How many possible values of $(p+q+r)$ are there satisfying $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$, where $p$, $q$ and $r$ are natural numbers (not necessarily distinct)?

The correct option is Three.

We are asked to find the number of possible values of $(p+q+r)$ such that:

$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1$

where $p$, $q$, and $r$ are natural numbers ($p,q,r\∈\mathrm{\&mathbb;}N$).

Without loss of generality, we can assume that $1\≤p\≤q\≤r$.

Step 1: Determine the possible values for $p$

Since $p\≤q\≤r$, we have $\frac{1}{p}\≥\frac{1}{q}\≥\frac{1}{r}$.

Thus:

$1=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\≤\frac{1}{p}+\frac{1}{p}+\frac{1}{p}=\frac{3}{p}$

This implies $p\≤3$.

Also, if $p=1$, then $\frac{1}{q}+\frac{1}{r}=0$, which is impossible for natural numbers $q$ and $r$.

Therefore, the possible values for $p$ are $p=2$ and $p=3$.

Case 1: $p=2$

Substituting $p=2$ into the original equation:

$\frac{1}{2}+\frac{1}{q}+\frac{1}{r}=1\⇒\frac{1}{q}+\frac{1}{r}=\frac{1}{2}$

Since $q\≤r$, we have $\frac{1}{2}\≤\frac{2}{q}\⇒q\≤4$.

Since $q\≥p=2$, if $q=2$, $\frac{1}{r}=0$ (impossible). So $q\>2$.

Thus, $q$ can be $3$ or $4$:

1. If $q=3$:

$\frac{1}{r}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\⇒r=6$

Here, $(p,q,r)=(2,3,6)$. The sum is $p+q+r=2+3+6=11$.

2. If $q=4$:

$\frac{1}{r}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\⇒r=4$

Here, $(p,q,r)=(2,4,4)$. The sum is $p+q+r=2+4+4=10$.

Case 2: $p=3$

Substituting $p=3$ into the original equation:

$\frac{1}{3}+\frac{1}{q}+\frac{1}{r}=1\⇒\frac{1}{q}+\frac{1}{r}=\frac{2}{3}$

Since $q\≤r$, we have $\frac{2}{3}\≤\frac{2}{q}\⇒q\≤3$.

Since $q\≥p=3$, the only possibility is $q=3$.

If $q=3$:

$\frac{1}{r}=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\⇒r=3$

Here, $(p,q,r)=(3,3,3)$. The sum is $p+q+r=3+3+3=9$.

Conclusion:

The possible distinct values of $(p+q+r)$ are $9$, $10$, and $11$.

Therefore, there are exactly 3 possible values for $(p+q+r)$.