Riddles and Puzzles: Extend the following to a valid equation

[fresh_42]

"They lied" is a very simple possible answer, but I guess that is not the intended answer.

Right. The real reason is completely natural and the easiest possible.

 

[fresh_42]

Let me phrase it with a mathematical anecdote my mentor once told me.
He listened to a dialogue between two students in the university's elevator.
One student was rather upset about the exam she just has had.
"This idiot asked me how often the constant function can be differentiated and I said once. This stupid question ruined a better grade."

So is the constant function exactly twice differentiable?

 

[DavidSnider]

72. Summer is over and there are two new students in the class who are hard to tell apart. "You are certainly twins?", asks the teacher.

"No," answer the two boys.

The teacher is surprised and looks again in her documents: both have the same birthday and the same parents. How can that be?D88

Spoiler: spoiler

Surrogacy?

 

[fresh_42]

No. Or let's say: It could have happened in ancient Greece.

Edit: No Disney fans here?

 

[fresh_42]

73. $45$ must be divided in four parts, such that we get the same number if we add $2$ to the first part, subtract $2$ from the second, multiply the third by $2$ and divide the fourth by $2$.

D89

 

[pbuk]

#73

Spoiler

8, 12, 5, 20

 

[fresh_42]

74. Enter the numbers $1,2,\ldots,10$ into the circles, such that the sums of all numbers along the three inner triangles are equal.

D89

 

[mfb]

So is the constant function exactly twice differentiable?

Triplets (or more) then.

 

[lpetrich]

Proof of #73

Spoiler

In symbolic form, $45 = n_1 + n_2 + n_3 + n_4$ where $n_1 + 2 = n$, $n_2 - 2 = n$, $2n_3 = n$, and $n_4/2 = n$ for some number n.

One can easily find n1 to n4 from these: $n_1 = n - 2$, $n_2 = n + 2$, $n_3 = n/2$, $n_4 = 2n$. Substituting into the equation for 45, I find
$$ 45 = (n-2) + (n+2) + (n/2) + 2n = 9n/2 $$
Its solution is $n = 10$ and the four parts are $8, 12, 5, 20$.

 

[fresh_42]

75. Two trains passed each other in opposite directions, one at a speed of 36 km / h, the other at 45 km / h. A passenger on the second train counted and found, that the first train to pass him needed six seconds.
How long was the train?

D90

 

[mfb]

74. Enter the numbers $1,2,\ldots,10$ into the circles, such that the sums of all numbers along the three inner triangles are equal.

The center is part of the sum for all triangles, we can fill it last with whatever number is remaining and ignore it from now on. For every pair of triangles there are two numbers that are shared between these triangles, and one number that is exclusive to the other triangle. We can find a solution if we can find three mutually exclusive triples that satisfy a+b=c where a+b are put in the shared numbers and c is put on the opposite side. That way the sum in every triangle will be the sum of all numbers apart from the three outer numbers.
Doesn't need much searching to find 1+9=10, 2+6=8, 3+4=7.
The leftover number for the center is 5 and the sum in every triangle is 30, which means the average per field is 5 as well. Nice symmetry around 5 despite the asymmetric start (average number we fill in is 5.5).

 

[fresh_42]

View attachment 246474

Very interesting. I have a solution with 38 instead of 30:

I didn't think that more than one solution is possible. Maybe I should have asked how many sums are possible.

 

[lpetrich]

For #74, I wrote a short program that does a brute-force search. The number of permutations searched was 10! = 3628800, and I found 6528 solutions. Because none of the solutions have any symmetries, the total number of inequivalent solutions is the total number divided by the size of the triangle symmetry group (6): 1088.

I wrote it in C++, and I used STL <algorithm> function next_permutation() to iterate through the possibilities. The Standard Template Library is *great*.

