Let the number be \(x\).
According to the problem, when \(x\) is divided by 5, 6, and 7, the remainders are 4, 5, and 6 respectively. This can be expressed as:
- \(x \equiv 4 \pmod{5}\)
- \(x \equiv 5 \pmod{6}\)
- \(x \equiv 6 \pmod{7}\)
Notice that in each case, the remainder is one less than the divisor. This means that \(x+1\) will be divisible by 5, 6, and 7. Therefore, \(x+1\) is a common multiple of 5, 6, and 7.
To find the least common multiple (LCM) of 5, 6, and 7, we find the prime factorization of each number:
- 5 = 5
- 6 = 2 × 3
- 7 = 7
The LCM is the product of the highest powers of all prime factors present in the numbers: \(LCM(5, 6, 7) = 2 × 3 × 5 × 7 = 210\).
So, \(x + 1 = 210\). Solving for \(x\), we get \(x = 210 - 1 = 209\).
Therefore, the least number which when divided by 5, 6, and 7 leaves remainders 4, 5, and 6 is 209.
Similar Questions
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Important Questions from LCM and HCF
The HCF and LCM of two numbers are 12 and 72, respectively. If the ratio of the two numbers is 2 ∶ 3, then the larger of the two numbers is:
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Joseph visits the club on every 5 th day, Harsh visits on every 24 th day, while Sumit visits on every 9 th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again?
What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?
Which of the following is a pair of co-primes?
