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The remainder must be less than the divisor- just set up an inequality from there:

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you dont seem to be from this planet
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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 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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