Which is the largest 3-digit number which when divided by 6 leaves the remainder 5 and when divided by 5 leaves the remainder 3?

Lcm of (6, 5) = 30Let’s assume the number be "30K + constant", where Constant is the remainder.let that number be ‘M’⇒ M/6 = 5 (remainder)M could be 5, 11, 17, 23, 29, ...⇒ M/5 = 3 (remainder)M could be 3, 8, 13, 18, 23, 28, ...The very first number common in both term is 23.⇒ M is 23 i.e. a constant term⇒ 30K + constant = 30K + 23The largest three digit number comes when K is 32⇒ 30 (32) + 23= 983∴ When 983 is divided by 11, leaves the remainder 4.

First find the smallest 3-digit number that is evenly divisible by both 5 and 6. Then add 1. Since 5 is prime and not a factor of 6, the smallest positive integer evenly divisible by both 5 and 6 must be 5 times 6 or 30. So the question becomes what is the smallest 3-digit number evenly divisible by 30? 1, 2, and 3 times 30 each produce a 2-digit number. Hence the smallest integer multiplier of 30 that produces a 3-digit product is 4, and the smallest 3-digit integer divisible by 30 is 120. Hence, 120 is the smallest 3-digit integer evenly divisible by both 5 and 6. And finally, 121 is the smallest 3-digit integer when divided by either 5 or 6 leaves a remainder of 1 in each case. John

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