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

[fresh_42]

30. Five old ladies are collecting buttons of all kind. Together they have 2000 pieces.
If we add Elise's buttons to Adele's, subtract Elise's buttons from Berta's, multiply Elise's buttons with Charlotte's, divide Doris' buttons by Elise's, we will always get the same number. Adele is happy that she soon has 100 buttons. How many are missing?

Correction: There are 2000 buttons without counting Elise's.

 

[pbuk]

I mean, nobody did #27 which is even easier

I'm not sure #27 is as easy as you think it is: it is fairly easy to convince yourself that 9 initial cells are not sufficient, but turning this conviction into a proof is another thing!

Spoiler: But i'll try...

1. Mold can spread into a new cell if adjacent cells in the same row and column are moldy. If there are only 9 initial moldy cells, this way mold can only spread into 9 rows and columns leaving 19 cells free.

2. Mold can also spread into a new cell if the two adjacent cells in the same row are moldy. This way a new column can be populated. But if there are initially 2 cells in the same row, then by the argument in 1. mold can only spread into 8 rows, so we have 'bought' another column at the cost of a row.

3. The argument in 2 applies with rows and columns interchanged.

Convinced? No, nor am I.

 

[fresh_42]

I'm not sure #27 is as easy as you think it is: it is fairly easy to convince yourself that 9 initial cells are not sufficient, but turning this conviction into a proof is another thing!

Spoiler: But i'll try...

1. Mold can spread into a new cell if adjacent cells in the same row and column are moldy. If there are only 9 initial moldy cells, this way mold can only spread into 9 rows and columns leaving 19 cells free.

2. Mold can also spread into a new cell if the two adjacent cells in the same row are moldy. This way a new column can be populated. But if there are initially 2 cells in the same row, then by the argument in 1. mold can only spread into 8 rows, so we have 'bought' another column at the cost of a row.

3. The argument in 2 applies with rows and columns interchanged.

Convinced? No, nor am I.

I'm a bit confused, to be honest. It can spread into an empty cell if at least two vertical or horizontal adjacent cells (of this new one) are already occupied.

The easiest way is indeed to count the perimeters.

 

[pbuk]

30. Together they have 2000 pieces. ... How many are missing?

Does this mean that they actually have some number less than 2,000?

Spoiler

Is this number 1,961 i.e. 39 are missing?

 

[fresh_42]

Does this mean that they actually have some number less than 2,000?

Spoiler

Is this number 1,961 i.e. 39 are missing?

I'm afraid it isn't solvable unless we observe that it has to be more than 2,000 buttons.
I overlooked a tiny condition. It are 2,000 buttons without Elise's.

Sorry, I definitely will have to post new questions at daytime.

 

[pbuk]

I'm afraid it isn't solvable unless we observe that it has to be more than 2,000 buttons.

For 1,961 buttons I had the solution A = 85, B = 119, C = 6, D = 1734, E = 17 and there are other solutions with fewer than 2,000. But with the new condition it is too easy

Spoiler

although Adele is now deluding herself with her proximity to 100!

 

[fresh_42]

For 1,961 buttons I had the solution A = 85, B = 119, C = 6, D = 1734, E = 17 and there are other solutions with fewer than 2,000. But with the new condition it is too easy

Spoiler

although Adele is now deluding herself with her proximity to 100!

Nice solution, but I don't get your spoiler.
We have $A+B+C+D=2000$ and $A+E=B-E=CE=D/E$ but I don't see why it's easy.

 

[pbuk]

but I don't see why it's easy.

Oh it is only easy for me because I had already tabulated the solution in my attempt to find a solution to the original problem - unless I am wrong of course!

Spoiler: #30

$A = 76; B = 114; C = 5; D = 1805; E = 19$

 

[fresh_42]

31. A famous mathematician (died $1871$) occasionally bragged that he was exactly $x$ years old in year $x^2$. Another one (possibly not a real person) said in $1925$ that he was exactly $a^2 + b^2$ years old in the year $a^4 + b^4,$ that he was exactly $2c$ years old in $2c^2$ and that he was exactly $3d$ years old in $3d^4$.

