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
The greatest three-digit number which is divisible by 14, 28, and 42 is:
What is the greatest number that will divide 209 and 347 leaving remainder 5 and 7 respectively?
A number is three times another number and their HCF is 8. What is the sum of the squares of the numbers?
The HCF of 2091, 3485 and 4879 is x. The sum of the digits of x is:
If three numbers are in ratio of 3 : 5 : 7 and their LCM is 2415, what is the difference between the second number and the first number?
