Question 1003340
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There is a number less than 3000 that when divided by 2 leaves a remainder of 1, when divided by 3 leaves a reminder of 2 when divided by 4 leaves a remainder of 3 and so on until 9. What is the number?
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Let N be that number.

Then  N-1  is divisible by  2;  is divisible by  3,  is divisible by  4;  is divisible by  5;  and so on until  9.


Thus  N-1  must be divisible by  {{{2^3*3^2*5*7}}} = 8*9*5*7 = 2520.


Now it is clear that the required number  N  less than  3000  does exist and is unique.

It is the number  2521.


Thank you for submitting this problem.