What are $x,a,b,c,d$ and do you know who the mathematician was?
Hint: The phrase is especially funny in this case!

 

[lpetrich]

I'll take on the first part of #31.

Spoiler

If that mathematician was x years old in year x², then his birth year was x(x-1). He must have been at least around 10 when he died, so he could have some mathematical ability, and at most around 122, the human longevity record. Since he died in 1871, that means 1749 to 1861. The only integer x that fits is x = 43, with x² = 1849. That makes him born in 1806, and age 65 when he died.

That mathematician was Augustus de Morgan.

 

[lpetrich]

Now the second part of #31.

Spoiler

Since he must have been between 10 and 122 years old in 1925, his birth year must have been between 1803 and 1915.

The d equation means that his birth year must have been 3d⁴-3d. The only birth year in that range for positive-integer d is 1860, for d = 5. That means that he was 15 years old in 1875.

Turning to the c equation, the only positive-integer value of c is c = 31, making him 62 years old in 1922.

Turning to the a and b equation, the birth year is B(a) + B(b) where B(x) = x⁴ - x². This function is monotonically increasing for x > 1, and B(1) = 0, B(2) = 12, B(3) = 72, B(4) = 240, B(5) = 600, B(6) = 1260, and B(7) = 2352. This means that both a and b are at most 6, and that at least one of a and b is at least 5. Considering values 5 and 6, 1860 - B(5) = 1260, and 1860 - B(6) = 600. This means that a and b are 5 and 6.

 

[lpetrich]

Now #30.

Spoiler

For Adele having a buttons, Berta having b buttons, Charlotte having c buttons, Doris having d buttons, and Elise having e buttons, the conditions are:

• a, b, c, d, and e are nonnegative integers
• a + b + c + d = 2000
• a < 100
• a + e = b - e = c*e = d/e = w

I was able to solve these equations using Mathematica's Reduce function. I found a single solution: a=76, b=114, c=5, d=1805 e=19, with Adele needing 24 more buttons.

 

[fresh_42]

32. Two friends are bored playing backgammon and decided to take the dice and roll dice. One takes the four normal dice with numbers $1,2,3,4,5,6$, the other one takes the die with numbers $2,4,8,16,32,64$. Who will win more often?

 

[Orodruin]

What is the win condition? Higher total in a roll?

 

[lpetrich]

I'll try to avoid brute force with #30.

Spoiler

From the fourth condition, a = w - e, b = w + e, c = w/e, d = w*e.

Let a + b + c + d = n. Then n = 2w + w/e + w*e = w*(e+1)²/e.

Since e and e+1 are relatively prime, w must be divisible by e, giving w = v*e. Thus,

a = (v-1)*e, b = (v+1)*e, c = v, d = v*e², n = v*(e+1)²

Here, n = 2000 = 2⁴*5³. This gives possible values for e+1 of +-1, +- 2, +-4, +-5, +-10, and +-20. Since e must be positive, that gives e = 1, 3, 4, 9, and 19, with corresponding values of v of 500, 125, 80, 20, and 5, and values of a of 499, 372, 316, 171, and 76. Of these, only the last one is less than 100.

 

[lpetrich]

I'll take on #32.

Spoiler

Let the first dice have probability distribution p(i) for roll-value i, and the second die have probability distribution q(j) for roll-value j. The total probability that the second die will have a higher number is
$$ \sum_{i<j} p(i) q(j) $$
For the four 1-to-6 dice, one can first find the probabilities for two dice, and then for four, by twice repeating the two-distribution self-convolution:
$$ p'(i) = \sum_j p(j) p(i-j) $$
This can be done with a generating-function approach, and it takes only a few lines of Mathematica code to do that.

I find these probabilities:

• First dice higher: 4127/7776 ~ 0.530736
• Equal: 161/7776 ~ 0.0207047
• Second dice higher: 109/243 ~ 0.44856

So the first player will more likely win.

 

[fresh_42]

What is the win condition? Higher total in a roll?

Yes.

