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

[fresh_42]

You cannot define "west" at the south pole.

 

[SSequence]

You cannot define "west" at the south pole.

Same is supposed to hold for "5km towards south" (at southpole)?

 

[fresh_42]

Same is supposed to hold for "5km towards south" (at southpole)?

Yes. That's why we have to circle the south pole.

 

[KnotTheorist]

Spoiler: 108 spoiler

The smallest 8 digit pandigital number divisible by 36 is: 10237896
Here is my python code that ran though the range (9999999, 99999999)
for i in range(9999999, 99999999): if i%36 == 0: if(len(set(str(i))))==8: print(i)

 

[fresh_42]

109. In an apartment house live some families. Of these, we know there are

• more children than parents
• more parents than boys
• more boys than girls
• more girls than families

No family is childless, each one has a different number of children. Every girl has at least one brother and at most one sister. One family has more children than all others combined.

How many families live in this house and how are they composed?

D110

 

[mfb]

Pen and paper approach for 108:

Spoiler

The number must be divisible by 9, which means the sum of its digits must be divisible by 9. There are just two digits left out. As 0+1+...+9 is divisible by 9 these two missing digits must add to 9. We can leave out 0 and 9, or 1 and 8, and so on, but we want to get the smallest number, so the best option is to leave out 4 and 5. That gives us the approach 1023xxxx with the remaining digits 6,7,8,9. It must be divisible by 4, it has to end in 68, 76 or 96. Out of that 96 leads to the smallest number, 10237896.

 

[bluej]

Spoiler: #108

10237896=36x284386
using 5-line perl

#!/usr/bin/perl
for ($a=10000000;$a<100000000;$a++) {
  ($a % 36 ==0) && (scalar(do{my %s; grep {!$s{$_}++}  ("$a" =~ /(\d)/g) }) == 8 ) && last;
}
print "$a=36x", $a/36;

 

[fresh_42]

110. My watch is broken. The small hand shows about ten and the big hand about two. Both hands form an identical angle to twelve.

When did my clock stop (exactly to the second)?

D114

 

[BvU]

104. How many different (connected) plane figures can be created with five squares?

Let's limit the playing field: 'connected' does not mean 'with a common point', but 'with a common side' and stacking is not allowed. If that's correct, then I doodle

Spoiler

14

 

[fresh_42]

Let's limit the playing field: 'connected' does not mean 'with a common point', but 'with a common side' and stacking is not allowed. If that's correct, then I doodle

Spoiler

14
View attachment 247657

I think we should skip the symmetries of our polyominoes, e.g. 2-5, 3-4. I should have said it, sorry.
Nevertheless, there are some missing, three if I had compared them correctly.

 

[KnotTheorist]

Spoiler: 110 Spoiler

I don't believe you get to an exact second.
I think the time is 10:09:13 +11/13 seconds

 

[fresh_42]

111. Between 10 and 100 people are present at a party. Some leave early. At the end of the party, everyone present shakes hands with everyone else. Overall, there were exactly half the number of handshakes as if all party guests had stayed until the end.

How many people were present at the party? How many left early?

D114

 

[jbriggs444]

111. Between 10 and 100 people are present at a party. Some leave early. At the end of the party, everyone present shakes hands with everyone else. Overall, there were exactly half the number of handshakes as if all party guests had stayed until the end.

How many people were present at the party? How many left early?

D114

Spoiler

So one triangular number equal double another when the larger triangle is between 10 and 100.
$ ./test.pl
Answer is: 21, 6
There were 210 handshakes before the people left and 105 after

#!/usr/bin/perl
#
use strict;

my @triangle;
for ( my $i=1; $i<=100; $i++ ) {
        $triangle[$i] = $i * ($i-1) / 2;
        if ( $i >= 10 ) {
                for ( my $j=1; $j<$i; $j++ ) {
                        if ( 2 * $triangle[$j] == $triangle[$i] ) {
                                print "Answer is: $i, ", $i-$j, "\n";
                                print "There were ", $triangle[$i], " handshakes before the people left and ", $triangle[$j], " after\n";
                        };
                };
        };
};

 

[KnotTheorist]

Spoiler: 111 Spoiler

Assuming each guest present at the end of the party shakes hands with every other guest, not including themselves, we need to find an integer solution to 2n(n-1) = (n+k)(n+k-1), where n is the number of people at the end of the party and k is the number of people who left.
Let n = 15, and k = 6
Thus we get there were 105 hand shakes. If everyone stayed, a total of 21, there would be 210 handshakes.

Thus 21 guests were there, 15 were there at the end, and 6 left early

 

[fresh_42]

112. John plans to hike on the Appalachian Trail. He starts somewhere in the middle and throws a coin to decide whether he shall go north or south. Then he walks to the next lodge and stays over night. Every morning he throws a coin again and wanders either north or south, according to the coin toss.

Back at home, he tells his friend, Ben, how he spent his vacation and that he finished his walk right where he started it. Ben now wants to know how big the likelihood was.

"The probability is exactly $95 \%$ of the probability this event would have had if my hike had been two days shorter," John replies.

How big is the likelihood that John finished his walk right where he started it and how long was John on the trail?

D114

 

[jbriggs444]

Spoiler: 112 Spoiler

We are looking at the middle terms in binomial distributions. We want the middle terms in two successive rows with an odd number of terms, one of which is to be 95% of the other.

Naturally, since the totals for a binomial distribution double at each step, this will mean that the term from the latter row will actually be 4*.95 = 3.8 times that in the previous.

I will brute-force it with Pascal's triangle and trust double precision float to be adequate. [The evaluation order leads to fairly well-conditioned arithmetic and the floating point range will not be stressed, though precision will].

e.g.

After two days, the ratio is the middle 2 on line 3 to the 1 on line 1.
After four days, the ratio is the middle 6 on line 6 to the 2 on line 3. etc.

[john.d.briggs@nhnaunxlfapb001 ~]$ ./test.pl
After 2 days ratio is 2
After 4 days ratio is 3
After 6 days ratio is 3.33333333333333
After 8 days ratio is 3.5
After 10 days ratio is 3.6
After 12 days ratio is 3.66666666666667
After 14 days ratio is 3.71428571428571
After 16 days ratio is 3.75
After 18 days ratio is 3.77777777777778
After 20 days ratio is 3.8
Probability of landing at starting spot after 20 days is 0.176197052001953
Probability of landing at starting spot after 18 days is 0.185470581054688

 

[fresh_42]

Spoiler: 112 Spoiler

We are looking at the middle terms in binomial distributions. We want the middle terms in two successive rows with an odd number of terms, one of which is to be 95% of the other.

Naturally, since the totals for a binomial distribution double at each step, this will mean that the term from the latter row will actually be 4*.95 = 3.8 times that in the previous.

I will brute-force it with Pascal's triangle and trust double precision float to be adequate. [The evaluation order leads to fairly well-conditioned arithmetic and the floating point range will not be stressed, though precision will].

e.g.

After two days, the ratio is the middle 2 on line 3 to the 1 on line 1.
After four days, the ratio is the middle 6 on line 6 to the 2 on line 3. etc.

[john.d.briggs@nhnaunxlfapb001 ~]$ ./test.pl
After 2 days ratio is 2
After 4 days ratio is 3
After 6 days ratio is 3.33333333333333
After 8 days ratio is 3.5
After 10 days ratio is 3.6
After 12 days ratio is 3.66666666666667
After 14 days ratio is 3.71428571428571
After 16 days ratio is 3.75
After 18 days ratio is 3.77777777777778
After 20 days ratio is 3.8
Probability of landing at starting spot after 20 days is 0.176197052001953
Probability of landing at starting spot after 18 days is 0.185470581054688

Of course, there is also an easy exact solution. Just calculate $P(n)$ and solve $\dfrac{P(n)}{P(n-2)}=0.95\,.$

 

[KnotTheorist]

Of course, there is also an easy exact solution. Just calculate $P(n)$ and solve $\dfrac{P(n)}{P(n-2)}=0.95\,.$

This is so much easier than what I was trying. I was trying to run enough simulations to satisfy the 95% using the incomplete gamma function and whatnot.

