What number has a remainder of 3 when divided by 5 a remainder of 6 when divided by 9 and a remainder of 8 when divided by 11?

Answer 1

The lowest number is 393. Here's how - the lowest number divided by 11, R8 is 19. In order to maintain this remainder, the possible numbers must increase by 11 (30,41,52,etc). However, to be divisible by 5 with a R3, the number must end in a 3 or 8. So 63 is the first number that would match your first and last criteria. And to maintain that relationship, that sequence must increase by 55 each time (11x5). At 63 divided by 9, there is no remainder. Each time you add 55 to your sequence, and divide by 9, you would be increasing your remainder by 1 because 6x9=54, 55-54=1. Therefore, in order to get a remainer of 6 when divided by 9, you'll have to add six 55's to 63. 63 + (55x6) = 393 From that point, every time you add 495, you'll match the criteria. So 393+495=888, that works. 888+495=1383, that works. etc. Why 495? Because that is the product of your divisors - 5x9x11=495.