 

[fresh_42]

33. Build with $\{\,0,1,2,4,+,-,\cdot,:\,\}$ all integers $1,2,\ldots,25$.

 

[mfb]

Do we have to use all numbers once, or at most once?
I'll start with 1-9:

Spoiler

4-2-1-0=1
4-2*1-0=2
4-2+1-0=3
4-2*1*0=4
4+2-1-0=5
4+2+1*0=6
4+2+1+0=7
4*2+1*0=8
4*2+1+0=9

 

[fresh_42]

Do we have to use all numbers once, or at most once?
I'll start with 1-9:

Spoiler

4-2-1-0=1
4-2*1-0=2
4-2+1-0=3
4-2*1*0=4
4+2-1-0=5
4+2+1*0=6
4+2+1+0=7
4*2+1*0=8
4*2+1+0=9

The solution I have uses all numbers exactly once, so this would be a nice to have constraint, yes.

 

[mfb]

Is exponentiation allowed? It doesn't have an extra symbol...
A few towards the end, assuming concatenation is okay:

Spoiler

20+4*1=19 = 21-4/2
20 is easy to make if we don't use 1 and 4, or if we can use exponents (20*1⁴)
21+4*0=21 = 40/2+1
22?
24-1+0=23 = 21+4/2
24+1*0=24 = 20+4*1
24+1+0=25 = 21+4+0

 

[fresh_42]

22 goes with exponentiation. I don't understand

20+4*1=19

but 19 can easily be done with all four digits.

Now we need: 10-19 and 22.

 

[mfb]

I don't understand 20+4*1=19 either. Copy&paste error, probably. The right side is wrong, too.

20-1⁴ = 19.
21+4⁰ = 22

Spoiler

10 is trivial if we don't use all numbers, and it is easy with square roots, but I don't find a proper solution for now.
12-4⁰=11
10+4-2 = 12 = 4*(2+1+0) = 12*4⁰
14-2⁰=13
14+0*2=14
2⁴-1+0 = 15 = 20-4-1
2⁴+1*0 = 16 = 10+4+2
2⁴+1+0 = 17 = 21-4+0
10+4*2 = 18
20-1⁴=19 from above

10 is missing.

Outside the range but why not:
2⁴+10 = 26

 

[fresh_42]

10 is missing.

Think of the prime factorization.

And now again one of the kind you do not like: What is the next number?
34.
The author of the first Russian mathematical book was born.
The oldest German wine bill dates to this year.
The Knights Templar was dissolved.
The first Scottish university was founded.
The Viri Mathematici, one of the first books about history of science was published.
Kepler's mother was arrested for suspected witchcraft.
The second most famous English mathematician of his time after Newton dies.
The Connecticut Asylum for the Education and Instruction of Deaf and Dumb Persons was founded.
Max Planck wins the Nobel Prize in physics.
This is the year we are looking for.

 

[mfb]

Think of the prime factorization.

Oh... how could I miss something so simple.
(4+1)*2+0=10
Needs brackets, however, unlike every other number.

 

[pbuk]

Think of the prime factorization.

And now again one of the kind you do not like: What is the next number?
34.
The author of the first Russian mathematical book was born.
The oldest German wine bill dates to this year.
The Knights Templar was dissolved.
The first Scottish university was founded.
The Viri Mathematici, one of the first books about history of science was published.
Kepler's mother was arrested for suspected witchcraft.
The second most famous English mathematician of his time after Newton dies.
The Connecticut Asylum for the Education and Instruction of Deaf and Dumb Persons was founded.
Max Planck wins the Nobel Prize in physics.
This is the year we are looking for.

Spoiler

This is the year we are looking for

 

[fresh_42]

Spoiler

This is the year we are looking for

?

 

[pbuk]

?

Alternatively (but equivalently):

Spoiler

This year is the year we are looking for.

 

[fresh_42]

Alternatively (but equivalently):

Spoiler

This year is the year we are looking for.

Have you figured out some of the other ones?

 

[pbuk]

Have you figured out some of the other ones?

Spoiler

Well I assume the first two are 1110 and 1211 because from the Knights Templar on they are 1312, 1413 ... Max Plank 1918, 2019

The Knights Templar is a large rather soulless pub in the Wetherspoons chain notable mainly for its urinal.

 

[fresh_42]

35. In a math exam the smallest natural number had to be determined with the following property: The first digit of this number is 6. If this digit would be placed at the end instead, then a new natural number was created whose value is one quarter of the original number.

Which natural number is it?

 

[DavidSnider]

35. In a math exam the smallest natural number had to be determined with the following property: The first digit of this number is 6. If this digit would be placed at the end instead, then a new natural number was created whose value is one quarter of the original number.

Which natural number is it?

This is really easy to brute force, is there another way to do it?

Spoiler: spoiler

615384 = 153846 * 4

 

[fresh_42]

This is really easy to brute force, is there another way to do it?

Spoiler: spoiler

615384 = 153846 * 4

Not sure what you mean by brute force, you probably haven't started at 61 and ran up to 615,384.

$6\cdot 10^k+y = 4\cdot(10y+6) \Longrightarrow 10^k=\frac{13}{2}y +4 \stackrel{2z=y}{\Longrightarrow}13z =10^k-4$ and then it is brute force by testing $k=0,1,\ldots ,5$ which means five divisions by $13$.

There are some other solutions, but they look a bit like numerology, i.e. they would require a proof why they work.

Another way is a long division $6abc \ldots \, : \,4 = abc\ldots$

 

[fresh_42]

36. Several horse-drawn coaches drive with the exact same number of people on a festival in the country. Halfway, ten coaches drop out, so each of the others has to pick up another person. Before the journey home, another fifteen coaches were broken, which means that each coach has three more people than at the departure in the morning.

How many people attended the festival?

 

[lpetrich]

I will take on #36:

Spoiler

I will let the initial number of coaches be c and the initial number of riders be r. The total number of people $n = c r$ initially. Halfway, c goes down by 10 and r goes up by 1, thus $n = (c-10)(r+1)$. At arrival, c had gone down by 10 and it goes down by 15 and r goes up by 3, and thus $n = (c-25)(r+3)$.

Subtracting the first equation from the second one gives $c - 10 r - 10 = 0$, and subtracting the third one from the second one gives $3 c - 25 r - 75 = 0$. Multiply the first of these and subtract it from the second one: $5 r - 45 = 0$. That gives $r = 9$, and substituting it back in gives $c = 100$.

Thus, the total number of people $n = c r = 900$. Initially, there were 100 coaches and 9 people per coach. After 10 of them broke down, there were 90 coaches and 10 people per coach, and after an additional 25 of them broke down, there were 75 coaches and 12 people per coach.

 

[fresh_42]

37. Replace all $O$ by odd and all $E$ by even numbers, so that the following calculation is correct:
\begin{align*}
OEO \cdot OO\\
\hline \\
OOO\\
EOEO\quad \\
\hline \\
OEEEO
\end{align*}

 

[mfb]

OOO+EOEOn = OEEEO (where "n" is null, 0 would be too similar to O), therefore we get a carry from the 100 digit to the 10,000 digit, which means the thousand digit must be a 9 and afterwards we get a 0.
OOO+E9EOn = OnEEO

Let's study OEO*O=OOO: The last digits must produce an odd carry (as E*O=E). As there is no carry to the thousand digit one of the leading digits has to be 1 or both have to be 3. The second factor cannot be 1 as OEO*1=OEO. This leaves two options:
1EO*O=OOO or 3EO*3=OOO
In the second case the first digit of the product is 9 and the last digit of the first factor must be 5 to produce an odd carry:
1EO*O=OOO or 3E5*3=9O5
To avoid a carry to the hundreds E must be 0 or 2:
1EO*O=OOO or 305*3=915 or 325*3=975

Explore the 3xx*3 options more for the other product:
305*O = E9EO?
This needs 3*O=E9 which doesn't work.
325*O = E9EO?
Testing 5,7,9 gives us 325*9 = 2925 as solution.

