MHB Maximum Value of Positive Integers in a Product-Sum Equation

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The discussion revolves around the equation pqrst = p + q + r + s + t, where p, q, r, s, and t are positive integers. The maximum value of the largest integer among them is determined to be 5. Participants share solutions and insights, with references to an article by Michael W. Ecker that explores the conditions under which the sum of positive integers equals their product. The article states that a positive integer v can be expressed as both a sum and a product if and only if v is not prime. Overall, the thread highlights collaborative problem-solving and the sharing of mathematical resources.
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Let $p,\,q,\,r,\,s,\,t$ be positive integers such that $pqrst=p+q+r+s+t$.

Find the maximum possible value of max $\{p,\,q,\,r,\,s,\,t\}$.
 
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anemone said:
Let $p,\,q,\,r,\,s,\,t$ be positive integers such that $pqrst=p+q+r+s+t$.

Find the maximum possible value of max $\{p,\,q,\,r,\,s,\,t\}$.

Hello.

I have not found another way to solve the problem:p

Suppose: p >1p=\dfrac{q+r+s+t}{qrst-1}

It follows:

qrst \cancel{=} 1 \rightarrow{}q+r+s+t \ge{}4

qrst \le{}4

If \ qrst >4 \rightarrow{} p <2

Example: p=2, \ q=2, \ r=2, \ s=1, \ t=1

Therefore:

max(p) \rightarrow{} min(q)

Solution:

p=5 \ , \ q=2 \ , \ r=1 \ , \ s=1 \ , \ t=1

p+q+r+s+t=10 \ and \ pqrst=10

Note: the variables I considerate interchangeable

Regards.
 
mente oscura said:
Hello.

I have not found another way to solve the problem:p

Suppose: p >1p=\dfrac{q+r+s+t}{qrst-1}

It follows:

qrst \cancel{=} 1 \rightarrow{}q+r+s+t \ge{}4

qrst \le{}4

If \ qrst >4 \rightarrow{} p <2

Example: p=2, \ q=2, \ r=2, \ s=1, \ t=1

Therefore:

max(p) \rightarrow{} min(q)

Solution:

p=5 \ , \ q=2 \ , \ r=1 \ , \ s=1 \ , \ t=1

p+q+r+s+t=10 \ and \ pqrst=10

Note: the variables I considerate interchangeable

Regards.

Thanks mente for your solution (yes, it has the correct answer) and I will wait a bit before posting the two solutions that I have to share with you and the community here...I believe others might want to take a stab at it as well.:)
 
Let $P = p\cdot q\cdot r\cdot s\cdot t$, and $S = p+q+r+s+t$, where $p,q,r,s,t \in \mathbb{N}$

WLOG I can choose a $p$-value and let $q,r,s,t$ all be equal to $1$. Once $p$ is fixed, I will look for the smallest possible increments of $P$ by varying $q$.

If $p = 1$ … the smallest possible values of $P$ are: $1,2,3, …$ and $S = 5,6,7,…$.

If $p = 2$ … the smallest possible values of $P$ are: $2,4,6,…$ and $S = 6,7,8,…$



It is easy to see, that $P \neq S$ for $p > 5$.

In general:

For a given $p$ the smallest possible values of $P$ are: $p, 2p, 3p, …$ and the corresponding

$S$-values are $p+4, p+5, p+6, …$.

Thus, I´m looking for integer solutions of $p$: $np = p + 3 + n, \;\; n = 1,2,3, …$

Or $ p = \frac{n+3}{n-1}$ for $n > 1$. There are exactly three integer solutions the largest of which is $5$ (for $n=2$).
 
I found an interesting article "When does a sum of positive integers equal their product?" by Michael W. Ecker (Math. Mag. 75 (2002) 41–47 – you can find it online if you have access to JSTOR). It gives the answer to the problem in this thread, and much more besides.


Sample result: Given a positive integer $v$, the equation $v = x_1 +x_2 + \ldots + x_n = x_1x_2\cdots x_n$ (with $n>1$) has a solution if and only if $v$ is not prime. The proof is easy enough that readers of this subforum ought to be able to find it for themselves. (Wait)
 
mente oscura said:
Hello.

I have not found another way to solve the problem:p

Suppose: p >1p=\dfrac{q+r+s+t}{qrst-1}

It follows:

qrst \cancel{=} 1 \rightarrow{}q+r+s+t \ge{}4

qrst \le{}4

If \ qrst >4 \rightarrow{} p <2

Example: p=2, \ q=2, \ r=2, \ s=1, \ t=1

Therefore:

max(p) \rightarrow{} min(q)

Solution:

p=5 \ , \ q=2 \ , \ r=1 \ , \ s=1 \ , \ t=1

p+q+r+s+t=10 \ and \ pqrst=10

Note: the variables I considerate interchangeable

Regards.

lfdahl said:
Let $P = p\cdot q\cdot r\cdot s\cdot t$, and $S = p+q+r+s+t$, where $p,q,r,s,t \in \mathbb{N}$

WLOG I can choose a $p$-value and let $q,r,s,t$ all be equal to $1$. Once $p$ is fixed, I will look for the smallest possible increments of $P$ by varying $q$.

If $p = 1$ … the smallest possible values of $P$ are: $1,2,3, …$ and $S = 5,6,7,…$.

If $p = 2$ … the smallest possible values of $P$ are: $2,4,6,…$ and $S = 6,7,8,…$



It is easy to see, that $P \neq S$ for $p > 5$.

In general:

For a given $p$ the smallest possible values of $P$ are: $p, 2p, 3p, …$ and the corresponding

$S$-values are $p+4, p+5, p+6, …$.

Thus, I´m looking for integer solutions of $p$: $np = p + 3 + n, \;\; n = 1,2,3, …$

Or $ p = \frac{n+3}{n-1}$ for $n > 1$. There are exactly three integer solutions the largest of which is $5$ (for $n=2$).

Hi mente oscura and lfdahl,

Thank you for the solution and for participating in my challenge thread!:) As I have mentioned before, $5$ is the correct answer, bravo!;)

Opalg said:
I found an interesting article "When does a sum of positive integers equal their product?" by Michael W. Ecker (Math. Mag. 75 (2002) 41–47 – you can find it online if you have access to JSTOR). It gives the answer to the problem in this thread, and much more besides.


Sample result: Given a positive integer $v$, the equation $v = x_1 +x_2 + \ldots + x_n = x_1x_2\cdots x_n$ (with $n>1$) has a solution if and only if $v$ is not prime. The proof is easy enough that readers of this subforum ought to be able to find it for themselves. (Wait)


Thanks, Opalg for your reply and sharing with us what you've found.:o

Solution(Suggested by other):
Suppose $t \ge s \ge r \ge q \ge p$.

We need to find the maximum of $t$.

Since $t<p+q+r+s+t \le 5t$, then $t<pqrst \le 5t$, i.e. $1<pqrs \le 5$.

Hence, $(p,\,q,\,r,\,s)=(1,\,1,\,1,\,2),\,(1,\,1,\,1,\,3),\,(1,\,1,\,1,\,4),\,(1,\,1,\,2,\,2),\,(1,\,1,\,1,\,5)$ which leads to maximum $t$ of 5.
 
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