# Challenge Riddles and Puzzles: Extend the following to a valid equation

• Featured

#### lpetrich

Proof. Start with n players. If the games in the tournament are two-player elimination ones, then for each game, one player drops out. So for g games, g players drop out, and for n' players to remain, (n-n') must drop out, giving (n-n') games.

In this problem, 116 players are playing in the tournament and 2 players will remain for the final game. This means 116 - 2 = 114 dropouts, and thus 114 games.

If the tournament has other rules, like round-robin phases, then it will have more games in it.

#### fresh_42

Mentor
2018 Award
51. A car has to carry an important person across the desert. There is no petrol station in the desert and the car has space only for enough petrol to get it half way across the desert. There are also other identical cars that can transfer their petrol into one another. How can we get this important person across the desert?

D68

#### jbriggs444

Homework Helper
51. A car has to carry an important person across the desert. There is no petrol station in the desert and the car has space only for enough petrol to get it half way across the desert. There are also other identical cars that can transfer their petrol into one another. How can we get this important person across the desert?
The relationship to the harmonic series is left to the imagination.

4 cars set out on the way across the desert.
1/8 of the desert width along, these cars will have used a total of one car's worth of petrol. One car is selected and empties its fuel tank into the remaining cars. The remaining cars all have full tanks. The driver is left to perish.

3 cars continue across the desert...
1/6 of the desert width further, these cars will have used a total of one car's worth of petrol. One car is selected and empties its fuel tank into the remaining cars.

2 cars continue across the desert...
1/4 of the desert width further, these cars will have used a total of one car's worth of petrol. One car is selected and empties its fuel tank into the remaining cars.

1 car continues across the desert and reaches the far side.

#### BvU

Homework Helper
The driver is left to perish.
Does it have to be so dramatic ? If other cars do something comparable from the other side of the desert a lot of lives can be saved -- and it won't be so difficult to get volunteer drivers for the return trip
(detailed calculation left to those who didn't know the trick and had to consult the spoilers...)

#### fresh_42

Mentor
2018 Award
52. A poor man and a rich man are talking about music.
The poor man says he has studied music and can find a song with any name in it.

The rich man says "OK, if you can find a song with my son's name in it, I will give you a thousand dollars. His name is Demarcus-Jabari."

The poor man gives his answer and is instantly \$1,000 richer.

What's the song?

D69

#### BvU

Homework Helper
Skip 52. Google itself is the spoiler !

#### fresh_42

Mentor
2018 Award
I will stop trying to be funny ...

#### BvU

Homework Helper
As long as you continue the series ... It's fun !

#### fresh_42

Mentor
2018 Award
53. Before the days of the Trans-Siberian Railway, the route from Moscow to Vladivostok had to be mastered by riders. The Russian Czar had two reliable persons who often served as couriers for him. The one makes the track in 40 days, the other was even faster, and makes it in 30 days.

One day the Tsar sent the first horseman with an important message, but soon it occurred to him that he would not be fast enough at the finish, and therefore he sent the second horseman with the same message, after the first rider was already 8 days on his way.

How many days did the first rider still need to Vladivostok, when he was overtaken by the second rider, and how many days will remain to get the message delivered?

D69

#### mfb

Mentor
The relationship to the harmonic series is left to the imagination.
In rocketry this concept is known as asparagus staging or propellant crossfeed. All boosters fire together using propellant from the tanks of a single booster (or two boosters to keep it symmetric). Once that is empty it drops away, reducing the dry mass of the rocket. This continues until all boosters are empty. A very efficient way to use boosters but really difficult to implement in practice, no rocket uses it.
- The Space Shuttle transferred fuel from the main tank (the orange one) to the Space Shuttle Orbiter but the orbiter didn't have a big tank on its own.
- Falcon Heavy was planned with propellant crossfeed, but the concept was dropped as too difficult and risky. Other improvements made the rocket without crossfeed more powerful than the original design with it.

#### fresh_42

Mentor
2018 Award
While we are still waiting for the Czar's courier to arrive in Vladivostok (53.), an easy one for pleasure:

54. I am air, I am water, I am electricity. What am I?

D72

#### jbriggs444

Homework Helper
While we are still waiting for the Czar's courier to arrive in Vladivostok (53.), an easy one for pleasure:

54. I am air, I am water, I am electricity. What am I?

D72
Current

#### BvU

Homework Helper
Czar courier number 2 departs on day 9
and needs 30 days, so he'll arrive on day 38.

