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Problem of the Week
Problem E and Solution
Three Lists


Ameya has two lists, List \(1\) and List \(2\), which each have six entries that are consecutive positive integers. The smallest entry in List \(1\) is \(a\) and the smallest entry in List \(2\) is \(b\), and \(a < b\).

Ameya creates a third list, List \(3\). The thirty-six entries in List \(3\) come from the product of each number in List \(1\) with each number of List \(2\). (There could be repeated numbers in List \(3\).)

Suppose that List \(3\) has \(49\) as an entry, has no entry that is multiple of \(64\), and has an entry larger than \(75\). Determine all possible pairs \((a, b)\).


We will start by considering what the first condition tells us about the values of \(a\) and \(b\), as it seems to be the most restrictive of the three conditions.

The first condition tells us that \(49\) must be the product of an integer from List \(1\) and an integer from List \(2\). Since \(49 = 7^{2}\), \(7\) is prime, and all integers in the two lists are positive, these integers must be either \(1\) and \(49\), or \(7\) and \(7\).

Note: It is not possible for \(49\) to be obtained in both of these ways at once because if a list contains \(49\), then it cannot also contain \(7\). However, knowing this will not be important for our solution.

We will find all possible values of \(a\) and \(b\) by considering the two cases separately:

Combining the two cases, we conclude that there are exactly three pairs, \((a,b)\), that satisfy all three conditions. They are \((1,49)\), \((2,6)\), and \((2,7)\).