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

[anemone]

Let $p,\,q,\,r,\,s,\,t$ be positive integers such that $pqrst=p+q+r+s+t$.

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

 

[mente oscura]

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.

Spoiler

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.

 

[anemone]

Hello.

Spoiler

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.:)

 

[lfdahl]

Spoiler

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 $n=2$).

 

[Opalg]

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)

 

[anemone]

Hello.

Spoiler

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.

Spoiler

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!;)

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.

Solution(Suggested by other):

Spoiler

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.