What is the least number which, when divided by 6, 8, 9, 12 and 14, leaves a remainder of 2 in each case, but is exactly divisible by 23?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 506
A number that leaves the same remainder 2 with each divisor is 2 more than a multiple of their LCM. Among these, the first one divisible by 23 is wanted.
\(6=2\times3\), \(8=2^{3}\), \(9=3^{2}\), \(12=2^{2}\times3\), \(14=2\times7\)
LCM \(=2^{3}\times3^{2}\times7=504\), so the number is \(504k+2\)
\(k=0\) gives 2, not divisible by 23
\(k=1\) gives 506, and \(506=23\times22\)
The least such number is 506.