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Bunuel
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Bunuel
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Bunuel
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ezhilkumarank
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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.
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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..?
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hirendhanak
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Simply amazing bunuel...... you are a genius....
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cool, BUenel, i had go arround with the formulations but can not find that A-2<B tks much
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Bunuel
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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..?

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
gmat-club-please-explain-83605.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 :)
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Awesome solution Bunuel. +1

Great question Kudos to GMAT club tests
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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!!
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nice and efficient !
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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!
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Nice approach. +1 kudos
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sanjoo
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

Answer is (D) . Cheers!
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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
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Bunuel
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.
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ankit41
Bunuel
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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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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Hello Everyone,

How is option B wrong?
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Bunuel
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siddharthmk
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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