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 \(b=a-1\)).
Answer: D.
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?