We call the distance Moscow - Vladiwostok D
And the overtaking takes place at a fraction f of D where $${f*D\over D/40 } + 8 = {f*D\over D/30}$$ i.e when f = 0.8, so after 32 days. The message arrives 6 days later.

$\$

#### BvU

Homework Helper
54. I am air, I am water, I am electricity. What am I
I am also contemporary

#### fresh_42

Mentor
2018 Award
55. After taking a quick look at the following addition: $6 + 10 + 16 + 26 + 42 + 68 + 110 + 178 + 288 + 466$,
the master answered without a second of hesitation: The result is $1210.$

What principle did he rely on?

D73

mfb

#### fresh_42

Mentor
2018 Award
Hint for #55:

$32,490+10,375+42,865+53,240+96,105+149,345+245,450+394,795+640,245+1,035,040=2,699,950$
can be calculated in less than a second.

#### jbriggs444

Homework Helper
Hint for #55:

$32,490+10,375+42,865+53,240+96,105+149,345+245,450+394,795+640,245+1,035,040=2,699,950$
can be calculated in less than a second.
Well, it has to be some observation about the sum of a fibonacci sequence. And by fiddling around, one can see that the sum is given by $2a_n + a_{n-1} - a_2$.

[I'd started with $2a_n + a_{n-1}$ as an approximation but realized we needed some correction term to account for where the series begins. The last two terms by themselves carry no information about how many terms precede them. The induction step worked fine, but the base case did not work out until the $-a_2$ was tacked on. There may be a more elegant way to think about that -- something to do with making the recurrence relation homogeneous].

The obvious way to proceed is by induction.

The base case for n=3 is easy. The three terms are $a_1$, $a_2$ and $a_3 = a_1 + a_2$. The sum is $2a_1 + 2a_2$ = $2a_n + a_{n-1} - a_{2}$

Now, suppose that the formula holds for n terms (n>=3). That is, suppose that $\sum_{i=1..n} a_i = 2a_n + a_{n-1} - a_2$

Then $\sum_{i=1..n+1}a_i = \sum_{i=1..n}a_i + a_{i+1} = ( 2a_n + a_{n-1} - a_2 ) + (a_{n-1} + a_{n}) = (a_n + a_{n+1} - a_2 ) + a_{n+1} = 2a_{n+1} + a_n - a_2$

And the induction is proven.

So back to the original problem, the sum of x, 10, ..., 288, 466 is 2x466 + 288 - 10 = 1210

An alternate way would have been to solve the recurrence relation, yielding the formula $a_n = a\varphi^n + b \varphi^{-n}$ and then apply the boundary conditions to find a and b based on $a_1$ and $a_2$. And then sum the two truncated geometric series. But that would have involved actual work.

$\varphi$ is, of course, the Golden Ratio, approximately 1.618033

#### fresh_42

Mentor
2018 Award
It is indeed far easier. Less than a second!
I didn't know this trick either, but it is a funny insight. Even if one from the category: useless knowledge.

#### fresh_42

Mentor
2018 Award
Are there still some members on it, or shall I post the solution?

#### pbuk

I thought #242 WAS the solution - surely there cannot be anything simpler?

#### fresh_42

Mentor
2018 Award
I thought #242 WAS the solution - surely there cannot be anything simpler?
Nope, see #243. The answer fits in one, maximal two lines (without proof, which is trivial).

#### pbuk

This will be surprising, please don't leave us in suspense any longer...

#### fresh_42

Mentor
2018 Award
The sum are the first ten numbers of a Fibonacci sequence, say $F_1,F_2,\ldots$ Now for any such sequence, i.e. for any basis pair of numbers, we have
$$\sum_{k=1}^{10} F_k = 11 \cdot F_7$$

#### fresh_42

Mentor
2018 Award
56. Marcy, Henry and Esther have decided to do something for their fitness and are therefore organizing a small jogging race each morning. In the past month of June, Marcy ran in front of Henry more often than he did before her, and Henry ran more often in front of Esther than she did before him.

Could it be that Esther ran in front of Marcy more often than Marcy did in front of Esther?

D76

#### BvU

Homework Helper
The sum are the first ten numbers of a Fibonacci sequence, say $F_1,F_2,\ldots$ Now for any such sequence, i.e. for any basis pair of numbers, we have
$$\sum_{k=1}^{10} F_k = 11 \cdot F_7$$
Less than a second ? Hats off !

(what's the icon for incredulity?)

"Riddles and Puzzles: Extend the following to a valid equation"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving