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

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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.
This is the year we are looking for :biggrin:
 
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Alternatively (but equivalently):
This year is the year we are looking for.
 
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Have you figured out some of the other ones?
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

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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?
 

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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?
615384 = 153846 * 4
 

fresh_42

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This is really easy to brute force, is there another way to do it?
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##
 
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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?
 
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I will take on #36:
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

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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*}
 
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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.
 

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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?
 
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Solution for #38:
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.
 

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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?
 
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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

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40. Which is the smallest natural number that can be written as exactly two different sums of cubes of natural numbers?
 
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You should probably add that the cubed numbers should be natural numbers as well, otherwise there is a trivial solution.
 

fresh_42

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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.
 

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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
 
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#41:
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.
 
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The last point is missing a crucial argument:
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.
 
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fresh_42

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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
 

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Too easy ? Or tighter criteria lost in translation ?

Code:
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
 

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