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

[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

 

[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