# Problem 5. (3 points)

Vasya chose four numbers and for each pair, he calculated the logarithm of the larger number to the base of the smaller one. This resulted in six logarithms. Four of them are 15, 20, 21, and 28. What values can the largest of all six logarithms take?
# Answer: $28 ; 420$

## Solution:

Let the four original numbers be $x \leqslant y \leqslant z \leqslant t$. Denote $a=\log _{x} y, b=\log _{y} z, c=\log _{z} t$. Then $\log _{x} z=a b, \log _{y} t=b c, \log _{x} t=a b c$, which means our six logarithms are $a, b, c, a b, b c$ and $a b c$. The largest of these is $a b c$, and this is what we need to find.

Notice that among our four logarithms, none is the product of the other two. This means that in each triplet $(a, b, a b),(b, c, b c),(a, b c, a b c),(a b, c, a b c)$, at least one number is missing. Each of the six numbers appears in exactly two of these triplets, so to "break" all the triplets, we need to remove two numbers that do not appear together in any triplet, that is, the numbers we do not know are either $a$ and $c$, or $b$ and $a b c$, or $b c$ and $a b$.

Accordingly, we have one of the quadruples $(b, a b, b c, a b c),(a, c, a b, b c)$ and ( $a, b, c, a b c$). The third option is impossible because none of our four numbers is the product of the other three. For a quadruple of numbers to correspond to the first or second option, it is necessary and sufficient that the product of two numbers equals the product of the other two. This condition is satisfied: $15 \cdot 28=20 \cdot 21$.

In the first case, we have $b \cdot a b c=a b \cdot b c$, and $a b c$ is the largest of our four numbers. In the second case, $a \cdot b c=b \cdot a c$ and $a b c$ is precisely the product we are looking for. Therefore, we have two
possible answers: 28 and 420.