Problem 973

Assume that the result 48 is obtained by multiplying the numbers on the first and second slip, the result 192 is obtained from the first and third slip, and the result 36 from the second and third slip.

Since $192=48 \cdot 4$, the number on the third slip must be four times the number on the second slip. Therefore, the number on the second slip, when multiplied by its quadruple, gives 36. This means that if we multiply this number by itself, we get a quarter of the previous result, i.e., 9. The number on the second slip is thus 3. On the third slip, there must be $3 \cdot 4=12$, and on the first slip, $48: 3=16$. The numbers on Eva's slips are therefore: 16, 3, and 12.

Another hint. Every natural number can be written as the product of two natural numbers in only a finite number of ways.

Another solution. Similarly to the previous solution, we assume that the result 48 is obtained by multiplying the numbers on the first and second slip, the result 192 is obtained from the first and third slip, and the result $36$ from the second and third slip.

The number 36 can be written as the product of two natural numbers in only five ways:

$$36=1 \cdot 36=2 \cdot 18=3 \cdot 12=4 \cdot 9=6 \cdot 6 \text{. }$$

From the problem, we know that the number on the third slip must be greater than the number on the second slip (the product of the numbers on the first and third slip is greater than the product of the numbers on the first and second). Therefore, we have only four possible pairs of numbers on the second and third slips. From the known values of the product of the numbers on the slips, we can calculate the number on the first slip in two ways, and if these results match, we have the solution.

(A dash indicates that the numbers cannot be divided without a remainder, i.e., the result is not a natural number.) We see that there is only one suitable option: the numbers on Eva's slips are 16, 3, and 12.