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I've been given the following puzzle

Let $a_1, a_{100}$ be given real numbers. Let $a_i=a_{i-1}a_{i+1}$ for $2\leq i \leq 99$. Further suppose that the product of the first $50$ is $27$, and the product of all the $100$ numbers is also $27$.

Find $a_1+a_2$.

I tried the following, looking at the sequence for a moment we see: $$ a_2=a_1\, a_3\\ a_3=a_2\,a_4\\ \vdots\\ a_{99}=a_{98}\,a_{100} $$ So $$a_2=\frac 1 {a_{99}} \prod_i a_i =\frac {27}{a_{99}}$$

Looking at the other elements, we find that $$a_3=\frac {a_2} {a_1}, a_4=\frac {a_2}{a_1 a_2}, a_5=\frac {a_2}{a_1a_2a_3},\dots , a_n=\frac {a_2}{\prod_{i=1}^{n-2} a_i}$$

Then, $$27=\prod_i a_i=\prod_{1\leq i\leq 100} \frac {a_2}{\prod_{k=1}^{i-1}a_i}$$

The last product I believe is much more complicated than what I should have gotten...

Has anyone ran into this puzzle before?

YoTengoUnLCD
  • 13,384

3 Answers3

9

Note that since $a_i=a_{i-1}a_{i+1}$, then, for $i\geqslant 3$, we also have $a_{i-1}=a_{i-2}a_i$. Plug this in the first formula, we have $a_i=a_{i-2}a_ia_{i+1}\Rightarrow a_{i-2}a_{i+1}=1\Rightarrow a_{i+3}=\frac{1}{a_i}.$

Then, the sequence is in the form of $a_1=a,a_2=ab,a_3=b,a_4=\frac{1}{a},a_5=\frac{1}{ab},a_6=\frac{1}{b},a_7=a,\dots$.

$a_{i+6}=a_i$. So $a_{50}=ab$ and $a_{100}=\frac{1}{a}$. Hence, $1=\prod_{i=51}^{100}=\frac{b}{a}\Rightarrow a=b$. Also, $27=\prod_{i=1}^{50}=a\cdot (ab)=a^3\Rightarrow a=3\Rightarrow a_1+a_2=3+9=12$.

Qingzhong Liang
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5

Such a sequence is periodic with period length 6, one cycle $$ A, B, \frac{B}{A}, \frac{1}{A}, \frac{1}{B}, \frac{A}{B} $$ Note that the product of all six is $1.$ Therefore the product of the first 96 is one, and 27 is $\frac{B^2}{A}.$ The product of the first 48 is one, and 27 is also $AB.$ That is $$ 27 A = B^2, $$ $$ 27 = AB.$$ From $$ 729 = 27AB = (27A)B = B^3$$ I get $729 = B^3 $ and $$ B = 9. $$ Then $$ A = 3. $$ The sum is $$ A + B =12. $$ The cycle of length six becomes $$ 3, \; 9, \; 3, \; \frac{1}{3}, \; \frac{1}{9}, \; \frac{1}{3}. $$ Putting two together shows $$ 3, \; 9, \; 3, \; \frac{1}{3}, \; \frac{1}{9}, \; \frac{1}{3}, \; 3, \; 9, \; 3, \; \frac{1}{3}, \; \frac{1}{9}, \; \frac{1}{3} $$

In the original notation, with cycles of six multiplying to one, we have $$ a_{49} a_{50} = a_1 a_2 = 3 \cdot 9 = 27. $$ Also $$ a_{97} a_{98} a_{99} a_{100} = a_1 a_2 a_3 a_4 = 3 \cdot 9 \cdot 3 \cdot \frac{1}{3} = 27. $$

Will Jagy
  • 139,541
  • This is very classy! Is there some kind of theory encompassing sequences like this? There's something similar in notes by Fomin and Reading, a sequence defined by $f_{n+1} = \dfrac{f_n + 1}{f_{n - 1}}$ that has period five; I'm just curious if these kinds of things fit into any particular genre, or are curiosities. – pjs36 Jun 01 '16 at 22:10
  • @pjs36 I got suspicious after taking logarithms $b_i = \log a_i$ and having the characteristic roots for the $b_i$ relation being cube roots of unity. When I simply wrote out the first few terms, beginning with unknown $A,B$ it fell into place. Indeed, the sequence is something similar to $$ 3^{2 \sin \left( \frac{(2n-1) \pi}{6} \right)} $$ – Will Jagy Jun 01 '16 at 22:29
2

Since the product of all the numbers is nonzero, the numbers are all nonzero, which means we can write them as $a_i=3^{b_i}$ (where $b_i$ is potentially complex, although we'll see it never is). The recursion translates to $b_{i+1}=b_i-b_{i-1}$, which in turn implies $b_{i+3}=-b_i$ for $i=1,2,\ldots,97$, so that the sequence of $100$ $b_i$'s is just

$$b_1,b_2,b_3,-b_1,-b_2,-b_3,b_1,b_2,b_3,\ldots,b_1,b_2,b_3,-b_1$$

The product of the first $50$ $a$'s corresponds to the sum of the first $50$ $b$'s, which, after cancellations, is just $b_1+b_2$, so we have

$$b_1+b_2=3$$

Similarly, the sum of all $100$ $b$'s reduces to $b_2+b_3$, so we have

$$b_2+b_3=3$$

Putting this together with $b_3=b_2-b_1$, we find $b_1=1$ and $b_2=2$, which means

$$a_1+a_2=3^1+3^2=12$$

Barry Cipra
  • 79,832