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

[fresh_42]

Am I to understand it does NOT spread to the smaller columns ? Please clarify this spreading (e.g. after D1 it can spread to D2 and to E1 -- not to C1 ?)

I don't know what you are quoting here, I said

the fire spreads to all horizontally or vertically neighboured hectares (no diagonals).

So $D1\to C1, D2, E1$ or $F3 \to F2,F4,G3,E3$.

 

[mfb]

Improved fire fighting plan:

Spoiler

Make two "stairs". Save D2, E2, C3, F3, B4, G4, A5, H5 in that order. The fire needs two cycles to danger the next step in the stair, that means you can fight on both sides at the same time. A total of 20 fields will burn down.

 

[fresh_42]

Improved fire fighting plan:

Spoiler

Make two "stairs". Save D2, E2, C3, F3, B4, G4, A5, H5 in that order. The fire needs two cycles to danger the next step in the stair, that means you can fight on both sides at the same time. A total of 20 fields will burn down.

 

[fresh_42]

122. You want to put groceries into the drawer. Between three round glasses of peanut butter is still some room left for another small bottle of olives. If the radius of the peanut butter glasses is $R$ and the radius of the glass of olives is $r$, how big may be the ratio $r/R$ at most, such that the olives fit in?

D122

 

[KnotTheorist]

Spoiler: 122 spoiler

Constructing an equilateral triangle by connecting the centers of the peanut butter jars, we need to find the distance between the center of the triangle and one of the corners. This will be equal to R+r.
Thus using basic trig, we get that r = R(2/√3 -1). Thus r/R = (2/√3 -1)

 

[fresh_42]

123. Show that there are infinitely many $11-$tuples of consecutive natural numbers such that none of them is prime.

D122

 

[mfb]

Spoiler: 123

Consider x to x+10 where x=2310n+2 for a positive integer n (2310 = 2*3*5*7*11).
x, x+2, ..., x+10 are even, x+1 is divisible by 3, x+3 is divisible by 5, x+5 is divisible by 7, x+7 is divisible by 3, x+9 is divisible by 11. This gives an 11-tupel without primes for every positive n.

Alternative proof: If there would be only a finite set then the asymptotic prime density would have to be at least 1/11, in contradiction to that density going to zero.

Both proofs can be extended to arbitrary length of the tuples.

 

[fresh_42]

124. Two fair dice are rolled each round; the result of a throw is the product of the thrown numbers. A game goes over 5 rounds.

Alice throws 5 more points in the second round than in the first round, 6 less in the third round than in the second, 11 more in the fourth round than in the third, and 8 less in the fifth round than in the fourth.

How many points did she score in each of the 5 rounds?

D123

 

[KnotTheorist]

124.
and 8 less in the fourth round than in the fourth.

D123

One of these should say fifth, which is it?

 

[fresh_42]

One of these should say fifth, which is it?

Sorry. It's corrected now.

 

[fresh_42]

125. Find an integer which can be written as a sum of squares in three different ways.

D124

 

[lpetrich]

Doing #124

Spoiler

The first step is to collect the numbers. The problem statement is R2 = R1+5, R3 = R2-6, R4 = R3+11, R5 = R4-8.

We must now find dice-roll values that satisfy all these equations. The complete list of possible dice-roll values is (1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36).

The possible values of (R1,R2) are ((1,6), (3,8), (4,9), (5,10), (10,15), (15,20), (20,25), (25,30)).

The possible values of (R1,R2,R3) are ((3,8,2), (4,9,3), (5,10,4), (10,15,9), (25,30,24)).

The possible values of (R1,R2,R3,R4) are ((5,10,4,15), (10,15,9,20))

The possible values of (R1,R2,R3,R4,R5) are only one: (10,15,9,20,12). That is the solution.

 

[KnotTheorist]

125. Find an integer which can be written as a sum of squares in three different ways.

D124

Spoiler: 125 spoiler

325 = 10² + 15² = 18² + 1² = 6² + 17²

 

[fresh_42]

126. One of us has the least odd value under Euler's $\varphi$ function different from $1$, and one of us is the least that doesn't occur in its range. The third of us tells how many colors are sufficient to color a map on a ring. Me? This is the product of them.

