Problem number six. Anna makes a puzzle where two squares of a common side do not contain the same number. Which piece should she use to complete her puzzle? To help us, let's make a drawing of a blank puzzle piece that would go in the empty space in the middle. The values on the squares of the puzzle are 1, 2, 3, 4, and 5. And even if we look at the answer choices, none of them contain any digits other than those five. And in fact, five is not used. If we look at the top square here, it has fives and a three next to it. So that would leave three different numbers that could go here. But if we look at the other end, we have a two, three, and a five right here, which means that on this end piece, there's only either a one or a four. If we look at the answer choices, none of them have a one at the end here. And only two of them have fours. So we already know the answer will be either be D or E. Cross out A, B, and C. Now let's look at this top square. There's a three and fives next to it, so it could have a one, two, or a four in it. Looking at the remaining answer choices, we have a two and a three. The three can't be in the square because there's a three on top here, so choice E doesn't work, which means the answer should be D. Let's just double check the other numbers. Can we have a three right here? Yes, we can. There's a one here, a four here, and then a two here. And then as long as this number isn't three, it can be one because here we have the four, not the one. There's a five on top and a two on the bottom. The answer is D.

puzzle

adjoining squares

number placement

solution analysis

elimination process