325*93 = 975 + 29250 = 30225
OEO*OO = OOO + EOEOn = OEEEO
That is a solution.

What about 1EO*O=OOO? For the second number the first digit must be 5, 7 or 9 and larger than the second digit. The overall product is at most 189*97 = 18333, therefore the first digit must be 1. 1EO*OO = 1nEEO
Rule out further solutions with a spreadsheet: There are 25 numbers 1EO and 9 numbers for OO. In addition the product must be larger than 10,000 but smaller than 11000. There is no product in that range that has the OEEEO pattern. 325*93 = 975 + 29250 = 30225 is the unique solution.

 

[fresh_42]

38. You find an old parchment that tells of an immense treasure on an island. The paper also contains the description of how to find the treasure:

"To find the treasure, first find the wooden gallows on the island. When you have found it, you will see the only two trees on the island, an oak and a beech. Maybe wait for the end of a thunderstorm, then go to the oak and count the steps. Once at the oak, turn 90 degrees clockwise and walk the same number of steps again. There you put a piece of wood in the ground. Now go from the gallows to the beech and count again the steps you need. At the beech, turn 90 degrees counterclockwise and repeat the same number of steps. Here you put a little piece of wood in the ground again. The treasure is now in the middle of the two woods. "

Since you want to get rich, you travel to the island. Once there, you will also find the two trees, only the gallows has obviously become a victim of bad weather, it is nowhere to be seen. How do you still find the treasure without plowing the whole island?

 

[lpetrich]

Solution for #38:

Spoiler

The positions of the gallows, the oak, the beech, and the treasure are ${\mathbf x}_G$, ${\mathbf x}_O$, ${\mathbf x}_B$, and ${\mathbf x}_T$.

The coordinate axes I take to be counterclockwise: 90d clockwise rotation does (1,0) -> (0,-1) and (0,1) -> (1,0), and 90d counterclockwise rotation does (1,0) -> (0,1) and (0,1) -> (-1,0). The rotation matrices are thus
$$R_{clockwise} = - R_{counterclockwise} = \epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$
Applied to this problem, the position after visiting the oak is ${\mathbf x}_O + \epsilon\cdot({\mathbf x}_O - {\mathbf x}_G)$ and the position after visiting the beech is ${\mathbf x}_B - \epsilon\cdot({\mathbf x}_B - {\mathbf x}_G)$. The average of these positions is
$${\mathbf x}_T = \frac12 ({\mathbf x}_O + {\mathbf x}_B) + \frac12 \epsilon\cdot({\mathbf x}_O - {\mathbf x}_B)$$
So one goes to halfway in between the oak and the beech trees, then turns 90d clockwise from the beech-to-oak direction, and then goes half of the separation between the beech and the oak.

 

[fresh_42]

39. A village community currently has $1111$ inhabitants - $1$ more than twice as many as $111$ years ago. Except of the village's mill, whose population is $11$ today as then, all districts (Lower Village, Central Village and Upper Village) have grown. Lower Village gained twice as many inhabitants as the smaller Upper Village, whereas Central Village increased its population by exactly $7/11$, so that today it has exactly $1/11$ of the inhabitants of Lower Village.

How many people live(d) in the districts today and then?

 

[lpetrich]

Spoiler

If the total population 111 years ago was $n_0$, then $2n_0 + 1 = 1111$, giving $n_0 = 555$.

The mill, upper village, central village, and lower village have populations $n_m, n_u, n_c, n_l$ now, and $n_{m0}, n_{u0}, n_{c0}, n_{l0}$ then, and those populations are related by these equations:
$$ n_m + n_u + n_c + n_l = 1111 \\ n_{m0} + n_{u0} + n_{c0} + n_{l0} = 555 \\ n_m = n_{m0} = 11 \\ n_l - n_{l0} = 2 (n_u - n_{u0}) \\ n_c = n_{c0} + \frac{7}{11} n_{c0} \\ n_c = \frac{1}{11} n_l $$
The last two equations give us $n_l = 11 n_c$ and $n_c = \frac{18}{11} n_{c0}$. Since all the population numbers must be nonnegative integers, $n_{c0}$ must be a multiple of 11, and for convenience, I set $n_{c0} = 11 n_1$. Then, $n_c = 18 n_1$ and $n_l = 198 n_1$.

