# Problem of the Week Problem E and Solution Three Lists

## Problem

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 $$(a, b)$$.

## Solution

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:

• Case $$1$$: $$49$$ was obtained in the third list by multiplying $$1$$ and $$49$$.

Since the number $$1$$ is in one of the lists, we must have either $$a = 1$$ or $$b = 1$$. The condition of $$a<b$$ means we must have $$a=1$$. This means that List $$1$$ must be $1,2,3,4,5,6$ and the number $$49$$ must appear somewhere in List $$2$$.

Therefore, List $$2$$ is one of the following six lists:

$44, 45, 46, 47, 48, 49$ $45, 46, 47, 48, 49, 50$ $46, 47, 48, 49, 50, 51$ $47, 48, 49, 50, 51, 52$ $48, 49, 50, 51, 52, 53$ $49, 50, 51, 52, 53, 54$

Notice that $$4\times 48=192 = 64\times 3$$. Since $$4$$ is in List $$1$$, and no number in the third list can be a multiple of $$64$$, then List $$2$$ cannot contain the number $$48$$. This leaves just one possibility for List $$2$$: $49, 50, 51, 52, 53, 54$

This case gives exactly one possibility for the pair $$(a,b)$$, namely $$(1,49)$$.

We can verify that the third list for the pair $$(a,b)=(1,49)$$ actually satisfies the second and third conditions. For the second condition, we note that $$64 = 2^{6}$$ and that we can get at most two factors of $$2$$ from a number in List $$1$$ and at most two factors of $$2$$ from a number in List $$2$$. It follows that any product in the third list will have at most $$4$$ factors of $$2$$, and hence cannot be a multiple of $$64$$. For the third condition, we note that $$2 \times 49 = 98$$ is in the third list and is greater than $$75$$.

• Case $$2$$: $$49$$ was obtained in the third list by multiplying $$7$$ and $$7$$.

In this case, we know that the number $$7$$ must appear in both List $$1$$ and List $$2$$. In order for this to happen we need to have $$2 \leq a\leq 7$$ and $$2 \leq b \leq 7$$. Since $$a < b$$, we actually must have $$3 \leq b \leq 7$$. (The smallest $$a$$ can be is $$2$$ and so $$b$$ must be at least one more than that.)

Since $$3 \leq b \leq 7$$, List $$2$$ must contain the number $$8$$. This means that to satisfy the second condition, List $$1$$ cannot contain the number $$8$$. Therefore, we must have $$a=2$$. This means that List $$1$$ must be $2,3,4,5,6,7$

Since $$7 \times 10 = 70$$ and $$7 \times 11 = 77$$, the third list can only satisfy the third condition if List $$2$$ contains a number at least as large as $$11$$. This means we cannot have $$b =3$$, $$b=4$$, or $$b=5$$, leaving the only possible values to be $$b = 6$$ or $$b=7$$. These values produce the following possibilities for List $$2$$:

$6,7,8,9,10,11$ or $7,8,9,10,11,12$

Therefore, this case gives two additional possibilities for the pair $$(a,b)$$, namely $$(2,6)$$ and $$(2,7)$$.

We can verify that the third list for each of the the pairs $$(a,b)=(2,6)$$ and $$(a,b)=(2,7)$$ satisfies the second and third conditions using an argument similar to the one given in Case $$1$$.

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 $$(2,7)$$.