Subfield: Number Theory


Question: Five distinct integers are to be chosen from the set $\{1,2,3,4,5,6,7,8\}$ and placed in some order in the top row of boxes in the diagram. Each box that is not in the top row then contains the product of the integers in the two boxes connected to it in the row directly above. Determine the number of ways in which the integers can be chosen and placed in the top row so that the integer in the bottom box is 9953280000 .

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Solution:
Suppose that the integers in the first row are, in order, $a, b, c, d, e$.

Using these, we calculate the integer in each of the boxes below the top row in terms of these variables, using the rule that each integer is the product of the integers in the two boxes above:

$a$

| $b$ | $c$ | $c$ |  | $d$ |
| :---: | :---: | :---: | :---: | :---: |
| $a b^{2} c$ | $b c$ | $c d$ | $c$ |  |
|  | $a b^{3} c^{3} d$ | $b c^{2} d$ | $c d^{2} e$ |  |$\quad d e$

Therefore, $a b^{4} c^{6} d^{4} e=9953280000$.



Next, we determine the prime factorization of the integer 9953280000 :

$$
\begin{aligned}
9953280000 & =10^{4} \cdot 995328 \\
& =2^{4} \cdot 5^{4} \cdot 2^{3} \cdot 124416 \\
& =2^{7} \cdot 5^{4} \cdot 2^{3} \cdot 15552 \\
& =2^{10} \cdot 5^{4} \cdot 2^{3} \cdot 1944 \\
& =2^{13} \cdot 5^{4} \cdot 2^{3} \cdot 243 \\
& =2^{16} \cdot 5^{4} \cdot 3^{5} \\
& =2^{16} \cdot 3^{5} \cdot 5^{4}
\end{aligned}
$$

Thus, $a b^{4} c^{6} d^{4} e=2^{16} \cdot 3^{5} \cdot 5^{4}$.

Since the right side is not divisible by 7 , none of $a, b, c, d$, $e$ can equal 7 .

Thus, $a, b, c, d, e$ are five distinct integers chosen from $\{1,2,3,4,5,6,8\}$.

The only one of these integers divisible by 5 is 5 itself.

Since $2^{16} \cdot 3^{5} \cdot 5^{4}$ includes exactly 4 factors of 5 , then either $b=5$ or $d=5$. No other placement of the 5 can give exactly 4 factors of 5 .

Case 1: $b=5$

Here, $a c^{6} d^{4} e=2^{16} \cdot 3^{5}$ and $a, c, d, e$ are four distinct integers chosen from $\{1,2,3,4,6,8\}$. Since $a c^{6} d^{4} e$ includes exactly 5 factors of 3 and the possible values of $a, c, d$, e that are divisible by 3 are 3 and 6 , then either $d=3$ and one of $a$ and $e$ is 6 , or $d=6$ and one of $a$ and $e$ is 3 . No other placements of the multiples of 3 can give exactly 5 factors of 3 .

Case 1a: $b=5, d=3, a=6$

Here, $a \cdot c^{6} \cdot d^{4} \cdot e=6 \cdot c^{6} \cdot 3^{4} \cdot e=2 \cdot 3^{5} \cdot c^{6} \cdot e$.

This gives $c^{6} e=2^{15}$ and $c$ and $e$ are distinct integers from $\{1,2,4,8\}$.

Trying the four possible values of $c$ shows that $c=4$ and $e=8$ is the only solution in this case. Here, $(a, b, c, d, e)=(6,5,4,3,8)$.

Case 1b: $b=5, d=3, e=6$ We obtain $(a, b, c, d, e)=(8,5,4,3,6)$.

Case 1c: $b=5, d=6, a=3$

Here, $a \cdot c^{6} \cdot d^{4} \cdot e=3 \cdot c^{6} \cdot 6^{4} \cdot e=2^{4} \cdot 3^{5} \cdot c^{6} \cdot e$.

This gives $c^{6} e=2^{12}$ and $c$ and $e$ are distinct integers from $\{1,2,4,8\}$.

Trying the four possible values of $c$ shows that $c=4$ and $e=1$ is the only solution in this case. Here, $(a, b, c, d, e)=(3,5,4,6,1)$.

Case 1d: $b=5, d=6, e=3$ We obtain $(a, b, c, d, e)=(1,5,4,6,3)$.

Case 2: $d=5$ : A similar analysis leads to 4 further quintuples $(a, b, c, d, e)$.

Therefore, there are 8 ways in which the integers can be chosen and placed in the top row to obtain the desired integer in the bottom box.

Final answer: 8