Exhaustion

327 /www/nrich/html/content/00/09/15plus4/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> Find the positive integer solutions of the equation <div style="text-align: center;">\[ \left(1 + \frac{1}{a}\right)\left(1 + \frac{1}{b}\right) \left(1+ \frac{1}{c}\right) = 2. \]</div> <div style="text-align: center;"> </div> <div style="text-align: center;"> </div> <div class="framework">This problem was taken from the Hungarian magazine KoMaL. There are many other challenging problems in English on <a href="http://www.cs.elte.hu/komal/index.e.html">the KoMal website</a>.</div></mdoxml> <mdoxml version="1.0"> 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 $c$. </mdoxml> <mdoxml version="1.0"> </mdoxml> <mdoxml version="1.0">Perhaps you might like to try the problem <a href="/801">Unit Fractions</a> first. What value does the expression take for small values of a? What value does the expression take for large values of a? When you try a particular value for a what can you say about b? Once you have narrowed down the possible values, you have to test all possible cases; this is known as proof by exhaustion.</mdoxml> 2 4 0 0 0 1 0 Exhaustion Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2 Algebra Inequality/inequalities Using, Applying and Reasoning about Mathematics Working systematically Using, Applying and Reasoning about Mathematics Proof by exhaustion Algebra Diophantine equations Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof