In the triangular array below, n different prime numbers p_1, p_2, \dots, p_n are written in the bottom row. Then in each box, we write the product of the two numbers below it. The number $K = p_1^{\alpha_1} p_2^{\alpha_2} \dotsm p_n^{\alpha_n}$ is written in the box at the very top. If \alpha_2 = 2, then how many numbers in the array are divisible by the number p_4?

Question

In the triangular array below, n different prime numbers p_1, p_2, \dots, p_n are written in the bottom row.  Then in each box, we write the product of the two numbers below it.  The number $K = p_1^{\alpha_1} p_2^{\alpha_2} \dotsm p_n^{\alpha_n}$ is written in the box at the very top.  If \alpha_2 = 2, then how many numbers in the array are divisible by the number p_4?

GPT-4o mini · Answer

To find how many numbers in the triangular array are divisible by $ p_4 $, we first need to understand how numbers are constructed in this array based on the given prime numbers $ p_1, p_2, \ldots, p_n $.

Starting with $ n $ different prime numbers, they are placed in the bottom row. Each subsequent row is filled by multiplying the two numbers that lie directly below it. This multiplication will be key in determining the divisibility by $ p_4 $.

Here's how we can break down the calculation step by step:

Understanding the Structure:
- The bottom row consists of $ n $ numbers, which are the primes: $ p_1, p_2, \ldots, p_n $.
- The next row, just above, will contain $ n-1 $ products: \[ p_1 p_2, , p_2 p_3, , p_3 p_4, \ldots, p_{n-1} p_n \]
- The next row contains $ n-2 $ products, and so forth.
Identifying the Boxes:
- The $ k $-th row from the bottom has $ n-k $ entries where each entry is a product of two primes from the row directly below it.
Counting Divisible Boxes:
- We need to check each box to see which products are divisible by $ p_4 $.
Bottom Row: The numbers in the bottom row (1st row) are $ p_1, p_2, p_3, p_4, \ldots, p_n $. Clearly, only $ p_4 $ is divisible by $ p_4 $.

2nd Row: The numbers are products of adjacent prime numbers:
- $ p_1 p_2 $
- $ p_2 p_3 $
- $ p_3 p_4 $ (This is divisible by $ p_4 $)
- $ p_4 p_5 $ (This is divisible by $ p_4 $)
- $ p_5 p_6 $
- Continue... until $ p_{n-1} p_n $
So in the second row, the products $ p_3 p_4 $ and $ p_4 p_5 $ contribute to our count.

3rd Row: The numbers here will be:
- $ (p_1 p_2)(p_2 p_3) = p_1 p_2^2 p_3 $
- $ (p_2 p_3)(p_3 p_4) = p_2 p_3^2 p_4 $ (This is divisible by $ p_4 $)
- $ (p_3 p_4)(p_4 p_5) = p_3 p_4^2 p_5 $ (This is divisible by $ p_4 $)
- $ (p_4 p_5)(p_5 p_6) $
- Continue until the end.
Thus, in the third row, both products involving $ p_4 $ contribute.
Continuing This Process: This pattern continues up the rows.
Summary:
- Bottom Row: 1 (only $ p_4 $)
- 2nd Row: 2 (from $ p_3 p_4 $ and $ p_4 p_5 $)
- 3rd Row: 2 (from $ (p_2 p_3)(p_3 p_4) $ and $ (p_3 p_4)(p_4 p_5) $)
- As you go upward, every time we can construct two new products that still involve $ p_4 $.

In general, for any $ n \geq 4 $, the number of boxes containing $ p_4 $ will be counted straightforwardly.

Putting this all together:

Total contributions through all rows can be summarized. Starting from the bottom which contributes $ 1 + 2 + 2 + \ldots $ until the apex.

Since each higher row progressively builds and keeps, the contribution will grow based on the positioning but strictly depends on not losing the involvement of $ p_4 $.

To directly answer the decisive queries:

The total number of elements in the array that are divisible by $ p_4 $ is ultimately linked back as $ 1 + 2 + 2 + \ldots $, making our answer:

\[ \boxed{5} \text{ or however matches until the sum ends} \]

In every layout, ensure to keep checking terms work till it exhausts updates up to the apex.