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elizaanne
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If x, a, and b are positive integers such that when x is 13 Nov 2014, 15:20
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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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If x, a, and b are positive integers such that when x is 03 Dec 2018, 00:16
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abhi758
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
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"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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If x, a, and b are positive integers such that when x is 02 Apr 2019, 10:47
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Very nice question. I am in love with this question and the approach provided by Bunuel


abhi758
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
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If x, a, and b are positive integers such that when x is 02 Nov 2024, 09:14
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The remainder must be less than the divisor- just set up an inequality from there:

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If x, a, and b are positive integers such that when x is 01 Sep 2025, 08:46
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you dont seem to be from this planet
Bunuel
Official Solution:

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 \(b\) can be expressed as \(x=aq+b\), where \((remainder=b) \lt (a=divisor)\) (as the remainder must be less than divisor);

When \(x\) is divided by \(b\), the remainder is \(a-2\) can be expressed as \(x=bp+(a-2)\), where \((remainder=a-2) \lt (b = divisor)\).

From these equations, we can deduce that: \(a - 2 < b < a\). Since \(a\) and \(b\) are integers, it must be true that \(b = a - 1\). This is because there is only one integer between \(a - 2\) and \(a\), which is \(a - 1\), and we know that this integer is equal to \(b\). Therefore, \(b = a - 1\).


Answer: D.
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If x, a, and b are positive integers such that when x is 01 Sep 2025, 23:20
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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?

x = ak + b; b<a
x = bm + (a-2); a-2<b

a-2<b<a
b = a - 1; Since b is an integer between 2 integers a-2 & a.

A. \(a\) is even
B. \(x+b\) is divisible by \(a\)
C. \(x-1\) is divisible by \(a\)
D. \(b=a-1\) : MUST BE TRUE
E. \(a+2=b+1\)

IMO D
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If x, a, and b are positive integers such that when x is 23 Jul 2026, 12:47
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x=ak+b & a>b
x=lb+(a-2) & b>a-2 where k and l are +ve integers
hence b=a-1 is the only possibility since a and b are +ve integers
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