Library / CEMC Gauss (Grade 7) /435 of 462

CEMC Gauss (Grade 7), 2016

Number theory Difficulty 3.7 AMC 10/12 Find the answer Canada

In the diagram, four different integers from 1 to 9 inclusive are placed in the four boxes in the top row.

The integers in the left two boxes are multiplied and the integers in the right two boxes are added and these results are then divided, as shown. The final result is placed in the bottom box. Which of the following integers cannot appear in the bottom box?

Pick one

Solution

We begin by naming the boxes as shown.

Of the five answers given, the integer which cannot appear in box $M$ is 20. Why?

Since boxes $F$ and $G$ contain different integers, the maximum value that can appear in box $K$ is $8\times9=72$ .

Since boxes $H$ and $J$ contain different integers, the minimum value that can appear in box $L$ is $1+2=3$ .

Next, we consider the possibilities if 20 is to appear in box $M$ .

If 3 appears in box $L$ (the minimum possible value for this box), then box $K$ must contain 60, since $60\div3=20$ .

However, there are no two integers from 1 to 9 whose product is 60 and so there are no possible integers which could be placed in boxes $F$ and $G$ so that the product in box $K$ is 60.

If any integer greater than or equal to 4 appears in box $L$ , then box $K$ must contain at least $4\times20=80$ .

However, the maximum value that can appear in box $K$ is 72.

Therefore, there are no possible integers from 1 to 9 which can be placed in boxes $F,G,H,$ and $J$ so that 20 appears in box $M$ .

The diagrams below demonstrate how each of the other four answers can appear in box $M$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.