Hints:

1. How large can the greatest of these numbers be?

2. Try organizing the numbers in increasing order to better understand the relationships.

3. Suppose x is the largest number. Then the sum of all five numbers can be no greater than 5x.

4. That means the product of the other four numbers must be less than or equal to 5, which severely limits the possibilities.

Solution:

When examining this problem, it helps to note that as the numbers increase, the sum grows much more slowly than the product. To get traction, we can relate the sum to the product.

Let the five numbers be A,B,C,D,E, and assume they are in non-increasing order, so that A is the largest (possibly tied with others).

Since A≥B≥C≥D≥E, we have:

A+B+C+D+E≤5A

But we are told:

A⋅B⋅C⋅D⋅E=A+B+C+D+E

Combining both:

A⋅B⋅C⋅D⋅E≤5A⇒B⋅C⋅D⋅E≤5

This gives us a major simplification: since B,C,D,E are positive integers, and their product is ≤ 5, there are very few possibilities. Let’s examine each case:

Case-by-case analysis (Values of B, C, D, E):

• 5,1,1,1

A⋅5=A+5+1+1+1⇒5A=A+8⇒4A=8⇒A=2

But this contradicts A≥5 . Invalid.

• 4,1,1,1

A⋅4=A+4+1+1+1⇒4A=A+7⇒3A=7⇒A=7:3​

Not an integer.  Invalid.

• 3,1,1,1

A⋅3=A+3+1+1+1⇒3A=A+6⇒2A=6⇒A=3

Valid. Solution: 3,3,1,1,1
Check:

3+3+1+1+1=9, 3⋅3⋅1⋅1⋅1=9

• 2,1,1,1

A⋅2=A+2+1+1+1⇒2A=A+5⇒A=5

Valid. Solution: 5,2,1,1,1
Check:

5+2+1+1+1=10, 5⋅2⋅1⋅1⋅1=10

• 1,1,1,1

A⋅1=A+1+1+1+1⇒A=A+4⇒0=4 

Contradiction. Invalid.

• 2,2,1,1

A⋅4=A+2+2+1+1⇒4A=A+6⇒3A=6⇒A=2 

Valid. Solution: 2,2,2,1,1 
Check:

2+2+2+1+1=8,2⋅2⋅2⋅1⋅1=8 

Conclusion

There are three valid solutions where the product equals the sum:

1. {3, 3, 1, 1, 1}

2. {5, 2, 1, 1, 1}

3. {2, 2, 2, 1, 1}

Therefore, the largest number that can appear in any valid solution is: 5​

​

Exploration:

A similar analysis can be extended to any number of values, though the complexity increases as the number of terms grows.

Let’s look at small cases first:

• Two numbers:
  2×2=2+2=4

• Three numbers:
  3×2×1=3+2+1=6

• Four numbers:
  4×2×1×1=4+2+1+1=8

These specific examples show a pattern:
In each case, the product equals the sum when the numbers include:

• The largest number n,

• A 2, and the remaining values are 1’s.

Beyond these specific answers, after doing a number of examples it is “clear” that when there are n numbers,
the largest number will always be “n” associated with the solution n x 2 x 1 x 1 x … x 1 = n + 2 + 1 + 1 + … + 1.

Conclusion

This pattern shows that for any number of terms n, there is at least one solution in which the largest number is exactly n, and the remaining values are 2 and several 1’s. This insight helps in predicting and constructing valid solutions across varying group sizes.