Find the largest four-digit number which, when divided by 12, 15, 18 and 21, leaves remainders 9, 12, 15 and 18 respectively.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8817
Each remainder is 3 less than its divisor (\(12-9=15-12=18-15=21-18=3\)), so the number plus 3 is divisible by all four divisors. The number is therefore 3 less than a multiple of their LCM.
\(12=2^{2}\times3\), \(15=3\times5\), \(18=2\times3^{2}\), \(21=3\times7\)
LCM \(=2^{2}\times3^{2}\times5\times7=1260\)
Largest four-digit multiple of 1260: \(1260\times7=8820\), since \(1260\times8=10080\) has five digits
Number \(=8820-3=8817\)
The required number is 8817.