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Problem 738

AMC 10/12, early questions

Combinatorics Difficulty 3.5 Multiple choice CEMC Fermat · Canada · 2024

Francesca put one of the integers $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , $9$ in each of the nine squares of the $3$ by $3$ grid shown. No integer is used twice. She calculated the product of the three integers in each row and wrote the products to the right of the corresponding rows. She calculated the product of the integers in each column and wrote the products below the corresponding columns. Finally, she erased the integers from the nine squares.Which integer was in the square marked $N\,$ ?

Pick one

Official solution

Of the row and column products, only $135$ and $160$ are divisible by $5$ . This means that $5$ must go in the square in the 2nd row, 3rd column. Of the row and column products, only $21$ and $56$ are divisible by $7$ . This means that $7$ must go in the square in the 1st row, 1st column. Of the row and column products, only $108$ and $135$ are divisible by $9$ . This means that $9$ must go in the square in the 2nd row, 2nd column. So far, this gives the following grid: [[IMAGE0]] In the 2nd row, the product is $135$ which means that the missing entry is $\frac{135}{5 \cdot 9} = 3$ . In the 1st column, the product is $21$ which means that the missing entry is $\frac{21}{7 \cdot 3} = 1$ . The 3rd row, whose product is $48$ , thus includes $1$ and two more integers between $1$ and $9$ . The only divisor pair of $48$ with both divisors less than $10$ is $48 = 6 \cdot 8$ .

Since $8$ is not a divisor of $108$ , then $N$ must be $6$ .

We can complete the square as follows:

Figure for this problem

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