D124

 

[fresh_42]

127. Someone replaced the round shooting targets at the local gun club by equilateral triangles with a side length of 10 inches. Show that among five hits will always be two closer to each other than five inches.

D125

 

[BvU]

129 Make a drawing !

Spoiler

Worst case for 1, 2, 3 is in the corners. dashed radii are < 5" (by at least half a bullethole size) So equilateral triangle ABC has sides < 5" and you can't find two points in there that are > 5"apart

 

[fresh_42]

128. The riches of the country are redistributed after the revolution. Nine subordinates and the former king - a total of ten people - can distribute 10 thalers per month. So every person gets a gold thaler paid out every month - even the king.

The distribution key may be changed if there is a majority for it. A subordinate agrees to a change, provided this increases his own monthly salary. He votes against it when his wages fall. If his monthly salary does not change, he abstains. The king is not allowed to vote, but he is the only one who can suggest how to change the distribution.

What is the maximum monthly paycheck the king can secure$^*)$?

Additional question: The people consists of a total of 1000 people including the king. Each month, 1000 thalers are paid, at the beginning exactly one to each person. Which maximum sum is achievable$^*)$ for the king now?

$^*)$ This means in the long run.

D125

 

[jbriggs444]

Spoiler: 128

7 and 997

Every successful change of distribution which begins with at least two subordinates having a non-zero distribution must also end with at least two subordinates having a non-zero distribution. If the result were to have only one subordinate with a non-zero distribution then there could have been at most one vote "for" and there would also have been at least one vote "against".

Consider a final redistribution that gives the king his maximum share. There must have been at least two subordinates receiving non-zero distributions. If both are victimized then three subordinates must vote in favor of the measure -- three thalers which the King cannot receive. If only one is victimized then at least two other subordinates must vote in favor of the measure. Again there are at least three thalers which the King cannot receive.

So we have an upper bound of n-3 = 7 thalers to the King.

What remains is to determine whether this upper bound is attainable. The following scheme will serve.

The first vote hands the King's thaler to subordinate number 1 and subordinate n's thaler to subordinate number 2.

After this, it is straightforward to take any subordinate with two or more thalers and redistribute them to two other subordinates, thus finally concentrating the wealth on at most two subordinates. Let n be the total distribution. Then these subordinates have k and n-k thalers respectively. Assume n is large enough that both n and n-k are greater than or equal to two (easily achievable for n=10 or n=1000).

The final moves go like:

K  S1 S2 S3 S4
0   k n-k 0 0
k-2 0 n-k 1 1
n-4 0 0   2 2
n-4 1 0   0 3
n-3 2 1   0 0

 

[mfb]

126. One of us has an odd value under Euler's $\varphi$ function, and one of us doesn't occur in its range. The third of us tells how many colors are sufficient to color a map on a ring. Me? This is the product of them.

D124

I'm confused by that one. $\varphi(x)$ is odd for x=1 and x=2 only, giving us two choices for the first number. No odd value apart from 1 occurs in its range (and various even numbers don't occur either), giving an infinite set of choices. The torus needs 7 colors.
That's a lot of options for the product.

 

[fresh_42]

I'm confused by that one. $\varphi(x)$ is odd for x=1 and x=2 only, giving us two choices for the first number. No odd value apart from 1 occurs in its range (and various even numbers don't occur either), giving an infinite set of choices. The torus needs 7 colors.
That's a lot of options for the product.

Yeah, I should have added smallest number which doesn't occur in the range, and excluded $1$. As $1$ wouldn't contribute to the product, I thought I could leave it.

 

[fresh_42]

129. A bag contains $111$ marbles in the colors red, green, blue, and yellow. If we take out $100$ marbles, we will have at least one marble of each color in our hand.

How many marbles do we have to take out at least, such that we will safely hold at least one marble of three colors in our hands?

D126

 

[BvU]

129

Spoiler

Each colour has at least 12 marbles in the bag. We can miss one colour and 11 from a second colour and still get three colours if we grab 88 marbles...

 

[fresh_42]

130. All, but less than 100 party guests, are distributed into teams of equal prime size for a game. Someone observed that the number of teams, except his, is also divisible by that prime. How many persons attended the party, if we assume that the observation is independent of the prime chosen?