 

[fresh_42]

For #74, I wrote a short program that does a brute-force search. The number of permutations searched was 10! = 3628800, and I found 6528 solutions. Because none of the solutions have any symmetries, the total number of inequivalent solutions is the total number divided by the size of the triangle symmetry group (6): 1088.

I wrote it in C++, and I used STL <algorithm> function next_permutation() to iterate through the possibilities. The Standard Template Library is *great*.

Have you calculated the possible sums in every solution?

 

[lpetrich]

Have you calculated the possible sums in every solution?

I have just done so. The number of solutions for each sum value is;
28: 96 16
29: 192 32
30: 480 80
31: 864 144
32: 1248 208
33: 768 128
34: 1248 208
35: 864 144
36: 480 80
37: 192 32
38: 96 16
with the total number and the number to within symmetries.

 

[mfb]

There is more symmetry. There is the factor 6 for the arrangement of the three triangles, but there is also a factor 8 from swapping numbers on the diagonals - different pattern but effectively the same solution. This is the reason all your numbers are divisible by 8. Removing this symmetry we end up with 136 options.

There is a more subtle symmetry: Replacing 1,2,...,10 by 10,9,...,1 leads to a solution as well (with triangle sum 66-x instead of x). That leads to the symmetric pattern of options you found. Removing this symmetry as well we end up with 68 options.

Is there a solution that also has the same sum for the big outer triangle?

 

[lpetrich]

There is more symmetry. There is the factor 6 for the arrangement of the three triangles, but there is also a factor 8 from swapping numbers on the diagonals - different pattern but effectively the same solution. This is the reason all your numbers are divisible by 8. Removing this symmetry we end up with 136 options.

How does that symmetry work? I find it difficult to picture.

 

[mfb]

Just swap e.g. 3 and 4 in the solution I posted. You can also swap 1 and 9 or 2 and 6.

 

[lpetrich]

So it's swap outer and middle along each diagonal.

Is there a solution that also has the same sum for the big outer triangle?

Indeed there is.

Inner-triangle length-equal solutions
28: 96 16 2
29: 192 32 4
30: 480 80 10
31: 864 144 18
32: 1248 208 26
33: 768 128 16
34: 1248 208 26
35: 864 144 18
36: 480 80 10
37: 192 32 4
38: 96 16 2
Total: 6528 1088 136
(raw, triangle symmetry, diagonal-interchange symmetry)

Equal to the outer-triangle length
31: 48 8
32: 120 20
34: 120 20
35: 48 8
Total: 336 56
(raw, triangle symmetry)

 

[fresh_42]

76. $n^3+ n^2 u +n^2 v+n^2w+nuv+nuw+nvw+uvw = 27,673,509,091$ with $u<v<w$.
What is $u\cdot n^2\,?$ (All numbers are non negative integers and $n$ maximal among all solutions.)

D91

 

[fresh_42]

77. Which $n$ times $m$ puzzles exist, that have as many boundary parts as interior parts?

D93

 

[lpetrich]

I will try #77, interpreting as "jigsaw puzzles".

Spoiler

Imagine a rectangle formed out of same-size squares, m on one side and n on the other. The problem states that the number of edge squares equals the number of interior squares. Since the total number is the number of both edge and interior squares, this means that the total number is twice the interior number.

In effect, $mn = 2 (m-2)(n-2)$.

Solving for m gives
$$ m = \frac{4(n-2)}{n-4} $$
with an interchange of m and n giving n as a function of m.

Since this problem is symmetric in m and n, we can impose the condition that m >= n without loss of generality. The next step is to try the possible values of positive-integer n and find which ones give positive-integer m. To get an idea of how feasible that is, I consider how m varies as a function of n. So I calculate
$$ \frac{dm}{dn} = - \frac{8}{(n-4)^2} $$
This means that increasing n makes m decrease.

Since m must be greater than 0, n must either be 1 or greater than 4. Since n must be at least 3 to allow both the interior and the edges to have a positive number of squares, this means n > 4.

