Introduction to the Problem

In this article, we explore a mathematical problem involving four positive integers a, b, c, and d. We are given that the product of these numbers, i.e., abcd, equals 10!. This problem requires us to find the smallest possible sum of these integers with the given constraint. The key to solving this problem lies in understanding the properties of prime factorization and the application of the AM-GM (Arithmetic Mean - Geometric Mean) inequality.

Understanding the Constraints

Given the equation abcd 10!, the first step is to factorize 10!:

10! 3628800

Factorization of 10! is as follows:

10! 2^8 x 3^4 x 5^2 x 7

For the sum of a, b, c, and d to be minimized, each of the factors should be as close to each other as possible. The AM-GM inequality states that the arithmetic mean is always greater than or equal to the geometric mean.

Application of AM-GM Inequality

Let's apply the AM-GM inequality:

AM ≥ GM

frac14;abcd ≥ (abcd)^(1/4)

To find the approximate value, we can calculate 10!^(1/4) as follows:

10!^(1/4) ≈ 44.27

This implies that each of a, b, c, and d should be close to this value.

Optimization Approach

By setting each of the factors approximately equal to 44, we can choose the closest possible values of 48, 45, 42, and 40. Here is the breakdown:

a 48 (3 x 2^4) b 45 (5 x 3^2) c 42 (7 x 3 x 2) d 40 (5 x 2^3)

The sum of these values gives us the smallest possible sum of a, b, c, and d. Let's calculate the sum:

48 45 42 40 175

This confirms that 175 is indeed the smallest possible sum under the given conditions.

Alternatives and Code Implementation

Another approach involves checking different combinations of factors of 10! to find the smallest sum. Given the equation a b c 29 and abcd 10!, we can use a computational approach to find the possible solutions. Here is a simple code snippet to find these solutions:

#include bits/stdc  .husing namespace std;int main() {    int c, b, a;    for (c  1; c  60; c  ) {        for (b  c 1; b  60; b  ) {            for (a  b 1; a  60; a  ) {                if (72 * a * b * c / (a   b   c)  29) {                    cout  a  " "  b  " "  c  endl;                }            }        }    }    return 0;}

The output of this code will give the possible triplets (a, b, c) that satisfy the given conditions. For instance, the output may include:

(36, 24, 3) (36, 8, 4) (24, 9, 4) (9, 8, 6)

Among these, the sum of (9, 8, 6) is the smallest, which is 23. However, since we need to multiply by d to get 10!, we should verify that each triplet can be part of a valid solution set.

Conclusion

The smallest possible sum of positive integers a, b, c, and d given that their product is 10! is 175. This solution leverages the AM-GM inequality and careful factorization. The code implementation provided helps in verifying alternative solutions and understanding the constraints better.