Mixing
![$mathbb{E}[(Y_{n+1} - {Y_n})Z] =0.$ implies $Y_n = mathbb{E}[Y_{n+1}|H_n]$?](https://lh3.googleusercontent.com/-XdUIqdMkCWA/AAAAAAAAAAI/AAAAAAAAAAA/4252rscbv5M/s72-c/photo.jpg?sz=32)
A game awards either $7$ points or $4$ points at a time. How many possible total scores can never be reached...
$begingroup$
In the Martian game of QZX. a JBL is worth 7 points and a KMD is worth 4 points. There is no other way to score. Games can continue indefinitely. How many positive integer scores can never be reached? (e.g., no team could ever have a total score of 2 points.)
A) 5 B) 7 C) 9 D) 11 E) Infinity many
I don't understand this game. Can anyone explain the rules of Martian game, please?
word-problem
edited Jan 22 at 13:26
Blue
49k870156
asked Jan 22 at 12:09
narnar
112
$endgroup$
add a comment |
$begingroup$
In the Martian game of QZX. a JBL is worth 7 points and a KMD is worth 4 points. There is no other way to score. Games can continue indefinitely. How many positive integer scores can never be reached? (e.g., no team could ever have a total score of 2 points.)
A) 5 B) 7 C) 9 D) 11 E) Infinity many
I don't understand this game. Can anyone explain the rules of Martian game, please?
word-problem
edited Jan 22 at 13:26
Blue
49k870156
asked Jan 22 at 12:09
narnar
112
$endgroup$
$begingroup$
That is just, if I understand correctly, a problem with a very poor statement. The question is: what integers can be written as sums of fours and sevens?
$endgroup$
– Mindlack
Jan 22 at 12:15
2
$begingroup$
Well, essentially we're considering numbers that are linear combinations of 7 and 4, so $$ text{score} = 7n + 4 m $$ where $n$ and $m$ are natural numbers. Note that $ngeq0$ and $mgeq 0$, otherwise this wouldn't be a very interesting question (since $gcd(7,4)=1$)
$endgroup$
– Matti P.
Jan 22 at 12:16
$begingroup$
Without knowing the inner workings of the game (how do you get the scores?) it is impossible to know, but if each sequence of those two scores is possible then we get all the (non. neg) linear combinations.
$endgroup$
– ploosu2
Jan 22 at 13:15
1
$begingroup$
This is basically the Frobenius Coin Problem, which asks for the largest amount (the "Frobenius number") that cannot be obtained from $n$ coin types. For $n=2$ coins of value $x$ and $y$, the number is $xy-x-y$; in this problem, $x=7$ and $y=4$, so the largest non-attainable amount is $17$. How many non-attainable amounts there are remains to be determined, but we know the answer isn't "infinitely many", and we know the number of cases to check is limited. (Related: McNugget Numbers.)
$endgroup$
– Blue
Jan 22 at 13:17
add a comment |
$begingroup$
In the Martian game of QZX. a JBL is worth 7 points and a KMD is worth 4 points. There is no other way to score. Games can continue indefinitely. How many positive integer scores can never be reached? (e.g., no team could ever have a total score of 2 points.)
A) 5 B) 7 C) 9 D) 11 E) Infinity many
I don't understand this game. Can anyone explain the rules of Martian game, please?
word-problem
edited Jan 22 at 13:26
Blue
49k870156
asked Jan 22 at 12:09
narnar
112
$endgroup$
In the Martian game of QZX. a JBL is worth 7 points and a KMD is worth 4 points. There is no other way to score. Games can continue indefinitely. How many positive integer scores can never be reached? (e.g., no team could ever have a total score of 2 points.)
A) 5 B) 7 C) 9 D) 11 E) Infinity many
I don't understand this game. Can anyone explain the rules of Martian game, please?
word-problem
word-problem
edited Jan 22 at 13:26
Blue
49k870156
asked Jan 22 at 12:09
narnar
112
edited Jan 22 at 13:26
Blue
49k870156
asked Jan 22 at 12:09
narnar
112
edited Jan 22 at 13:26
Blue
49k870156
edited Jan 22 at 13:26
Blue
49k870156
edited Jan 22 at 13:26
Blue
49k870156
49k870156
asked Jan 22 at 12:09
narnar
112
asked Jan 22 at 12:09
narnar
112
asked Jan 22 at 12:09
narnar
112
112
$begingroup$
That is just, if I understand correctly, a problem with a very poor statement. The question is: what integers can be written as sums of fours and sevens?
