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

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fresh_42

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149. What time is it if until 9 o'clock so many minutes are missing, as 40 minutes ago three times as many minutes after 6 o'clock had elapsed?

D134
 
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146. A newly combined class has ##33## students. Each of them introduces themselves by first name and family name. The kids recognize, that some have the same first name and some even the same family name. So every kid writes on the blackboard how many others have the same first name, and how many the same family name, not counting themselves. At the end there are ##66## numbers on the board, and every number ##0,1,2,\ldots, 9,10## occurs at least once.

Are there at least two kids in class with the same first and family name?

D133
Quite straight forward:
Instead of using "others", let's add one and count how many students are there for a given name. All numbers from 1 to 11 occur at least once now. That means there must be names A,B,..K that appear 1,2,..,11 times, although we don't know which names are first and family names. 11+10+...+1 = 66, which means we covered all names already (as every kid has two names in this counting), all students have first and last name out of these 11 names.
Without loss of generality let K be a surname. If no kids share both names then all 11 kids with this surname would need a different first name - but there can't be 11 first names, some kids need to share both surname and first name.
We can do more. There must be at least 4 surnames to reach a sum of 33 (e.g. 11+10+9+3), or at most 7 first names. We must have at least 4 name collisions with K, and more if we consider all surnames.
 

fresh_42

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150. Determine ##\{\,(p_n,p_{n+1},p_{n+2})\in \mathbb{P}\,|\,p^2_n+p^2_{n+1}+p^2_{n+2}\in \mathbb{P}\,\}## where ##\mathbb{P}=\{\,p_1<p_2<\ldots<p_n<\ldots\,\}## is the set of all primes.

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BvU

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149. Quiet on this front !
180 min between 6 and 9 -- three times x + 40 + x = 180 ##\Rightarrow## x = 35. It is now 8:25
 
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150. Determine ##\{\,(p_n,p_{n+1},p_{n+2})\in \mathbb{P}\,|\,p^2_n+p^2_{n+1}+p^2_{n+2}\in \mathbb{P}\,\}## where ##\mathbb{P}=\{\,p_1<p_2<\ldots<p_n<\ldots\,\}## is the set of all primes.
Apart from 3, squared primes always have a remainder of 1 modulo 3. If you add three of them you get a number that is divisible by 3 (but larger than 3) and cannot be a prime. That means we only have to test cases where 3 is one of the primes:
##2^2 + 3^2 + 5^2 = 38## - not a prime
##3^2 + 5^2 + 7^2 = 83## - a prime
The set consists only of (3,5,7).

What is the minimum number of name collisions for #146, by the way?
Extending the previous answer: Let's ignore the names that occur 1-3 times, this removes at most 6 children. We are left with 8 names. This can give at most 16 unique first+last name combinations for 27 students, which means we need at least 11 name collisions (three children sharing the name count as two name-collisions). Is this the best lower bound?
 

fresh_42

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I don't know. I haven't looked into it. This thread here is more of the "quick and dirty" kind. And I'm notoriously bad at counting.
 

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151. Consider a circular disc of radius R cut into six equal segments like a pizza. Now we inscribe a circle of radius r in such a segment such that it touches all three boundaries. What is the ratio r/R?

D147
 
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I put some names in a spreadsheet and found a solution with only 11 name collisions. We can't do better, we can get 11, that must be the minimal number.
K J
K J
K J
K J
K C
K E
K F
K F
K F
K H
K H
I J
I J
I J
I C
I E
I F
I H
I H
I H
G J
G A
G C
G E
G F
G H
G H
D J
D E
D F
D H
B J
B E
With some trial and error I found a solution with only 8 different names (25 collisions), but I don't know if that is minimal (maximal). The idea is to match up last and first names, e.g. B and D as 2 and 4 last names with F as 6 first names. The better that matches the fewer unique names we get.
 

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151.
1567859416592.png


##\theta = \pi/6 \Rightarrow x = 2r\ \ ## and ##\ \ x+r=R \Rightarrow r/R = 1/3 ##
 
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fresh_42

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152. Find a natural number which can be written as sum of two positive squares in at least two different ways.

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DrClaude

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##8^2 + 9^2 = 1^2 + 12^2 = 145##
 

fresh_42

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##8^2 + 9^2 = 1^2 + 12^2 = 145##
Btw, the smallest is ##65##.

153. On a circle are ##n## points that form a polygon. It is assumed that no more than two diagonals of the polygon intersect at any point (within the polygon).

The sides and diagonals of the polygon divide the circular area into partial areas. Taking into account the special cases ##n = 1## and ##n = 2##, we see:
\begin{array}{c|c|c}
n&\text{ areas }&\\
\hline \\
1& 1&\\
2& 2& \text{semicircles}\\
3 & 4&\text{ triangle plus segments} \\
4&8&\text{quadrilateral divided by diagonals plus segments}\\
\ldots &\ldots & \ldots
\end{array}
How many areas do we get for ##n=11\,?##

D149
 

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154. A plantation has the melon harvest transported to the station. On a truck are at the beginning exactly 1 ton of melons, which are known to consist of 99% water. The long drive through a hot desert landscape lets some of the water evaporate, so that they only consist of 98% of water on arrival at the station.

How much does the load of the truck weigh now?


D150
 
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fresh_42

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154. 41 Kids arrive at the summer camp. There are just enough empty rooms, with three, four and five beds - of at least one each, of the four-bed rooms even more than one - and there are more three-bed rooms than rooms with four or five beds. How many rooms of each kind are there?

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Integers are room/kid numbers, words are room size: We need at least 1 five, 2 four, 4 three, this alone houses 25 kids already, leaving 16 to distribute. We need at least 4 more rooms, which means we need at least 2 more rooms with three => 10 kids left to distribute. 2 five is a solution, for a total of 3 five, 2 four, 6 three. Can we have more rooms with three? 3 of them instead of 2 leave 7 to distribute, doesn't work. 4 of them leave 4 to distribute, that works and is the last option that does: 1 five, 3 four, 8 three.
 

fresh_42

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155. Find all six digit numbers with the following property: If we move the first (highest) digits at the end, we will get three times the original number.

D151
 

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