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

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The last point is missing a crucial argument:
If we would only hear these numbers they could both lie and it could be any even number. It is important that the second person says (correctly) "the person saying 47 is a liar", identifying that person as someone saying the truth.
The problem condition gets around that issue with the asserter of 50 calling the asserter of 47 a liar. If the number is not 50, then the asserter of 50 must be lying, implying that the number must be 47. But the number cannot be odd, thus the asserter of 50 must be a truth teller.
 

fresh_42

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Too easy ? Or tighter criteria lost in translation ?

Code:
n: 1        1
n: 2        2
n: 3        3
n: 4        4
n: 5        5
n: 6        6
n: 7        7
n: 8        8
n: 9        9
n: 10        10
n: 11        12
n: 12        18
n: 13        20
n: 14        21
n: 15        24
n: 16        27
n: 17        30
n: 18        36
n: 19        40
n: 20        42
n: 21        45
n: 22        48
n: 23        50
n: 24        54
n: 25        60
so my 5 cents is on the number 42
Yes, easy. I thought for a moment to phrase the question as
 

fresh_42

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43. Two ferries leave mutual river banks at the same time, and travel at a constant, but not necessarily equal speed to the other side. They meet 800m from the east bank. They continue to the other bank. On the way back they meet 400m from the west bank.

How wide is the river?

D52
 

BvU

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'leave mutual river banks at the same time' can be applied twice (opposing ferries always depart simultaneously) and that realistic approach gets you nowhere.

Or the boats turn around instantaneously (thus not ferrying at all) and you get two equations in three unknowns (##w, v_1, v_2##). Fortunately we only need ##v_2/v_1## and we get 1400 m from $$ \begin{align} {w-800\over v_1} & = {800\over v_2} \\ {2w-400\over v_1} & = {w+400\over v_2} \end{align} $$
 

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They leave at once. 1,400 m is not correct.
 
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BvU

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redo the math and shoot
2000 m
o:) it's different...
 

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44. Find a sphenic Carmichael and Harshad number with happy prime factors which is not a narcissistic Armstrong number but the product of two factors palindromic to each other.

D52
 
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44. Find a sphenic Carmichael and Harshad number with happy prime factors which is not a narcissistic Armstrong number but the product of two palindromic numbers.
Can you clarify that you do not mean the product of two numbers forming a palindromic pair such as 19 and 91?
 

fresh_42

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Can you clarify that you do not mean the product of two numbers forming a palindromic pair such as 19 and 91?
Yes, that was it. Corrected.
 
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fresh_42

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Seems as if nobody wants to pick up the hint ...
Can you clarify that you do not mean the product of two numbers forming a palindromic pair such as 19 and 91?
for
44. Find a sphenic Carmichael and Harshad number with happy prime factors which is not a narcissistic Armstrong number but the product of two factors palindromic to each other.

D52
... because the answer is again the taxicab number
$$
1729 = 7\cdot 13 \cdot 19 = 91 \cdot 19 = 9^3+10^3=1^3+12^3=(1+7+2+9)\cdot 91 \\
=1296\cdot 1^3+396\cdot 1^2+36\cdot 1+1 \text{ and } a^{1728} \equiv 1 \operatorname{mod}1729 \text{ for } (a,1729)=1
$$
 

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45. The sultan has ordered a new harem door. However, as he does not quite trust the five harem watchers and their boss, he makes the following request: The door may only be opened by the chief and any of the harem watchmen -or- any three of the harem watchmen.

The boss and the harem watchman can get as many keys to as many locks as necessary. None of the harem watchmen ever gives their keys away.

What is the smallest number of locks and keys for the door?

D55
 

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46. Construct a valid equation ##2019 = 0 \circ 1 \circ 2 \circ 3 \circ 4 \circ 5 \circ 6 \circ 7 \circ 8 \circ 9## with ## \circ \in \{\,+\, , \,-\, , \,\cdot\, , \,:\,\}^9## and any number of parentheses.

D62
 
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I knew a question similar to 45 already, but as no one replied so far:
Let's focus on the 3 out of 5 requirement first. Every lock needs at least three keys, otherwise there is a combination of three watchmen where the key is missing. More than three keys would be unnecessary - every combination of three watchmen will have one of the keys already. If three watchmen are missing then these three need to have all keys for a given lock. There are (5 choose 3)=10 ways 3 out of 5 watchmen can be chosen, we make 10 locks with three keys each and give the three keys to a unique group of watchmen.
The boss cannot have a key for every single lock, but leaving out one of the 10 locks we have would lead to combinations where boss+watchman cannot open the door. We need an 11th lock. The boss gets keys to all the 10 locks discussed before and every watchman gets a key to the 11th lock.
Overall we have 10 locks with 4 keys each and 1 lock with 5 keys.
New puzzle:
0+1+2+(-3-4+5+6)*7*8*9 = 2019
Found by the observation that 2019/(8*9) is very close to 28=4*7, which leads to this result.

Playing a bit more with it:
By changing 0+1+2 to 0*1+2, 0-1+2 and similar we can produce all numbers from 2013 to 2019.

If we use the number 2 as part of the factor 4 it gets a bit more complicated and we only get the 1 to play around:
0-1+2*3*(4+5)*6*7*8/9=2015
0*1+2*3*(4+5)*6*7*8/9=2016
0+1+2*3*(4+5)*6*7*8/9=2017
 

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My solution was ##0+1 \cdot 2 + (3+45) \cdot 6\cdot 7 -8 +9## which immediately creates ##2020##, too.
Ooops. My quickly constructed rule forbids ##45##... should have added a blank or emptyset.
 

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47. Given a watch with an old fashioned clock face. At what times [hh12 : mm : ss.ssssss] do both hands overlap?

D63
 
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My solution was ##0+1 \cdot 2 + (3+45) \cdot 6\cdot 7 -8 +9## which immediately creates ##2020##, too.
Ooops. My quickly constructed rule forbids ##45##... should have added a blank or emptyset.
With concatenation we can make all numbers from 2011 to 2020. New:
##0+12 + (3+45) \cdot 6\cdot 7 -8 -9 = 2011##
##0-1-2 + (3+45) \cdot 6\cdot 7 +8 -9 = 2012##
 
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Praise computers because they can search so many more options. I omitted the 0 as the search space is too large even without it and 0 is trivial to include afterwards. I also didn't consider concatenations or leading minus signs. All possible ways to combine two adjacent terms 8 times in a row with one of 4 operations still give 8!*48 = 2.6 billion options. The script found 36 numbers with (1+2) as first step, a subset with only 83 million options. While I was manually filling in the gaps the script reached (1-2) and quickly found 9 more solutions. Ultimately I just needed them for four more numbers to complete the list:
2000=((((1+2)*(3+4))-5)*(6+(7*(8+9))))
2001=((((1+2)+3)*((4-5)+((6*7)*8)))-9)
2002 = 0-1+2+3+(4+5*6*7+8)*9 (manually derived from 2004)
2003 = 0*1+2+3+(4+5*6*7+8)*9 (manually derived from 2004)
2004=(((1+2)+3)+(((4+((5*6)*7))+8)*9))
2005 = 0+1+2*3+(4+5*6*7+8)*9 (manually derived from 2004)
2006=((((1+2)/3)+((4+5)*(6+7)))*(8+9))
2007=((((((1+2)*3)-4)+((5*6)*7))+8)*9)
2008 = 0+1-2-3+4*(5-6+7*8*9) (manually derived from 2026)
2009 = 0*1*2-3+4*(5-6+7*8*9) (manually derived from 2026)
2010=(((1+2)+3)*((4*((5*6)+(7*8)))-9))
2011=(((((1+2)+3)+4)*(((5*6)*7)-8))-9)
2012=((((1+2)-3)+4)*((5-6)+((7*8)*9)))
2013=(((1+2)/3)+(4*((5-6)+((7*8)*9))))
2014=(((1+2)+((3+4)*5))*((6+(7*8))-9))
2015=((((((1+2)*3)-4)*5)+6)*((7*8)+9))
2016=((((((((1+2)+3)+4)+5)+6)+7)*8)*9)
2017=(((1+2)/3)-((((4*(5-6))*7)*8)*9))
2018=(((1+2)+3)+(4*((5-6)+((7*8)*9))))
2019=((((1+2)+3)*((4-5)+((6*7)*8)))+9)
2020=((((1+2)-3)-4)*((5-6)-((7*8)*9)))
2021=(((((1+2)*3)+4)+(5*6))*((7*8)-9))
2022=(((1+2)+3)*((((4-5)+(6*7))*8)+9))
2023=(((((((1+2)+3)*4)*5)+6)-7)*(8+9))
2024 = 0-1+2+3-4*(5-6-7*8*9) (manually derived from 2026)
2025=(((((((1+2)-3)+4)+5)+6)*(7+8))*9)
2026=(((1+2)+3)-(4*((5-6)-((7*8)*9))))
2027 = 0+1+2*3-4*(5-6-7*8*9) (manually derived from 2026)
2028=(((((((1+2)+3)+4)*5)*6)*7)-(8*9))
2029=(((((1+2)+3)+4)*(((5*6)*7)-8))+9)
2030=(((1-2)-3)*(4-(((5/6)+(7*8))*9))) (one of the minus solutions)
2031=(((((1+2)+3)*4)*(((5+6)*7)+8))-9)
2032=(((1-2)-3)*((4*(5-6))-((7*8)*9))) (one of the minus solutions)
2033=((((((((1+2)*3)*4)+5)*6)+7)*8)+9)
2034=((((((1+2)+3)*4)+((5*6)*7))-8)*9)
2035 = 0-(1+2)*3+4*(5*(6+7)*8-9) (manually derived from 2045)
2036=(((1-2)-3)*(4+(((5-6)-(7*8))*9))) (one of the minus solutions)
2037=(((((1+2)*3)*4)*((5/6)+(7*8)))-9)
2038 = 0-1-2-3+4*(5*(6+7)*8-9) (manually derived from 2045)
2039 = 0*1-2-3+4*(5*(6+7)*8-9) (manually derived from 2045)
2040=((((1+2)+3)-(4*((5/6)-(7*8))))*9)
2041=((((((1+2)*3)*4)+5)*((6*7)+8))-9)
2042 = 0-1+23-4*(5-6-7*8*9) (manually derived from 2026, using concatenation)
2043=((((((1+2)*3)-4)*(5+(6*7)))-8)*9)
2044=((((1+2)-3)+4)*(((5*(6+7))*8)-9))
2045=(((1+2)/3)+(4*(((5*(6+7))*8)-9)))
2046=(((((1+2)-3)+4)*((5/6)+(7*8)))*9)
2047=(((((((1+2)+3)*4)*(5+6))-7)*8)-9)
2048=(((1-2)+3)+((4*((5/6)+(7*8)))*9)) (one of the minus solutions)
2049=(((((1+2)+3)*4)*(((5+6)*7)+8))+9)
2050=(((((1+2)+3)+4)*5)*(((6*7)+8)-9))
As 2042=(((1-2)-3)+((4*((5/6)+(7*8)))*9)) it is possible to get all numbers from 2000 to 2050 without using concatenation.

Also interesting due to its chain character:
2033=((((((((1+2)*3)*4)+5)*6)+7)*8)+9)

2016, simplified: 2016=(1+2+3+4+5+6+7)*8*9

Edit: After a few hours of running the script now moved on to (2+3) and (2-3) as first step. All numbers found.
2002=( (1/(2+3))*4+5*6 )*(7*8+9)
2005=(1+((((((2+3)*4)+(5/6))+7)*8)*9))
2038=(1+(((2-3)+(4*((5/6)+(7*8))))*9))
The solution to 2002 is something typical for a computer: Starting with 1/5 without a multiplication by 5? What an exotic approach. The factor 5 comes from (7*8+9) later.
 
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fresh_42

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48. An avid chess player sits in front of a pack of Toffifay and tries to empty this box. Can he manage to eat all the caramel if he makes it a condition to gradually empty the pack like a knight in chess?

There are 2 different pack sizes 4x5 (20 pieces) and 3x5 (15 pieces).


245383
245384


Is that possible?

What would be the smallest possible rectangular package shape that would allow for complete emptying of the box?

Which the smallest possible square?

D67
 
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This problem of candy in a rectangular grid is the Knight's Tour problem in disguise, where a knight must visit each square exactly once. A tour can be open or closed, a closed tour being one where the knight can move once between the two ends of a tour. I will first find the general solutions, then specialize to the statement of the problem here. I will use size convention m*n, where m <= n. The general solutions are stated in these three papers:

Knight's Tour Revisited by Paul Cull and Jeffrey De Curtins, 1978
Every board with m >= 5 has a general tour, and every board with m >=5 and at least one of m and n even has a closed tour.

Which Rectangular Chessboards Have a Knight's Tour? by Allen J. Schwenk, Mathematics Magazine, Vol. 64, No. 5 (Dec., 1991), pp. 325-332
A board has a closed tour unless at least one of the following conditions holds:
  1. both m and n are odd
  2. m = 1, 2, or 4
  3. m = 3 and n = 4, 6, or 8

Knight's Tour, by Kevin McGown, Ananda Leininger. Advisor: Paul Cull
A board has a general tour if
  1. m = 3 and n = 4, 7, or higher
  2. m = 4 and n = 5 or higher

Some of the are rather complicated, with some of them involving constructing solutions and assembling larger solutions from them. Some of them are simple, however, like both dimensions being odd meaning no closed tours, and one dimension being 1 or 2 meaning no tours of any kind.

For a 3*5 board, there are no tours. Here is a proof. For 3*5, the square coordinates range from (1,1) to (3,5). Consider squares (2,1) and (2,5). Square (2,1) has tour neighbors (1,3) and (3,3), but square (2,5) has those tour neighbors also. So once one reaches (1,3) and goes to (2,1), one must go to (3,3), making (2,5) inaccessible.

