All numbers can be represented as a prime factorisation:
p^i * q^j * ... (a product of some primes to some powers)
It is not possible for the solution to have a number with all powers being zero, because that is 1, and any number is a multiple of 1. So all solutions must have some non-zero powers.
It is not possible for the solution to have a number that has more than one prime in its factorisation, which can be seen as follows.
If the number is a multiple of all other numbers except some chosen p^i * q^j for non-zero i and j, then of course it is a multiple of p^i and q^j, and, p^i and q^j being co-prime, it also is a multiple of p^i * q^j.
So the solution can only consist of numbers like p^i for some prime p and non-zero i.
Since the seat numbers are adjacent, one of them is even. Given the conclusions above, it also means it must be not just an even number, but necessarily a power of 2.
If the number were divisible by the biggest power of 2 that is less than 200, it is also divisible by any smaller power of 2. So we have to pick one seat with the biggest power of 2 that is less than 200, which is 128.
The other seat must necessarily be a power of a single prime, too. So we check 127 and 129. 129 is not a prime, 127 is, so we have a unique solution.
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winterfish
no subject
Date: 2019-10-12 09:15 am (UTC)p^i * q^j * ... (a product of some primes to some powers)
It is not possible for the solution to have a number with all powers being zero, because that is 1, and any number is a multiple of 1. So all solutions must have some non-zero powers.
It is not possible for the solution to have a number that has more than one prime in its factorisation, which can be seen as follows.
If the number is a multiple of all other numbers except some chosen p^i * q^j for non-zero i and j, then of course it is a multiple of p^i and q^j, and, p^i and q^j being co-prime, it also is a multiple of p^i * q^j.
So the solution can only consist of numbers like p^i for some prime p and non-zero i.
Since the seat numbers are adjacent, one of them is even. Given the conclusions above, it also means it must be not just an even number, but necessarily a power of 2.
If the number were divisible by the biggest power of 2 that is less than 200, it is also divisible by any smaller power of 2. So we have to pick one seat with the biggest power of 2 that is less than 200, which is 128.
The other seat must necessarily be a power of a single prime, too. So we check 127 and 129. 129 is not a prime, 127 is, so we have a unique solution.