• n = 5 -> m = 12
• n = 6 -> m = 8
• n = 7 -> m = 20/3 = 6.6666...

That last one violates m >= n, so we stop there.

That means that the only two solutions are

• (5,12) -- # squares = 60, # edge, interior squares = 30
• (6,8) -- # squares = 48, # edge, interior squares = 24

 

[fresh_42]

78. When a cyclist had covered two-thirds of his way, a tire burst. For the rest of the way he needed on foot twice as much time as for the previous ride on the bike.
How many times faster was he cycling than he was walking?

D94

 

[fresh_42]

79. Find all solutions of the following addition:
\begin{align*}
&\quad &{}xx \\
&\quad &{}yy \\
+&\quad &{}zz \\
\hline
& \quad & xyz
\end{align*}
D95

 

[pbuk]

#79

Spoiler

Apart from the trivial 00 + 00 + 00 = 000, 11 + 99 + 88 = 198 is the only solution in base 10.

 

[pbuk]

#79 proof

Spoiler

1. The last digit of the solution $= z \implies x + y = 10$
2. From the addition of the most significant digits (plus the carry from the least significant) we have
3. $x + y + z + 1 = 10x + y \implies 9x = z + 1$ so we must have $x =1, z = 8$
4. From 1 and 3 we have $1 + y = 10 \implies y = 9$

 

[lpetrich]

#78

Spoiler

We want to find v(cycling)/v(walking). Velocity = distance / time, and I must calculate the appropriate distances and times.

Distance cycling = (2/3) (total distance), meaning that distance walking = (1/3) (total distance), and (distance cycling) = 2 (distance walking).

Time walking = 2 (time cycling), yielding time cycling = (1/2) (time walking).

Thus, velocity cycling = (distance cycling) / (time cycling) = (2)/(1/2) (distance walking) / (time walking) = 4 (velocity walking).

The cyclist had biked four times faster than he walked.

#79

Spoiler

Let us break down the sums by digits, and let us use ones and tens carries a and b. Thus,
$$ x + y + z = z + 10 a \\ x + y + z + a = y + 10 b \\ b = x $$
The third equation gives us the second carry, $b = x$, and the first one gives us $x + y = 10 a$. This only nonzero multiple of 10 that any digits add up to is 10, and thus $a = 1$ and $x + y = 10$. Inserting into the second equation gives $10 + z + 1 = y + 10 x$ or $11 + z = 10 + 9 x$ or $1 + z = 9 x$. The only possible value of x that can solve that equation is 1, and that solution gives y = 9 and z = 8.

The solution: $x = 1, y = 9, z = 8$, giving us
$$
\begin{align*}
&\quad &{}11 \\
&\quad &{}99 \\
+&\quad &{}88 \\
\hline
& \quad & 198
\end{align*}
$$

 

[fresh_42]

80. What is the ratio between the (sum of all) colored areas and the (area of the) entire square?

D95

 

[pbuk]

76. $n^3+ n^2 u +n^2 v+n^2w+nuv+nuw+nvw+uvw = 27,673,509,091$ with $u<v<w$.
What is $u\cdot n^2\,?$ (All numbers are non negative integers and $n$ maximal among all solutions.)

Spoiler

Brute force appears to yield $n = 131, u = 0, v = 2,000, w = 99,000 \implies u \cdot n^2 = 0$ as the solution with maximal $n$, but finding this has not given me any greater insight!

 

[fresh_42]

81. Bart and Lisa are sitting in front of a huge heap of skittles. Since both of them want to eat as many as possible, they decide to play a game. Bart has to write two (different) numbers (positive integers) on two pieces of paper. Then Lisa turns around one of the papers and guesses, whether the other number is higher or lower. If she is right, then she will get 10 skittles, otherwise Bart will get them.

Is there a strategy for Lisa which improves her chances in comparison to a 50:50 guess? And can Bart counter this strategy?

D95