Can Seven Integers with a Subtraction of 3 Result in a Product 13 Times Larger?

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The discussion centers on the mathematical problem of whether seven positive integers, when each reduced by 3, can yield a product that is exactly 13 times greater than the product of the original integers. Participants confirm that this scenario is indeed possible and express interest in identifying multiple solutions. While one solution is acknowledged, the exact number of solutions remains uncertain, prompting further inquiry into alternative methods to derive additional solutions.

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Take seven positive integers and subtract 3 from each of them. Can the product of the resulting numbers be exactly 13 times the product of the original numbers?
 
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Evgeny.Makarov said:
Take seven positive integers and subtract 3 from each of them. Can the product of the resulting numbers be exactly 13 times the product of the original numbers?
Point of confusion:
If you take value away from a given set of numbers then the product of the first set will be larger than the second set automatically. The only way out of it is to have a "1" in the original set, which would put a "-2" in the second set, doubling the absolute value. That doesn't sound to me like it is a feature you wanted in the problem. Am I missing something?

-Dan
 
Dan, your understanding is correct. And since the situation you are describing is inevitable, it should be considered.
 
My solution

none of the numbers in 1st set can be 3 then 2nd product is zero and all cannot be >3 as product becomes smaller.
some of the numbers have to be 1 or 2. and number of numbers 1 or 2 has to be even or product is -ve.
as product of second set of numbers is multiple of 13 so one of the numbers shall be 13 or a multiple of the same.
so we have one number in 1st set 16 and corresponding number in 2nd set 13.
we need one or more numbers so as to mulitply 2nd set by 16 and pair of (1,-2) gives multiplication by -2 so if we take 4
times product by 16 is compensated
so 1st set (16,1,1,1,1,x,y) and 2nd set(13,-2,-2,-2,-2,x-3,y-3). we need to choose x,y such that xy = (x-3)(y-3) and x= 1, y = 2
gives x-3 = -2 and y-3 = -1 so the criteria is met
so numbers = (16,1,1,1,1,1,2)

Hence ans is Yes
 
Nice job. I know of one more solution, but I am not sure of how many there are.
 
Evgeny.Makarov said:
Nice job. I know of one more solution, but I am not sure of how many there are.

I could not find the 2nd solution. I would like to know it.
 
$$1^4\cdot29\cdot61\cdot64\mapsto(-2)^4\cdot26\cdot58\cdot61$$.
 

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