Problem
ALG-B1-M02-P028 Seven Factors and a Hidden Coefficient
In a product of seven positive integers, each factor was decreased by \(4\). Could the new product become exactly \(80\) times the original one? If yes, give an example and justify it.
Try taking five factors equal to \(1\), one factor equal to \(3\), and denoting the last factor by \(a\).
The task is not about brute force; it is about controlling the sign. If we take five ones and one three, then after decreasing by \(4\) we get the factors \(-3,-3,-3,-3,-3,-1\), whose product is positive and equals \(243\).
Let the original numbers be \(1,1,1,1,1,3,a\). The original product is \(3a\), while the new product is \(243(a-4)\).
We want an \(80\)-fold increase:
\[243(a-4)=80\cdot3a.\]
Thus \(243a-972=240a\), so \(a=324\). Check: the original product is \(972\), the new product is \(243\cdot320=77760\), and \(80\cdot972=77760\). Therefore such a set exists.
The level is higher: one must choose not only the number of ones, but also an extra small factor so that the sign and coefficient work.