The first four equations become
$$ n_u + 216 n_1 = 1100 \\ n_{u0} + 11 n_1 + n_{l0} = 544 \\ 198 n_1 - n_{l0} = 2(n_u - n_{u0}) $$
The first one gives $n_u = 1100 - 216 n_1$, and substituting it into the third equation and rearranging gives us
$$ 2n_{u0} - n_{l0} = 2n_u - 198 n_1 = 2200 - 630 n_1 \\ n_{u0} + n_{l0} = 544 - 11 n_1 $$
This system of equations gives us
$$ n_{u0} = \frac13 (2744 - 641 n_1) \\ n_{l0} = \frac13 (-1112 + 608 n_1) $$
Since all the numbers must be nonnegative integers, we find $\frac{1112}{608} \le n_1 \le \frac{2744}{641}$ or approximately $1.82895 \le n_1 \le 4.28081$. This means that $n_1$ must be 2, 3, or 4. Of these three values, only 4 gives integer numbers, and the complete solution for the three districts is
$$ n_u = 236, n_c = 72, n_l = 792, n_{u0} = 60, n_{c0} = 44, n_{l0} = 440 $$

 

[fresh_42]

40. Which is the smallest natural number that can be written as exactly two different sums of cubes of natural numbers?

 

[mfb]

You should probably add that the cubed numbers should be natural numbers as well, otherwise there is a trivial solution.

 

[fresh_42]

You should probably add that the cubed numbers should be natural numbers as well, otherwise there is a trivial solution.

Done. It isn't really a mathematical question anyway.

 

[DavidSnider]

Spoiler: spoiler

Hardy-Ramanujan number

 

[fresh_42]

41. A few people sit at a round table. Some are liars, others always tell the truth. Everyone claims over his seat neighbor that he is a liar. One woman says, "There are 47 people sitting at this table." Then a man angrily slaps his fist on the table and says, "That's not true, she's a liar. There are 50 people sitting at the table."

How many people are sitting at the table?

D51

 

[lpetrich]

#41:

Spoiler

Since everybody says that their neighbors are liars, that has an interesting result. If A has neighbors B and C, and if A is a truth teller, then B and C are liars. But if A is a liar, then B and C are truth tellers. Thus, the liars and truth tellers alternate around the table. Since the table is round, the sequence must wrap around. That is only possible if there are an even number of people at the table. Otherwise, when one completes a trip around the table, the first person's truthfulness would flip.

The only two numbers proposed are the woman's, 47, and the man's, 50. Of these two numbers, 47 is odd and 50 is even, and since the number of people must be even, it must be 50.

 

[mfb]

The last point is missing a crucial argument:

Spoiler

The only two numbers proposed are the woman's, 47, and the man's, 50. Of these two numbers, 47 is odd and 50 is even, and since the number of people must be even, it must be 50.

If we would only hear these numbers they could both lie and it could be any even number. It is important that the second person says (correctly) "the person saying 47 is a liar", identifying that person as someone saying the truth.

 

[fresh_42]

42. A Harshad or Niven number is a number which is divisible by the sum of its digits. E.g. the taxicab number $1729$ is a Harshad number, because $1+7+2+9= 19 | 1729 = 7\cdot 13\cdot 19 \,.$
Which is the twentieth Harshad number?

D52

 

[BvU]

Too easy ? Or tighter criteria lost in translation ?

Spoiler

n: 1        1
n: 2        2
n: 3        3
n: 4        4
n: 5        5
n: 6        6
n: 7        7
n: 8        8
n: 9        9
n: 10        10
n: 11        12
n: 12        18
n: 13        20
n: 14        21
n: 15        24
n: 16        27
n: 17        30
n: 18        36
n: 19        40
n: 20        42
n: 21        45
n: 22        48
n: 23        50
n: 24        54
n: 25        60

so my 5 cents is on the number 42