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Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135. ### Dodgy Proofs

What is wrong with these dodgy proofs?

# Exhaustion

##### Age 16 to 18 Challenge Level:

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 $c$.