For a 4*5 board, there is a tour. The third paper has a construction of it: (1,2), (3,1), (4,3), (3,5), (1,4), (2,2), (4,1), (3,3), (4,5), (2,4), (3,2), (4,4), (2,5), (1,3), (2,1), (4,2), (3,4), (1,5), (2,3), (1,1)

The smallest board with a tour is a 3*4 one. For smaller dimension 1, the knight has no move in the board. For smaller dimension 2, each corner square has only one possible move out of it, meaning that a tour must start or end at a corner. However, there are 4 corners, making a tour impossible. For 3*3, the center square is inaccessible from the other. That leaves 3*4, and a tour for that one is (1,1), (2,3), (3,1), (1,2), (2,4), (3,2), (1,3), (2,1), (3,3), (1,4), (2,2), (3,4). All other boards with tours have sizes greater than it, at least 3*7 or 4*4.

The smallest square board with a tour is a 5*5 one. That is because a 4*4 one does not have a tour, and because a tour does exist for 5*5, as stated in the first paper. The third paper has a proof that the 4*4 one does not have a tour. It involves domino tilling for 4*n boards, where squares (1,even), (2,even), (3,odd), and (4,odd) have the same color, as do squares (1,odd), (2,odd), (3,even), (4,even). The paper has a proof of a theorem that the knight must visit all of one color in sequence, then all of the other color in sequence, though the proof is rather complicated. With that proof, there is a simple proof that 4*4 does not contain a tour. After doing the first color, we are left with the second color. The only tours possible in it are (1,1), (2,3), (4,4), (3,2) and (2,1), (1,3), (3,4), (4,2), both of them closed. So a complete tour is not possible.

In summary:
  • 3*5 - no tour
  • 4*5 - tour
  • smallest with tour: 3*4
  • smallest square with tour: 5*5
 

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49. Little John silently thinks of two natural numbers ##n## and ##m##. Then he adds the numbers ##n, n + 1, n + 2, \ldots , n + m## and loudly proclaims the sum. For example, if he thinks of ##x = 4## and ##y = 3##, he says aloud at the end: ##22 (= 4 + 5 + 6 + 7)##.

Which natural numbers will he never proclaim?


D68
 
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The sum is (1/2)*(m+1)*(2n+m).

For even m, m = 2k, the sum is (2k+1)*(n+k)
For odd m, m = 2k+1, the sum is (k+1)*(2n+2k+1)

If natural numbers include 0, then one can set m = 0 and make the sum n. Since n is a natural number, the sum will also be a natural number.

But if natural numbers do not include 0, then the problem becomes more complicated.

For m = 1, the sum is (2n+1), covering every odd number at least 3
For m = 3, the sum is 2*(2n+3), covering every even number at least 10

So that leaves us with 1, 2, 4, 6, and 8 that cannot be sums.

We must look at additional values of m.

For m = 2, the sum is 3*(n+1), removing 6 from those values.
For m = 4, the sum is 5*(n+2), which is not very informative.
For m = 5, the sum is 3*(2n+5), also not very informative.
Higher m's likewise do not yield new information.

For the un-proclaimable numbers are 1, 2, 4, and 8.
 
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That has a mistake.
For m = 3, the sum is 2*(2n+3), covering every even number at least 10
How do you cover e.g. 12 with that? You only cover numbers that give a remainder 2 when divided by 4.

My solution:
The sum is S(n,m) = n(m+1) + m(m+1)/2.

All powers of 2 are unreachable:
For even m S(n,m) is a multiple of (m+1), namely (n+m/2)(m+1). m+1 is odd, therefore the product cannot be a power of 2.
For odd m let m=2k-1, then S(n,2k-1)=(2n+2k-1)k. The first factor is odd, therefore the product cannot be a power of 2.

All other numbers are reachable:

m=1 leads to a sum of 2n+1, we can reach all odd numbers starting at 3.

For even numbers (which are not powers of 2) let x be our target number. Define integers k,c such that x=c2k with odd c>1.
  • If c>2k+1 then m=2k-1 covers x as S(n,2k-1) = 2k n + (2k-1)2k-1
  • If c<2k+1 then m=c-1 covers it: m is even, therefore S=(n+m/2)(m+1) where both numbers are integers, the second factor is c, and n can be chosen to make the left factor 2k.
  • As c is odd it cannot be equal to 2k+1
 

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50. A tennis club is planning a tournament. 116 players have signed in. The final is planned for Sunday 4 pm and the tournament director wants to know how many matches he has to schedule before. A match lasts on average 2 hours. Breaks between matches should not be considered here.

D68
 
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BvU

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You mean matches ? 'Game' has a different meaning in tennis 😉

If every loser is sent home, there have to be 114 matches before the final
 

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