D126

 

[mfb]

Spoiler: 130

Let the team size be p. There are k groups. N=kp<100. We know that k-1 is divisible by p, let's write this as k-1=cp. Then N=(cp+1)p. As 100>N>p² the only possible primes are 2, 3, 5 and 7.

For p=7 the only option is c=1, we have 8 teams of 7 members each, 56 members.
For p=5 we have the options c=1,2,3, leading to 6,11,16 teams of 5 members each, 30, 55, 80 members.
For p=3 we get 4, 7, 10, ..., 31 teams of 3 members, 12, 21, 30, ..., 93 members.
For p=2 we get 3,5,...,49 pairs, 6,10,...,30,...,98 members.

There is no prime power, therefore all party sizes have a choice how to split them into prime groups. While 66 appears in both p=2 and p=3 you could split 66 people into 6 groups of 11 members.
To split 30 in teams with a prime number of members we can only make teams of 2, 3 and 5, in all cases someone can make the described observation. That must be the solution.

 

[fresh_42]

To split 30 in teams with a prime number of members we can only make teams of 2, 3 and 5, in all cases someone can make the described observation. That must be the solution.

Yes, and the next Giuga number is 858.

 

[fresh_42]

131. There are 300 apples. The heaviest apple is at most three times as heavy as the lightest one.
Can you divide the apples into groups of four, so that no group of four weighs more than 3/2 times as much as any other group of four?

D126

 

[fresh_42]

132. A sick, old man feels that only a few hours will remain. He definitely wants to settle his inheritance before he dies. So he calls for his sons and says: "I am distributing my money now, each of you should get the same share."

Then he gives one of the gold coins to the eldest son and exactly one seventh of the remaining coins. The second oldest son gets two coins and exactly one seventh of the rest. The third oldest son gets three coins and exactly one seventh of the rest.

The man has just counted the gold pieces for his third oldest son when he suddenly dies. Until then, he had neither distributed all the gold pieces nor considered all sons with their inheritance.

How many sons did the man have?

D127

 

[mfb]

131. There are 300 apples. The heaviest apple is at most three times as heavy as the lightest one.
Can you divide the apples into groups of four, so that no group of four weighs more than 3/2 times as much as any other group of four?

D126

This has a gigantic safety margin.

Spoiler

Sort the apples by weight, let the mass of the lightest one (#1) by 1. Put apple 1, 5, 9, ... in one group, 2, 6, 10, ... in the second and so on. Group 1 has a mass of at least 75. The mass difference between group 1 and 2 is m(2)-m(1) + m(6)-m(5) + ... <= m(300)-m(1) < 2. The mass ratio is smaller than 77/75. Similar for the other groups. Note that, by construction, the groups have ascending masses. In general it is possible to make the mass ratio even smaller by exchanging some apples between the groups.

 

[fresh_42]

This has a gigantic safety margin.

Spoiler

Sort the apples by weight, let the mass of the lightest one (#1) by 1. Put apple 1, 5, 9, ... in one group, 2, 6, 10, ... in the second and so on. Group 1 has a mass of at least 75. The mass difference between group 1 and 2 is m(2)-m(1) + m(6)-m(5) + ... <= m(300)-m(1) < 2. The mass ratio is smaller than 77/75. Similar for the other groups. Note that, by construction, the groups have ascending masses. In general it is possible to make the mass ratio even smaller by exchanging some apples between the groups.

I don't follow your logic. Why is $m(300)-m(1)$ an upper bound and not $4$ times of it? It is no telescope sum. And why is $m(300)-m(1) < 2$ and not $m(300)-m(1)\leq 2m(1)$. One can norm $m(1)=1$ but then we wouldn't need $m(1)$ anymore.

 

[mfb]

It is a telescope sum: m(300)-m(1) = (m(300)-m(299)) + (m(299)-m(298)) + ... + (m(2)-m(1))
All these terms on the right side are non-negative, taking a subset of them cannot be larger than m(300)-m(1).
I defined m(1)=1 before but didn't use it in that one equation for consistency.

 

[fresh_42]

It is a telescope sum: m(300)-m(1) = (m(300)-m(299)) + (m(299)-m(298)) + ... + (m(2)-m(1))
All these terms on the right side are non-negative, taking a subset of them cannot be larger than m(300)-m(1).
I defined m(1)=1 before but didn't use it in that one equation for consistency.

How do you finish? We must show $G_{75}/G_1 \leq 1.5$. What we have is
$$
\dfrac{G_{75}}{G_1}= \dfrac{G_{75}}{G_{74}}\cdot \dfrac{G_{74}}{G_{73}} \cdot \ldots \cdot \dfrac{G_{2}}{G_1} \leq \left( \dfrac{77}{75}\right)^{74} \approx 7
$$
since the telescope argument works only for consecutive numbers and thus consecutive groups.

Edit:
We have $4$ apples per group, so we get $G_{i+1}/G_i \leq 6/4 = 3/2$. And $(3/2)^{74}$ is rather big. I mean your solution is correct ($75$ and $4$ exchanged), but I don't see the reason.

 

[mfb]

Oh... I misread the problem statement. I divided the apples into four groups, not groups of four.

 

[jbriggs444]

Spoiler: 131

Yes, such a division is possible.

Consider an arbitrary division into 4-apple groups. If the heaviest group is more than 3/2 times as heavy as the lightest, rearrange the apples in those two groups to reduce the disparity to less than that. Repeat as needed.

Let us borrow from @mfb and use a telescoping sum to show how this can be done. Let the apples in the two groups be labelled $a_1$ through $a_8$ in order of increasing weight. We reassign $a_1$, $a_3$, $a_5$ and $a_7$ to the "light" group and $a_2$, $a_4$, $a_6$ and $a_8$ to the "heavy" group.

The discrepancy between the two resulting group weights is at most $a_2-a_1\ +\ a_4-a_3\ +\ a_6-a_5\ +\ a_8-a_7$ which is less than or equal to $a_8-a_1$ which is less than or equal to $2a_1$ which is less than or equal to $\frac{1}{2}$ the final weight of the light group.

Edit: @mfb's distribution is a special case of this since every pair of his apple groups $(g_n, g_m)$ with n < m will have apples $a_1$ through $a_8$ in the prescribed sorted order.

 

[fresh_42]

Looks like the worst programming code I've ever seen, but I can't find a loophole. My solution takes the distribution as suggested by @mfb : $\{\,1,76,151,226\,\},\{\,2,77,152,227\,\},\{\,3,78,153,228\,\},\ldots,\{\,75,150,225,300\,\}$ and works with a bunch of inequalities. Not necessarily nicer, however, from a programmer's perspective ...

That's an interesting example to demonstrate the difference between math, constructive math, and computer science.

 

[fresh_42]

133. At a fountain sits a lion made of stone. If the water only flows out of the mouth, the well is full in 24 hours. If the water only flows from the eyes, the well is full in 48 hours. In how many hours is the well filled when the water flows from mouth and eyes at the same time?

D129

 

[DrClaude]

Spoiler: 133

Water from the mouth fills 1/24th of the well in one hour; water from the eyes 1/48th. Therefore, we need to solve
$$
t \left(\frac{1}{24} + \frac{1}{48} \right) = 1
$$
and find $t=16$ hours.

 

[BvU]

132.

Spoiler

Let there be N sons
Son N gets the rest N (what's left over when N-1 has received his share)
Son N-1 gets N-1 plus 1/7 of the rest N-1

So rest N plus 1/7 of rest N-1 is divisible by 7

Simplest case: rest N is 6, rest N-1 is 12 etc: Six sons, 36 coins

 

[fresh_42]

134. How many times can the number 4 appear at the end of a square? It appears once in $2^2=4$ and twice in $12^2=144$, but what is the maximum?

D129

 

[mfb]

Spoiler: 134

"444" is possible, "4444" is not. The last N digits repeat every 10^N squares. To check the last 4 digits it is sufficient to consider the first 10,000 squares.

>>> for i in range(1,1000):
...   if 444 == (i*i % 1000):
...     print(str(i)+"^2="+str(i*i))
...
38^2=1444
462^2=213444
538^2=289444
962^2=925444
>>> for i in range(1,10000):
...   if 4444 == (i*i % 10000):
...     print(str(i)+"^2="+str(i*i))
...

no result

Generalizing a bit:

>>> def check(N,k):
...    for i in range(1,10**N):
...      if k*(10**N-1)/9 == ((i*i)%(10**N)):
...        print(str(i)+"^2="+str(i*i))

It turns out 4 is the only one that can appear more than once at the end of a square.

 

[fresh_42]

135. The numbers from 1 to 100 are noted on a large sheet of paper. Someone is stripping off 25 numbers of it. Now another person should select and delete 25 more among the remaining numbers, in such a way that the arithmetic mean of the remaining 50 numbers is the same as the arithmetic mean of the 100 numbers that were originally listed on the sheet of paper. Is that possible?

D130

 

[DrClaude]

Spoiler: 135

Of course this is possible. Pairing 1 with 100, 2 with 99, and so on, each pair has the same arithmetic mean as the arithmetic mean of the original numbers. Therefore, removing such a pair doesn't change the arithmetic mean.

The procedure then is to simply remove the 25 numbers that are paired with the 25 that were removed. Should a pair of numbers be already removed, then simply remove a random pair instead.

 

[fresh_42]

136. Tom's novel has 342 pages. Every day, he reads exactly the same number of pages. And that works until the last day he finishes reading the book without changing the number.

Tom starts on a Sunday. The following Sunday, he sits with the novel on the sofa as his phone rings. Tom looks again briefly into the book: He has made exactly 20 pages since the morning.

How many more pages will Tom read that day?

D130

 

[DrClaude]

Spoiler: 136

18

Since Tom reads the same number of pages each day, including the final day, in a day he can only read a number of pages that is a factor of 342:
1 2 3 6 9 18 19 38 57 114 171 342

We can eliminate all numbers < 20, since this is the number he has already read that second Sunday. Since this is his 8th reading day and still hasn't finished the book, he reads at most 342 / 8 = 42,75 pages a day. Thats leaves 38 as the only possible number. Since he has already read 20 that day, he will read 18 more.

 

[fresh_42]

137. If we choose a number $n \equiv 0 \mod 3$, and recursively add the cubes of its digits, will we come to an end?

D130

 

[DrClaude]

Spoiler: 137 (not a full answer)

Playing around with it, I found that 153 is a fixed point, as $1^3 + 5^3 +3^3 = 153$. The examples I have checked all seem to end up on that fixed point, but I have not yet been able to prove that it should always be so.

 

[fresh_42]

138. There are two disks with the radii of three and nine centimeters. They touch each other. A non-elastic band is wrapped around the discs, holding the two discs together without a gap between them.

How long does this band have to be?

D130

 

[mfb]

Spoiler: 137 (not a full answer)

Playing around with it, I found that 153 is a fixed point, as $1^3 + 5^3 +3^3 = 153$. The examples I have checked all seem to end up on that fixed point, but I have not yet been able to prove that it should always be so.

An N-digit number is at least 10^N-1 but the sum of cubes can be at most N*9³. This means all 5-digit numbers or larger must get smaller with this process as 5*9³ = 3645. From there on we can simply check all numbers below 10,000 divisible by 3.

 

[BvU]

138.

Spoiler

With a sketch a bit more to scale you see $\theta = \pi/6$ so A is $6\sqrt 3$ and the arcs are $4\pi/3 * 9$ and $2\pi/3 * 3$ for a total of $14\pi + 12\sqrt 3$

 

[fresh_42]

138.

Spoiler

View attachment 248484
With a sketch a bit more to scale you see $\theta = \pi/6$ so A is $6\sqrt 3$ and the arcs are $4\pi/3 * 9$ and $2\pi/3 * 3$ for a total of $14\pi + 12\sqrt 3$

If you were my student I would have answered: no.

Spoiler

The usual dialogue goes as follows:
Student: "Hm, I can see no mistake. Can you give me a hint?"
Me: "Look at your result!"
Student: "But these are the numbers I calculated."
Me: "There is nothing wrong with the numbers."
Student: "Then what else is wrong?"
Me: "You have got $14\pi + 12\sqrt 3$. But what? Bushes? Miles? Trees?

 

[fresh_42]

139. Given the numbers $\{\,1,2,3,4,5,6\,\}$. You can always selected any two of them and add $1$ to each. How do you have to proceed to end up with six equal numbers?

D131