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

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fresh_42

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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

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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
 
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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.
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

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This has a gigantic safety margin.
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.
 
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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

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

jbriggs444

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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.
 
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fresh_42

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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

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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?

Braunschweig_Loewenbrunnen_um_1880.jpg


D129
 

DrClaude

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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

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132.
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

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

Code:
>>> 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:
Code:
>>> 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

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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

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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

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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

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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.
 
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fresh_42

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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?

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DrClaude

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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

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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?

1566403943578.png


D130
 
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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 10N-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

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138.
1566427642956.png

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

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138.
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. :biggrin:
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? :biggrin::biggrin::biggrin:
 
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fresh_42

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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
 

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