 

[fresh_42]

113. We have a target to shoot arrows at with $6$ rings, labeled $16,17, 23, 24, 39$ and $40$ in the center. Given we always hit what we want, how many different combinations of which rings will add up to exactly $100$ points?

D115

 

[BvU]

I think we should skip the symmetries of our polyominoes, e.g. 2-5, 3-4. I should have said it, sorry.
Nevertheless, there are some missing, three if I had compared them correctly.

These can only be morphed into one another if you allow a third dimension... a restriction I stuck to and I should have mentioned (i.e. I consider a lefthand glove different from a righthand glove)
I completely missed these two

Spoiler

and am really curious about a possible last escapee

 

[fresh_42]

These can only be morphed into one another if you allow a third dimension... a restriction I stuck to and I should have mentioned (i.e. I consider a lefthand glove different from a righthand glove)
I completely missed these two

Spoiler

View attachment 247679

and am really curious about a possible last escapee

$1 \\ 2 \\ 3 \; \;4 \; \;5$

 

[KnotTheorist]

113. We have a target to shoot arrows at with $6$ rings, labeled $16,17, 23, 24, 39$ and $40$ in the center. Given we always hit what we want, how many different combinations of which rings will add up to exactly $100$ points?

D115

Spoiler: 113 Spoiler

I wrote a short script to check my math and it is not possible to add up any combination of these numbers to get 100. So the answer is 0.

 

[fresh_42]

Spoiler: 113 Spoiler

I wrote a short script to check my math and it is not possible to add up any combination of these numbers to get 100. So the answer is 0.

There is a solution.

 

[KnotTheorist]

There is a solution.

I totally forgot about repeating the same ring.

Spoiler: 113 Spoiler

I believe the answer is 10. The only way to add up to 100 is 16, 16, 17, 17, 17. So we need to find all possible orders to hit the 16 ring twice and the 17 ring 3 times.

 

[fresh_42]

I totally forgot about repeating the same ring.

Spoiler: 113 Spoiler

I believe the answer is 10. The only way to add up to 100 is 16, 16, 17, 17, 17. So we need to find all possible orders to hit the 16 ring twice and the 17 ring 3 times.

Well, four times 17, not three, but yes, this is the only solution.

 

[KnotTheorist]

Well, four times 17, not three, but yes, this is the only solution.

Yes, sorry. In that case it will be 15 combinations.

 

[fresh_42]

114. How many matches need to be removed at least, in order to destroy all squares?

D115

 

[mfb]

Spoiler: 114

I can prove that it must be at least 9, but I only find (many) solutions with 10, two of them are below.
Proof: You need to destroy 16 small squares, taking away one piece can destroy at most 2. But if you remove 8 matches that each destroy two small squares then the largest square from the perimeter is still there.

 

[fresh_42]

It can be done in $9$.

 

[mfb]

Found a solution.

Spoiler

 

[fresh_42]

115. There is a cube, constituted of 9 smaller ones, like Rubik's. On all three sides of every corner cube is a same number attached, either three times $+1$ or $-1$. Finally every center field carries $+1$ or $-1$, too. The respective value is calculated as the product of the four corners that are located on the respective side surface.

Is it possible, that we choose these fourteen numbers in a way that they sum up to $0$ and how?

D116

 

[mfb]

There is a cube, constituted of 9 smaller ones, like Rubik's.

3x3x3 = 27?

