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

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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

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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
 
33,031
8,809
Pen and paper approach for 108:
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.
 
12
1
10237896=36x284386
using 5-line perl
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;
 
Last edited:

fresh_42

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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

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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
14
1565015923780.png
 
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fresh_42

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

fresh_42

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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

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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
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
Perl:
#!/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";
                        };
                };
        };
};
 
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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

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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

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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.
Code:
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

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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.
Code:
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\,.##
 
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.
 

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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

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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
1565043654919.png
and am really curious about a possible last escapee :nb)
 

fresh_42

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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 and am really curious about a possible last escapee :nb)
##1 \\
2 \\
3 \; \;4 \; \;5##
 
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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
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

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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.
 
There is a solution.
I totally forgot about repeating the same ring.
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

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I totally forgot about repeating the same ring.
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.
 

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