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Math Help - Divisibility puzzle.

  1. #1
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    Divisibility puzzle.

    A teacher writes a positive integer N on the board and asks her 20 students which numbers divide it.

    Student 1 says 1, student 2 says 2, ....., student 20 says 20. Only 2 students make mistakes, and they are consecutive. Which 2 students make the mistakes and what is the smallest possible value of N?

    Well, the mistake can't be 1-10, and it can't be a number that can be written as n=ab, a>1, b>1, (a,b)=1, and so this leaves the only two consecutive possibilities as 16 and 17. My problem is in finding the smallest possible N. Do you just multiply all of the remaining numbers that aren't multiples of each other? How can I show it is the smallest?
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  2. #2
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    Close. Do not simply multiply all the remaining numbers together. Instead, take the GCD of all of them. For example, 8|N, so 4|N by default, so 4 does not need to be "multiplied in" otherwise 32|N making 16|N giving you a contradiction.
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  3. #3
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    Ah thanks. I was stupidly trying to do it the opposite way (i.e., I tried it with multiplying 4 in but not 8 since 8 is a multiple of 4, but it clearly didn't work). Thanks!
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  4. #4
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    So would it just be 11*13*14*18*19*20?
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  5. #5
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    Oops, no no no, I meant:

    2*3*2*5*7*2*3*11*13*19?
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  6. #6
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    That is correct. It may make more sense to think of it as the product of all prime powers p^n<20, or N=2^3*3^2*5*7*11*13*19
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