Riddles and Puzzles: Extend the following to a valid equation

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In summary, the task is to determine the correct labeling of the urns (WW, WB, BB) by drawing balls from each urn without looking and using the information that the urn labels have been switched.
  • #561
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
 
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  • #562
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.
 
  • #563
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
 
  • #564
#144
13211311123113112211
 
  • #565
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
 
  • #566
fresh_42 said:
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
679
 
  • #567
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
 
  • #568
fresh_42 said:
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
Python:
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)
 
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  • #569
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
 
  • #570
147.
There have been 752 *3 /3 = 2256 matches
2256 = 48 * 47 so there were 49 players who each played 47 matches
 
  • #571
There have been ##752 + \frac{1}{2}\cdot 752## matches.
 
  • #572
Yeah allright. 47*48/2 Keep nitpicking -- we do too :smile:

Takes two to tango they say
 
  • #573
BvU said:
Yeah allright. 47*48/2 Keep nitpicking -- we do too :smile:

Takes two to tango they say
Yes, but there also only 48 players for 47 matches each. :cool:

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
 
  • #574
We're gliding here, aren't we ? This is high school stuff ! 50 kg
 
  • #575
BvU said:
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.
 
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  • #576
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
 
  • #577
fresh_42 said:
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.
 
  • #578
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
 
  • #579
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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  • #580
fresh_42 said:
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?
 
  • #581
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.
 
  • #582
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
 
  • #583
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.
 
  • #584
151.
1567859416592.png


##\theta = \pi/6 \Rightarrow x = 2r\ \ ## and ##\ \ x+r=R \Rightarrow r/R = 1/3 ##
 
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  • #585
152. Find a natural number which can be written as sum of two positive squares in at least two different ways.

D147
 
  • #586
##8^2 + 9^2 = 1^2 + 12^2 = 145##
 
  • #587
DrClaude said:
##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
 
  • #588
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
 
Last edited:
  • #589
Been there, done that !
fresh_42 said:
## ##
 
  • #590
fresh_42 said:
Btw, the smallest is 65.
Darn, I missed that one.
 
  • #591
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
 
  • #592
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.
 
  • #593
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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