Riddles and Puzzles: Extend the following to a valid equation

[fresh_42]

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]

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.

Spoiler

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]

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]

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]

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]

Spoiler: 55 Useless Knowledge

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]

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]

Spoiler: 55 Useless Knowledge

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?)

 

[mfb]

I need more than a second to verify that this is indeed a Fibonacci sequence - to check that every number is the sum of the previous two.

Not a solution for 56:

Spoiler

I wrote an Insights article about it. Well, not exactly this problem, but I'll leave it to others.

 

[fresh_42]

Less than a second ? Hats off !

(what's the icon for incredulity?)

For the multiplication, not the rule.

 

[jbriggs444]

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?

The problem statement was a bit jarring. One would ordinarily expect the winner of a race where the runners start even to have been in front of each competitor exactly one more time than he or she was behind.

But a situation where a runner barely draws even and then falls behind without passing is possible and would allow the number of times ahead or behind to increment beyond one.

Spoiler

Marcy could play the "let him catch up but then pull away" game with Henry, racking up arbitrarily many "is ahead" events relative to Henry while Esther was either well ahead or well behind.

Henry could play the "let her catch up but then pull away" game with Esther, racking up arbitrarily many "is ahead" events relative to Esther while Marcy was either well ahead or well behind.

Esther could play the "let her catch up but then pull away" game with Marcy, racking up arbitrarily many "is ahead" events relative to Marcy while Henry was either well ahead or well behind

It is possible that "more often" is intended to be interpreted as "for a duration of more than half the race". Rock, paper, scissors works for that interpretation as well.

 

[fresh_42]

It is possible that "more often" is intended to be interpreted as "for a duration of more than half the race". Rock, paper, scissors works for that interpretation as well.

The poor wording is probably due to Google translate. I didn't change all of it. "run before / after" means "crosses the finish line ahead of / after". Sorry.

Correction 56.:

So the final results count:

#MH > #HM and #HE > #EH

At the finish line:
Marcy was ahead of Henry more often than he was ahead of her, and Henry was ahead of Esther more often than she was ahead of him.

Could it be that Esther was ahead of Marcy more often than Marcy was ahead of Esther?

 

[jbriggs444]

So the final results count:

#MH > #HM and #HE > #EH

Ok, so the race is run repeatedly and the daily results are tallied. My mistake for not realizing that.

Spoiler

Say there are 30 races. Rename Henry, Marcia and Esther to A, B and C to preserve sanity.
In 10 of them, A > B > C
In 10 of them B > C > A
In 10 of them C > A > B

A beats B 20 to 10
B beats C 20 to 10
C beats A 20 to 10

 

[fresh_42]

57. Two shepherds rest in a meadow. One has 5 pieces of cheese and the other 3 pieces. A hiker comes over and asks if he can eat the cheese with them. The two agree. At this common meal, all three people eat the same amount of cheese. After the meal when all cheese was eaten, the hiker gets up and gives $8 as compensation for the cheese.

How should this amount be divided among the herdsmen so that their contribution of 5 or 3 pieces of cheese is taken into account fairly?

D76

 

[BvU]

Spoiler

Each eats $2{2\over 3}$ pieces of cheese, so fiver contributes $2{1\over 3}$ and threepee ${1\over 3}$.
Ratio is $7:1$: fiver gets 7 bucks , the other guy only one.

 

[fresh_42]

58. How many matches are minimally needed to construct four equilateral triangles.

D76

 

[BvU]

Spoiler

I need 9

 

[jbriggs444]

Spoiler

Can do it with six.

 

[fresh_42]

Spoiler

Can do it with six.

View attachment 245851

Ok, ugly but possible.

Can someone do it without crossings?

 

[BvU]

#259 ? or are we allowed to chop them up ?

 

[fresh_42]

#259 ? or are we allowed to chop them up ?

Nope.

 

[jbriggs444]

#259 ? or are we allowed to chop them up ?

Oh, *doh*. Obvious.

