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

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fresh_42

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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.

Last edited:

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!
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

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

fresh_42

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Does this mean that they actually have some number less than 2,000?
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
although Adele is now deluding herself with her proximity to 100!

fresh_42

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

fresh_42

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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.
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.
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.
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

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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

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

lpetrich

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

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fresh_42

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

mfb

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Do we have to use all numbers once, or at most once?
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

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Do we have to use all numbers once, or at most once?
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

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

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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

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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

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

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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

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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.

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