The largest four-digit number which when divided by 6, 14 and 7 leaves remainder 5 in each case is:
9959
A number leaving the same remainder 5 on division by 6, 14 and 7 must be of the form (multiple of LCM) + 5.
Find the LCM of 6, 14 and 7. Since \(6 = 2 \times 3\), \(14 = 2 \times 7\) and \(7 = 7\), the LCM is \(2 \times 3 \times 7 = 42\).
Find the largest multiple of 42 that is at most 9999. Dividing, \(9999 \div 42 = 238\) remainder 3, so the largest multiple is \(42 \times 238 = 9996\).
Adding the remainder 5 gives \(9996 + 5 = 10001\), which exceeds four digits, so step down one multiple: \(9996 - 42 = 9954\).
Then \(9954 + 5 = 9959\), which is a four-digit number and leaves remainder 5 with each divisor.
Hence, the required number is 9959.
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Important Questions from LCM and HCF
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Calculate the HCF of \(\frac{12}{5}\) , \(\frac{14}{15}\) and \(\frac{16}{17}\) .
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?