Let the corners be identified as (x,y,z) where a,b,c= 0 or 1.
Let (0,0,0)=(0,1,0)=(0,0,1)=-1 and use 1 for all other corners. There is one face (x=0) showing -1 from three -1 corners, two faces (y=0 and z=0) showing +1 from two -1 corners, two faces (y=1 and z=1) showing -1 from one -1 corner and one face (x=1) showing +1 from no -1 corner. Sum: -3+5-1+2-2+1 = 0

 

[fresh_42]

3x3x3 = 27?

I avoided that since I feared endless discussions whether there is a cube at the center or not ...

Let the corners be identified as (x,y,z) where a,b,c= 0 or 1.
Let (0,0,0)=(0,1,0)=(0,0,1)=-1 and use 1 for all other corners. There is one face (x=0) showing -1 from three -1 corners, two faces (y=0 and z=0) showing +1 from two -1 corners, two faces (y=1 and z=1) showing -1 from one -1 corner and one face (x=1) showing +1 from no -1 corner. Sum: -3+5-1+2-2+1 = 0

I don't get it. There is only one number per corner, eight corners in total and six numbers at the center of the surfaces, namely the product of the four numbers in the corners, respectively.

 

[mfb]

There is only one number per corner, eight corners in total and six numbers at the center of the surfaces, namely the product of the four numbers in the corners, respectively.

That's how I understood the puzzle. I made a calculation error, however. -3+5-1+2-2+1 = 2 not 0.

Correction: It is impossible. Consider what happens if we change one corner from -1 to 1. We flip the values of all three adjacent faces. If all three where -1 before they are now all 1, we increased the sum by 8. If only two of them were -1 then we increase the sum by 4 (in total we change three -1 to 1 and one 1 to -1). If one of them was -1 then we keep the sum the same, and if they were all 1 then we reduce the sum by 4. Every corner change keeps the sum the same modulo 4. "All corners are 1" leads to a sum of 14, which has a remainder of 2 when divided by 4. All configurations are reachable via corner changes from this state, therefore they all have a remainder of 2 when divided by 4 - a sum of zero is impossible.

 

[fresh_42]

116. What is the smallest number that actually needs all nine cubes to write it as a sum of cubes?

D117

 

[fresh_42]

117. Find all $4$ even natural numbers $n$, such that $n=p+(n-p)$ is a sum of two primes for every possible prime $n/2 \leq p < n$.
E.g.: $10=5+5=7+3$

D118

 

[fresh_42]

118. If we multiply a two digit number such that the three digit result is obtained by moving a $0$ between the digits, which possibilities do we have?

D119

 

[mfb]

118:

Spoiler

We transform xy to x0y, which means we add 90*x. xy must be a factor of 90*x. Note that 90=2*3*3*5. Trivially x0 becomes x00 after multiplication by 10, I don't list these in the following:
x=1: xy=15 (becomes 105 after multiplying by 7), 18 (becomes 108 after multiplying by 6)
x=2: -
x=3: -
x=4: 45 (becomes 405 after multiplying by 9)
x=5: -
x=6: -
x=7: -
x=8: -
x=9: -

Hmm.. just casework.

 

[fresh_42]

118:

Spoiler

We transform xy to x0y, which means we add 90*x. xy must be a factor of 90*x. Note that 90=2*3*3*5. Trivially x0 becomes x00 after multiplication by 10, I don't list these in the following:
x=1: xy=15 (becomes 105 after multiplying by 7), 18 (becomes 108 after multiplying by 6)
x=2: -
x=3: -
x=4: 45 (becomes 405 after multiplying by 9)
x=5: -
x=6: -
x=7: -
x=8: -
x=9: -

Hmm.. just casework.

Yes. As several members use coding to solve problems ...

116. was very easy without the need of code. 117. has three easy solutions and one which is a bit more to check. I guess nobody has coded a primality check.

 

[mfb]

I was thinking about it, but a program to run over some small numbers feels boring.

