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'Exhaustion' printed from https://nrich.maths.org/
Eduardo and Arvind both looked for
solutions by restricting their search to the case $a< b< c$
and working out the limitations this gave them for values of $a,b$
and $c$. Both of them found a solution $a=3, b=4, c=5$ by this
method.
Kai rewrote $\left(1+\frac{1}{a}\right)$ as
$\frac{a+1}{a}$:
I first realised $\left(1+\frac{1}{a}\right)$ is the same as
$\frac{a+1}{a}$. So I set about trying to find to consecutive
non-prime numbers. I would then divide them into 2 and try
to find two other numbers
$\left(1+\frac{1}{b}\right)$ and $\left(1+\frac{1}{c}\right)$ that
multiply together to get this number. I found 8 and 9 first and put
that in: $2\div \frac{9}{8} = 2\times \frac{8}{9} = \frac{16}{9}$.
This number has a non-prime numerator and denominator. As
$\frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$, there is a
solution: $a = 8, b = 3, c = 3$
Kai used a similar method to find the
solution $a=15, b=2, c=4$.
Ming Chen and Thomas both used exhaustive
methods to find all the possibilities. Here is Thomas's
solution:
I tried to solve this by the process of exhaustion, using several
known facts.
First, a, b, and c are integers greater than zero: $a, b, c > 0
\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c} > 0$
so $1 + \frac{1}{a}, 1 + \frac{1}{b}$ and $1 + \frac{1}{c}$ are all
greater than $1$.
We can assume $a \leq b \leq c$ (because the three brackets can be
written in any order), which implies that $a> 1$, because if
$a=1$, $1 + \frac{1}{a}=2$. $1 + \frac{1}{b}$ and $1 + \frac{1}{c}$
are both strictly greater than $1$, so the product would be greater
than $2$.
Next, we find the greatest value for $a$.
If $a$ is the smallest, then $1 + \frac{1}{a}$ must be the largest
term. This can also be written $\frac{a+1}{a}$.
The largest possible value of $a$ is the case, $a = b= c$.
Any whole number value of $a$ greater than 3 will result in a
product less than 2: $(\frac{4}{3})^3 = 2.37...$ and
$(\frac{5}{4})^3 = 1.95...$
Thus we have proved that $2 \leq a \leq 3$, so $a$ must be either
$2$ or $3$.
Assuming that $a$ is 2, we have
$(1+\frac{1}{c})\times(1+\frac{1}{b}) = \frac{4}{3}$.
Using the same logic, $(\frac{b+1}{ b})^2 \geq 4/3$, thus the
greatest value for $b$ must be $6$, as $(\frac{7}{6})^2 = 1.36..
> \frac{4}{3}$ and $(\frac{8}{7})^2 = 1.31.. <
\frac{4}{3}$.
Similarly, $b > 3$, as $\frac{c+1}{c} > 1$, and for $b \leq
3$ we would need $c \leq 1$ for the product to be
$\frac{4}{3}$.
We have now proved $4 \leq b \leq 6$.
Now we just solve for $c$, using $\frac{3}{2}\times(1 +
\frac{1}{b}\times c = 2$,with $b=4,5,6$ giving us the triples $(2,
4, 15) (2, 5, 9) (2, 6, 7)$, and their permutations (as we assumed
$a \leq b \leq c$).
Finally, assuming that $a$ is now $3$,
$(1+\frac{1}{c})\times(1+\frac{1}{b}) = \frac{3}{2}$, and we follow
the same logic as above to restrict the value of $b$ to $3$ or $4$.
Then we solve for $c$ accordingly, giving us $(3, 3, 8), (3, 4, 5)$
and their permutations.
Therefore, we have proved that there are $5$ unique triples of $a,
b, c$ required to achieve
$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=2$
and there are in total $5 \times (3!)$ or 30 total triples by
interchanging values between $a, b,$ and $c$.