If the product of three consecutive positive integers is $15600$ then the sum of the squares of these integers is

We need to find three consecutive positive integers whose product is 15600, then find the sum of their squares.

Let's call our three consecutive integers: $(n-1)$, $n$, and $(n+1)$

So we have: $(n-1) \times n \times (n+1) = 15600$

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To solve this systematically, we find the prime factorization of 15600:

$15600 = 156 \times 100 = 156 \times 10^2 = 156 \times (2 \times 5)^2$

$156 = 4 \times 39 = 4 \times 3 \times 13 = 2^2 \times 3 \times 13$

Therefore: $15600 = 2^2 \times 3 \times 13 \times 2^2 \times 5^2 = 2^4 \times 3 \times 5^2 \times 13$

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Here's the key insight: Since 15600 ends in two zeros, it's divisible by 100 = 4 × 25.

For three consecutive integers, exactly one of them must be divisible by each prime factor's highest power. Since we need a factor of 25 = 5², one of our three consecutive integers must be divisible by 25.

If one number is a multiple of 25, we can factor it out to simplify our search.

$15600 ÷ 25 = 624$

This means: $25 \times \text{(two other consecutive integers)} = 15600$

So the two integers adjacent to our multiple of 25 must multiply to give 624.

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We need to find which multiple of 25 works. Let's say our middle number is 25.

Then we need: $24 \times 26 = ?$

$24 \times 26 = 624$

Perfect! This confirms our three consecutive integers are: 24, 25, 26

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Now we find: $24^2 + 25^2 + 26^2$

$24^2 = 576$

$25^2 = 625$

$26^2 = 676$

$24^2 + 25^2 + 26^2 = 576 + 625 + 676 = 1877$

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When dealing with consecutive integers and their products, we look for:

Prime factorization of the given product

Special factors like perfect squares (25, 49, etc.)

Systematic verification of our answer

Answer: 1877