gmatprep-96712.html
remainder-problem-divisible-by-86839.html
gmat-prep-2-remainder-86155.html
remainder-94472.html
remainder-problem-84967.html

COULD or MUST be true questions:
ds-number-theory-101025.html?hilit=must%20true
number-properties-question-101150.html?hilit=must%20true
must-be-true-101575.html?hilit=must%20true
gmat-prep-question-101282.html?hilit=must%20true
ab-2-c-is-even-101751.html?hilit=must%20true
mgmat-inequalities-101732.html?hilit=must%20true#p788920
division-and-inequalities-87707.html?hilit=could%20true%20following#p666131

Hope it helps.

Priceless info.Ton thanks Manager  Joined: 19 Apr 2010
Posts: 168
Schools: ISB, HEC, Said

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Awesome solution Bunuel. +1

Great question Kudos to GMAT club tests
Manager  Joined: 06 Nov 2009
Posts: 174
Concentration: Finance, Strategy

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Awesome!!! I didnt get the fastest way, but through number plug-in I was able to figure out the solution. But your solution is so much better!!
Manager  Joined: 06 Jun 2009
Posts: 233
Location: USA
WE 1: Engineering

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nice and efficient !
Manager  Joined: 16 Jul 2009
Posts: 161

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Bunnel, thanks so much for the compilation! By 'these problems' I meant 'Must be true' questions in which at times you have more than 1 correct answers. My apologies for the lack of clarity their. Your compilation should be enough to practice. Thanks again!
Intern  Joined: 20 Jul 2010
Posts: 30

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Nice approach. +1 kudos
Senior Manager  B
Joined: 11 May 2011
Posts: 269
Re: If x, a, and b are positive integers such that when x is div  [#permalink]

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2
sanjoo 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 divided by A, remainder is B, this implies A > B
When divided by B, remainder is A-2, this implies B > A -2

Combining both,
B < A < (B + 2)
Since, A and B are integers,
A = B + 1

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Manager  Joined: 28 Feb 2012
Posts: 103
GPA: 3.9
WE: Marketing (Other)
Re: If x, a, and b are positive integers such that when x is  [#permalink]

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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 $$b=a-1$$).

Indeed very nice explanation, but for me, for the person who is not that strong in quants sometimes difficult to keep all that concepts in my head and i am jumping to different approaches. Whenever i see must be true questions i plug in some numbers and see which answer works, since it is must be true questions any numbers should work equally. For example in this problem: lets says x=5, a=3 then b=2, so check all the answers and we see that only d works, but if there will be two answers that work try different numbers till we get only one. It could be time consuming, but when we are asked simple expressions it is easy to find numbers that work well.

Bunuel, do you think there are any pitfalls that i should be aware of?
Senior Manager  Joined: 23 Mar 2011
Posts: 392
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Re: If x, a, and b are positive integers such that when x is  [#permalink]

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My approach of plug in numbers was certainly not the best approach, in a zest of GMAT I dont know why I am loosing to think simple....this was a simple algebra which I complicated with numbers
Math Expert V
Joined: 02 Sep 2009
Posts: 64216
Re: If x, a, and b are positive integers such that when x is  [#permalink]

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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All Must or Could be True Questions to practice: search.php?search_id=tag&tag_id=193

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

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Way bunuel approch the problem is fantastic.

I dont know whether i would be able to think the way he does . . tat too in time pressure +1 kudos
Intern  Joined: 15 Dec 2007
Posts: 11
Re: If x, a, and b are positive integers such that when x is  [#permalink]

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1
This problem can be solved by plugging numbers. For instance if x=5, a=3, b=2 we get:
...when x is divided by a, the remainder is b... > 5/3=1+2/3
...and when x is divided by b, the remainder is a-2... > 5/2=2+1/2

Both work. If we plug numbers into the answers, only D will work.

Please give kudos, if you like this.
Intern  Joined: 11 Aug 2011
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Location: India
GMAT 1: 620 Q46 V30
GPA: 3
WE: Engineering (Other)

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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. Re: m16 #35   [#permalink] 06 Nov 2013, 13:10

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