Abe challenges Bee to determine a

*3-digit positive integer* N.

It is known that the

*number formed by the last two digits of N* when divided by 9, yields a remainder of 3.

Abe makes the following statements, precisely one of which is false:

- N divided separately by each of 2, 4, 6, and 8 yields a remainder of 1.
- N divided separately by each of 5 and 7 yields a remainder of 2.
- N divided separately by each of 5 and 11 yields a remainder of 3.

Determine the value of N from the above statements and given clues.

1) Statements 2 and 3 are contradictory, so 1 is true. Therefore, N = 24x + 1

2) From statement 0, N = 100y + 9z + 3.

3) If statement 3 is true, then (100y + 9z +3) mod 11 = 3 = (100y + 9z + 3) mod 5

Then 9z = 0 mod 5 and y = 2z mod 11.

But 9z = 0 mod 5 implies z = 0 or 5, and this fails if y = 2z mod 11.

So statement 3 is false.

4) So statement 2 is true. Then (100y + 9z +3) mod 5 = 2

So z = 1 mod 5. z = 1 or 6.

If z = 1, then N is even, and cannot be 24x + 1

So z must be 6.

Then 100y + 57 = 1 mod 24.

4y = 16 mod 24

y = 4

So the final answer is 457.

Checking, 457 mod 7 = 2

*Edited on ***February 16, 2016, 11:56 am**