Spoiler

Tetrahedron

 

[fresh_42]

59. The village of Brownlee has exactly 100 inhabitants. The oldest was born in 1900 and all inhabitants were born a different year, but all on January 1st. In 1999, the sum of the four digits of John's year of birth is equal to his age.

How old is John?

D77

 

[lpetrich]

#59:

Spoiler

Since the oldest was born in 1900, and since John's age was measured in 1999, we have
$$ 1900 \le \text{(John's birth year)} \le 1999 $$
That means that John birth year has digits 1, 9, t, and u, where t and u are between 0 and 9 inclusive. That gives it value $1900 + 10t + u$. His age in 1999 is thus 1999 - (birth year) = $99 - 10t - u$. The sum of the digits is $10 + t + u$, Their equality is thus
$$ 99 - 10t - u = 10 + t + u \\ 89 = 11t + 2u $$
Trying the possible values of u, {0 ,1,2 ,3, 4, 5, 6, 7, 8, 9}, gives possible values of 11t {89, 87, 85, 83, 81, 79, 77, 75, 73, 71}. The only integer solution for t is 7, for u = 6.

Thus, John's birth year is 1976, making him 23 years old in 1999.

The sum of his age's digits is 1+9+7+6 = 10+13 = 23.

 

[fresh_42]

60. The King on a chessboard is in the lower left corner (A1). He can move from field to field, but due to a lost bet he can only move according to the following rules:

• a field to the right (B1)
• a field upwards (A2)
• a field diagonally to the top right (B2)

How many different ways are there into the upper right corner (H8)?

D77

 

[BvU]

Direct answer: No moves possible: there are no G8, H9 or G9

But you want to know how any different ways there are to reach H8 from A1, right ?

And the rules are to be taken as OR rules (not in sequence: right, then up, then diagonal, right etc) ? Must be quite a heap of possible routes ...

 

[fresh_42]

Direct answer: No moves possible: there are no G8, H9 or G9

But you want to know how any different ways there are to reach H8 from A1, right ?

And the rules are to be taken as OR rules (not in sequence: right, then up, then diagonal, right etc) ? Must be quite a heap of possible routes ...

Less than 100,000. I am personally rather bad at combinatorics, I always count wrong. This question has chances to remain unanswered, or to figure out whom I can call if a question about combinatorics occurs in the set theory forum. The solution I have distinguishes between diagonal moves and others, i.e. counting without diagonals on an $n \times n$ board, then $k$ diagonals, and then partitioning ways into these two possibilities.

 

[jbriggs444]

Less than 100,000. I am personally rather bad at combinatorics, I always count wrong. This question has chances to remain unanswered, or to figure out whom I can call if a question about combinatorics occurs in the set theory forum. The solution I have distinguishes between diagonal moves and others, i.e. counting without diagonals on an $n \times n$ board, then $k$ diagonals, and then partitioning ways into these two possibilities.

Spoiler

Sensible. So one can catalogue the ways of doing it with 0, 1, 2, 3, 4, 5, 6 and 7 diagonal moves:
$$\frac{14!}{7! 7! 0!} + \frac{13!}{6! 6! 1!} + \frac{12!}{5! 5! 2!} + \frac{11!}{4! 4! 3!} + \frac{10!}{3! 3! 4!} + \frac{9!}{2! 2! 5!} + \frac{8!}{1! 1! 6!} + \frac{7!}{0! 0! 7!}$$
$$= 3432 + 12012 + 16632 + 11550 + 4200 + 756 + 56 + 1$$
$$=48639$$

Or one could brute force it with a kind of Pascal's triangle variant. w(x+1,y+1)=w(x,y) + w(x+1,y) + w(x,y+1)

     1     1     1     1     1     1     1     1
     1     3     5     7     9    11    13    15
     1     5    13    25    41    61    85   113
     1     7    25    63   129   231   377   575
     1     9    41   129   321   681  1289  2241
     1    11    61   231   681  1683  3653  7183
     1    13    85   377  1289  3653  8989 19825
     1    15   113   575  2241  7183 19825 48639