116: 23 = 8+8+7*1
It is interesting that only small numbers require so many cubes. 23 and 239 are the only ones needing 9. 454 is the last one needing 8. We know large numbers can need at least 4 and need at most 7 but it is not known what the limit is.

 

[fresh_42]

119. A deck of 3,040 cards is shuffled and both players, Alice and Bob, get $1,520$ cards. Each card has a number printed on it, from $1$ to $3,040$. The first player who can lay down a card such that the sum of all played cards is divisible by $3,041$ wins the game. Alice begins.

Is there a winning strategy for one of the players?

D119

 

[mfb]

119:

Is there a winning strategy for one of the players?

Spoiler

The sum of all cards is 3040*3041/2 = 1520*3041 and divisible by 3041. At the very latest the second player wins with their last card. We have a game with perfect information and no randomness (after dealing the cards) that has a finite length cannot end in a draw. For every way the cards are shuffled one of the players must have a winning strategy.

Yeah, while technically an answer it feels like cheating.

We can group the cards in 1520 pairs that each sum to 3041.. If A and B have one card of each pair then B wins with his first card. In general A must start with a card where she has the partner.
As only the sum mod 3041 matters we can relabel 1521...3040 to -1520...-1. A starts with card a1 where she also has -a1. B responds with b1 where -b1-a1 must not be in A's hand (mod 3041 of course). A responds with a2 such that -b1-a1-a2 is not in B's hand. Hmm... not seeing a long-term strategy yet.

Is there a somewhat easy strategy to follow?

 

[jbriggs444]

#119
With only 3004 cards dealt out of 3040, this is not a perfect information game.

 

[fresh_42]

#119
With only 3004 cards dealt out of 3040, this is not a perfect information game.

Sorry, typo.

 

[fresh_42]

Is there a somewhat easy strategy to follow?

Yes.

Spoiler

Alice cannot win with her first move. After that, there will always be exactly one card to win the game, and Bob can see in his cards, whether Alice has it or not. Since he has one more card than Alice, Alice can't have the winning card for all of Bob's cards, hence there is always a card he can play accordingly. Since, as you mentioned, the last card will definitely win, Bob has a winning strategy.

 

[fresh_42]

120. We have $64$ hectares of a square forest, say they are numerated like the chess board. At twelve o'clock a fire starts on $D1$. Firefighters can secure one hectare from burning every hour. Every next hour, the fire spreads to all horizontally or vertically neighboured hectares (no diagonals). How many hectares can be saved at best?

D120

 

[fresh_42]

121. Let $x <y$ be two positive integers. We have two transformations $S(x,y)=(x+1,2y)$ and $T(x,y)=(2x,y+1)$. If we start with such an arbitrary pair $(x,y)$, is there always a number $z$ and a path such that $$S^{n_1}T^{m_1}S^{n_2}T^{m_2}\ldots S^{n_k}T^{m_k}(x,y)=(z,z)$$
for some $n_i,m_j \in \mathbb{N}_0$?

D121

 

[mfb]

120. We have $64$ hectares of a square forest, say they are numerated like the chess board. At twelve o'clock a fire starts on $D1$. Firefighters can secure one hectare from burning every hour. Every next hour, the fire spreads to all horizontally or vertically neighboured hectares (no diagonals). How many hectares can be saved at best?

D120

If I understand the question correctly then the best I found is covering the E column one by one, saving half of the field. If you let it go into E then it will spread both vertically and towards the larger columns and you can only keep one side from spreading in both directions.

 

[fresh_42]

If I understand the question correctly then the best I found is covering the E column one by one, saving half of the field. If you let it go into E then it will spread both vertically and towards the larger columns and you can only keep one side from spreading in both directions.

You can save more than 50% for burning.

 

[BvU]

it will spread both vertically and towards the larger columns

Am I to understand it does NOT spread to the smaller columns ? Please clarify this spreading (e.g. after D1 it can spread to D2 and to E1 -- not to C1 ?)