## riddle of the seats

An arithmetic quick riddle from The Riddler:

If an integer n is a multiple of every integer between 1 and 200, except for two consecutive ones, find those consecutive integers.

Since the highest power of 2 less than 200 is 2⁷=128 and since 127 is a prime number, the number

$2^6\times \prod_{i=0,i\ne 63}^{99} (2i+1)$

should work in that it contains all odd integers but 127, and all even numbers, but 128. Of course a smaller number that avoids duplicates by only considering the 44 primes other than 127 and 2 to a power that keep them less than 200 is also valid. Which gives a number of the order of 1.037443 10⁸⁵.

### One Response to “riddle of the seats”

1. The same puzzle appeared as a question on math.stackexchange a few weeks later.

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