The largest four-digit number which when divided by 15, 12 and 14 leaves remainder 5 in each case is:
9665
A number that leaves the same remainder 5 on division by 15, 12 and 14 must be of the form \(\text{LCM}(15,12,14) \times k + 5\).
Find the LCM: \(15 = 3\times5\), \(12 = 2^2\times3\), \(14 = 2\times7\), so \(\text{LCM} = 2^2\times3\times5\times7 = 420\).
The largest multiple of 420 not exceeding 9999: \(420 \times 23 = 9660\) (since \(420 \times 24 = 10080 > 9999\)).
Add the remainder: \(9660 + 5 = 9665\), which is still a four-digit number.
Hence, the required largest four-digit number is 9665.
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Important Questions from LCM and HCF
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