Question:What is the smallest number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7, when divided by 7 leaves a remainder of 6 and so on until when divided by 2 leaves a remainder of 1?
Answer:
The smallest such number is 2519.
The easiest way is to find the Least Common Multiple (LCM) of 2, 3, 4, 5, 6, 7, 8 and 9.
And subtract 1 from it.
The LCM of 2, 3, 4, 5, 6, 7, 8 and 9 is given by 2520. Hence, the required number is 2519.