- 10,011
- 6,751
I know, my bad. But 15+6 is too easy.Moving the goal posts eh ... ?
I know, my bad. But 15+6 is too easy.Moving the goal posts eh ... ?
19. Construct ##21## with symbols from ##\{\,1,5,6,7,*,/,+,-,(,)\,\}## but use the digits at most once.
There are of course some 'cheating' answers e.g. 1,023,457,986_{23} = 1,808,433,654,721_{10} which is prime (probably not the smallest such prime though).15. What is the smallest prime which includes all ten digits exactly once?
How must the prisoner distribute the balls so that his chances to survive are as high as possible?
I don't think I understand what this is asking. "This sentence contains n0 times the 0". "The" 0?Correct, with an overall chance of almost ##3/4##.
21. Which are the numbers (as symbols) ##n_0,\ldots ,n_9## such that the following statement is true:
'This sentence contains ##n_0## times the ##0##, ##n_1## times the ##1##, ##n_2## times the ##2, \ldots ## and ##n_9## times the ##9##'?
(And don't force me to phrase it without backdoors, you know what I mean.)
'This sentence contains 1 times the 0 and 2 times 1.' is true. The question is: how do we have to chose the ##n_i## that the sentence with all digits is true.
What do you mean by this? We need a vector ##\vec{n} \in \mathbb{Z}^{10}## such that11,10,9,8,7,6,5,4,3,2 -> 0, 10, 18, 24, 28, 30, 30, 28, 24, 18
That would need e.g. 11 zeros in the sentence "This sentence has 11 times '0', 10 times '1', ...", but the sentence with your numbers plugged in doesn't have that many (it just has 2).11,10,9,8,7,6,5,4,3,2 -> 0, 10, 18, 24, 28, 30, 30, 28, 24, 18
Ahh I see. "This sentence contains n0 occurances of '0', n1 occurances of '1', "That would need e.g. 11 zeros in the sentence "This sentence has 11 times '0', 10 times '1', ...", but the sentence with your numbers plugged in doesn't have that many (it just has 2).
Maybe it is clearer if we add "digit"?
"This sentence contains n_{0} times the digit '0', n_{1} times the digit '1', ..."
Correct, with an overall chance of almost ##3/4##.
21. Which are the numbers (as symbols) ##n_0,\ldots ,n_9## such that the following statement is true:
'This sentence contains ##n_0## times the ##0##, ##n_1## times the ##1##, ##n_2## times the ##2, \ldots ## and ##n_9## times the ##9##'?
(And don't force me to phrase it without backdoors, you know what I mean.)
rrrrrr, gggggg, bbbbb - 3, unicolor
rrrrrg - 1 - 6 - mirror symmetry through beads
rrrrgg rrrgrg rrgrrg - 3 - 18 - mirror symmetry through edge, mirror symmetry through beads, 180 degree symmetry (6 each)
rrrggg rrgrgg rgrgrg - 3 - 9 - mirror symmetry through beads, no symmetry, 120 degree symmetry (3 each)
For 2 different kinds of beads, 33, for at most 2, 36.
rrrrgb rrrgrb rrgrrb - 3 - 9 - no symmetry, no symmetry, mirror symmetry through beads (3 each)
rrrggb rrgrgb rgrrgb grrrgb rrggrb rgrgrb - 6 - 36 - no symmetry, no symmetry, no symmetry, mirror symmetry through beads, no symmetry, mirror symmetry through beads (6 each)
rrggbb - 1 - 1 - no symmetry
rgrgbb rggrbb - 2 - 6 - no symmetry, mirror symmetry through edge
rgrbgb - 1 - 3 - mirror symmetry through beads
rgbrgb - 1 - 1 - 180 degree symmetry
The easiest way to solve all such problems is to use the successor function ##S## of Peano arithmetic. For instance, the first one can be solved as1. Extend the following to a valid equation, using only mathematical symbols!
Example: ##1\; 2\; 3 \;=\; 1 \longrightarrow - (1 \cdot 2) + 3 = 1##. Solutions are of course not unique.
##9\;9\;9\;=\;6##
##8\;8\;8\;=\;6##
##7\;7\;7\;=\;6##
##6\;6\;6\;=\;6##
##5\;5\;5\;=\;6##
##4\;4\;4\;=\;6##
##3\;3\;3\;=\;6##
##2\;2\;2\;=\;6##
##1\;1\;1\;=\;6##
##0\;0\;0\;=\;6##
Sorry, yes, forgotten a condition. Corrected now.Is this missing something? There are many options that are trivial to find.
Sorry, I had forgotten the condition of 100 pieces in total.They must all add up to $100, and one should buy at least one of each.
- Box of pralines: $7
- Bag of chips: $3
- Chocolate bar: $0.50
I will estimate the minimum number of items purchased using a "greedy" algorithm. First, the minimum purchases, one of each: $7 + $3 + $0.50 = $10.50, giving $89.50 left over. One can have at most 12 boxes of pralines, costing $84 and giving $5.50 left over. One can have at most one bag of chips, giving $2.50 left over. One can have 5 chocolate bars, costing $2.50, with $0.00 left over.
Thus, one should purchase 13 boxes of pralines, 2 bags of chips, and 6 chocolate bars.
Relaxing that minimum-number constraint means that the solution is no longer unique, since
- 1 box of pralines = 2 bags of chips + 2 chocolate bars
- 3 boxes of pralines = 7 bags of chips
- 1 box of pralines = 14 chocolate bars
- 1 bag of chips = 6 chocolate bars
23. John wants to buy sweets for his party. He has $100 to spend. A box Belgian pralines costs $7, chips cost $3, and for 50 cents he gets a chocolate bar. What does he have to buy, if he wants at least one of each and spend the whole $100 to buy exactly 100 pieces in total?