Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder?
Q1. Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder?
Answer: 5035
Explanation: LCM(6,7,8,9,10) = 2520; each remainder is exactly 5 less than its divisor, so the number is of the form 2520k − 5. Testing k=2 gives 5035, which is divisible by 19 (5035 ÷ 19 = 265).