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

[BvU]

139. This time you are the

Spoiler

start with a total of 21 and add 2 at each turn -- no way to get an even number

 

[fresh_42]

139. This time you are the

Spoiler

start with a total of 21 and add 2 at each turn -- no way to get an even number

It is hard to compete with codes, smart people and quick stuff ... Should have taken a bit longer.

 

[fresh_42]

140. Can you decode the password?

D131

 

[BvU]

It is hard to compete with codes, smart people and quick stuff ... Should have taken a bit longer.

How come you have such a high like percentage ?

 

[fresh_42]

141. Can you separate the bulls with two square fences?

D131

 

[TeethWhitener]

#141

Spoiler

 

[fresh_42]

142. 1600 bananas are distributed among 100 monkeys. Individual monkeys can also go out empty-handed. Prove that there are always four monkeys with the same number of fruits!

D131

 

[BvU]

142.

Spoiler

Cheapest is to give three monkeys nothing, three monkeys one banana each, three two each etc. By the time you have given the 97th monkey sixteen bananas you are out of bananas.

 

[fresh_42]

142.

Spoiler

Cheapest is to give three monkeys nothing, three monkeys one banana each, three two each etc. By the time you have given the 97th monkey sixteen bananas you are out of bananas.

I don't get it. 16 bananas would be given for the 51st monkey. And 3(0+1+2+...+16)=408 leaves me with 49 monkeys and 1192 bananas. Applying your scheme, I will have 112 bananas left before I turn to monkey 97. Number 96 received 31 bananas.

 

[BvU]

Forgot the three that get nothing - poor sods. Monkey 99 gets 32 bananas and THEN there's only 16 of them yellow thingies left.

The idea was good - the execution slightly less ...

 

[fresh_42]

143. How many numbers are there that contain their length as a digit? Example: 1024 has the length 4 and also contains the number 4. Is there a solution without a brute force code?

D131

 

[DrClaude]

Spoiler: 143

Take $n$ to be the length of the number. There are $9 \times 10^{n-1}$ numbers of length $n$: 9 possibilities for the first digit, since it cannot be zero (*), and 10 for each subsequent digit. The count of numbers that do not have the digit $n$ in them is $8 \times 9^{n-1}$ (same as the previous formula, but with one less possibility at each position). Therefore, there are
$$
\sum_{n=1}^{9} \left( 9 \times 10^{n-1} - 8 \times 9^{n-1} \right)
$$
numbers that contain their length as a digit. Using a calculator, I get this sum as 612579511.

Edit: (*) zero can be the first digit for $n=1$, but the formula obtained works also in that case.

 

[fresh_42]

144. I know you - and me - don't like them very much, but from time to time ...
What is the next row?
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
...

D132

 

[TeethWhitener]

#144

Spoiler

13211311123113112211

 

[fresh_42]

145. What is the smallest natural number with a multiplicative persistence of $5$ in the decimal system?
The multiplicative digit root is the repeated process of multiplying all digits, e.g.
$$
3784 \longrightarrow 672 \longrightarrow 84 \longrightarrow 32 \longrightarrow 6
$$
Tme multiplicative persistence is the number of arrows, iterations. For $3784$ it is $4$.

D132

 

[DavidSnider]

145. What is the smallest natural number with a multiplicative persistence of $5$ in the decimal system?
The multiplicative digit root is the repeated process of multiplying all digits, e.g.
$$
3784 \longrightarrow 672 \longrightarrow 84 \longrightarrow 32 \longrightarrow 6
$$
Tme multiplicative persistence is the number of arrows, iterations. For $3784$ it is $4$.

D132

Spoiler: spoiler

679

 

[fresh_42]

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

 

[DavidSnider]

145. What is the smallest natural number with a multiplicative persistence of $5$ in the decimal system?
The multiplicative digit root is the repeated process of multiplying all digits, e.g.
$$
3784 \longrightarrow 672 \longrightarrow 84 \longrightarrow 32 \longrightarrow 6
$$
Tme multiplicative persistence is the number of arrows, iterations. For $3784$ it is $4$.

D132

Spoiler: In honor of Bill's recent Insight Article :)

def mp(n,o):   
    digits = [int(c) for c in str(n)]
    prod = 1
    for d in digits:
        prod = prod * d

    if prod < 10:
        return o
    
    return mp(prod,o+1)

n = 10
while mp(n,1) < 5:
    n+=1

print(n)

 

[fresh_42]

147. In a chess tournament every player has a match with everyone else. The winner will receive a green card, the loser a red one, and in case of a remis, they will receive a yellow card each. At the end of the tournament there have been distributed exactly 752 cards of each color. How many competitors have been in the competition?

D133

 

[BvU]

147.

Spoiler

There have been 752 *3 /3 = 2256 matches
2256 = 48 * 47 so there were 49 players who each played 47 matches

 

[fresh_42]

There have been $752 + \frac{1}{2}\cdot 752$ matches.

 

[BvU]

Yeah allright. 47*48/2 Keep nitpicking -- we do too

Takes two to tango they say

 

[fresh_42]

Yeah allright. 47*48/2 Keep nitpicking -- we do too

Takes two to tango they say

Yes, but there also only 48 players for 47 matches each.

Well, that was an easy one. The only hurdle was to avoid double counting. With the usual points instead of cards it would have been even easier.

148. 100 kilograms of fruits lie to dry in the sun. The water content is initially at 99 percent. If the proportion is only 98 percent, how heavy are the fruits?

D133

 

[BvU]

We're gliding here, aren't we ? This is high school stuff ! 50 kg

 

[fresh_42]

We're gliding here, aren't we ? This is high school stuff ! 50 kg

Yeah, and I even skipped a couple of others being either obvious or solvable by code. But 146. is more challenging and still open! Or the relatively easy 140.

 

[fresh_42]

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

 

[mfb]

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:

Spoiler

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]

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.

D135

 

[BvU]

149. Quiet on this front !

Spoiler

180 min between 6 and 9 -- three times x + 40 + x = 180 $\Rightarrow$ x = 35. It is now 8:25

 

[mfb]

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.

Spoiler

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]

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.

 

[fresh_42]

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

 

[mfb]

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.

Spoiler

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.

 

[BvU]

151.

Spoiler

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

 

[fresh_42]

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

D147

 

[DrClaude]

Spoiler: 152

$8^2 + 9^2 = 1^2 + 12^2 = 145$

 

[fresh_42]

Spoiler: 152

$8^2 + 9^2 = 1^2 + 12^2 = 145$

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

 

[fresh_42]

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 let's 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

 

[BvU]

Been there, done that !

## ##

 

[DrClaude]

Btw, the smallest is 65.

Darn, I missed that one.

 

[fresh_42]

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?

D150

 

[mfb]

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]

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