$endgroup$
– Mindlack
Jan 22 at 12:15
2
$begingroup$
Well, essentially we're considering numbers that are linear combinations of 7 and 4, so $$ text{score} = 7n + 4 m $$ where $n$ and $m$ are natural numbers. Note that $ngeq0$ and $mgeq 0$, otherwise this wouldn't be a very interesting question (since $gcd(7,4)=1$)
$endgroup$
– Matti P.
Jan 22 at 12:16
$begingroup$
Without knowing the inner workings of the game (how do you get the scores?) it is impossible to know, but if each sequence of those two scores is possible then we get all the (non. neg) linear combinations.
$endgroup$
– ploosu2
Jan 22 at 13:15
1
$begingroup$
This is basically the Frobenius Coin Problem, which asks for the largest amount (the "Frobenius number") that cannot be obtained from $n$ coin types. For $n=2$ coins of value $x$ and $y$, the number is $xy-x-y$; in this problem, $x=7$ and $y=4$, so the largest non-attainable amount is $17$. How many non-attainable amounts there are remains to be determined, but we know the answer isn't "infinitely many", and we know the number of cases to check is limited. (Related: McNugget Numbers.)
$endgroup$
– Blue
Jan 22 at 13:17
add a comment |
$begingroup$
That is just, if I understand correctly, a problem with a very poor statement. The question is: what integers can be written as sums of fours and sevens?
$endgroup$
– Mindlack
Jan 22 at 12:15
2
$begingroup$
Well, essentially we're considering numbers that are linear combinations of 7 and 4, so $$ text{score} = 7n + 4 m $$ where $n$ and $m$ are natural numbers. Note that $ngeq0$ and $mgeq 0$, otherwise this wouldn't be a very interesting question (since $gcd(7,4)=1$)
$endgroup$
– Matti P.
Jan 22 at 12:16
$begingroup$
Without knowing the inner workings of the game (how do you get the scores?) it is impossible to know, but if each sequence of those two scores is possible then we get all the (non. neg) linear combinations.
$endgroup$
– ploosu2
Jan 22 at 13:15
1
$begingroup$
This is basically the Frobenius Coin Problem, which asks for the largest amount (the "Frobenius number") that cannot be obtained from $n$ coin types. For $n=2$ coins of value $x$ and $y$, the number is $xy-x-y$; in this problem, $x=7$ and $y=4$, so the largest non-attainable amount is $17$. How many non-attainable amounts there are remains to be determined, but we know the answer isn't "infinitely many", and we know the number of cases to check is limited. (Related: McNugget Numbers.)
$endgroup$
– Blue
Jan 22 at 13:17
$begingroup$
That is just, if I understand correctly, a problem with a very poor statement. The question is: what integers can be written as sums of fours and sevens?
$endgroup$
– Mindlack
Jan 22 at 12:15
$begingroup$
That is just, if I understand correctly, a problem with a very poor statement. The question is: what integers can be written as sums of fours and sevens?
$endgroup$
– Mindlack
Jan 22 at 12:15
2
2
$begingroup$
Well, essentially we're considering numbers that are linear combinations of 7 and 4, so $$ text{score} = 7n + 4 m $$ where $n$ and $m$ are natural numbers. Note that $ngeq0$ and $mgeq 0$, otherwise this wouldn't be a very interesting question (since $gcd(7,4)=1$)
$endgroup$
– Matti P.
Jan 22 at 12:16
$begingroup$
Well, essentially we're considering numbers that are linear combinations of 7 and 4, so $$ text{score} = 7n + 4 m $$ where $n$ and $m$ are natural numbers. Note that $ngeq0$ and $mgeq 0$, otherwise this wouldn't be a very interesting question (since $gcd(7,4)=1$)
$endgroup$
– Matti P.
Jan 22 at 12:16
$begingroup$
Without knowing the inner workings of the game (how do you get the scores?) it is impossible to know, but if each sequence of those two scores is possible then we get all the (non. neg) linear combinations.
$endgroup$
– ploosu2
Jan 22 at 13:15
$begingroup$
Without knowing the inner workings of the game (how do you get the scores?) it is impossible to know, but if each sequence of those two scores is possible then we get all the (non. neg) linear combinations.
$endgroup$
– ploosu2
Jan 22 at 13:15
1
1
$begingroup$
This is basically the Frobenius Coin Problem, which asks for the largest amount (the "Frobenius number") that cannot be obtained from $n$ coin types. For $n=2$ coins of value $x$ and $y$, the number is $xy-x-y$; in this problem, $x=7$ and $y=4$, so the largest non-attainable amount is $17$. How many non-attainable amounts there are remains to be determined, but we know the answer isn't "infinitely many", and we know the number of cases to check is limited. (Related: McNugget Numbers.)
$endgroup$
– Blue
Jan 22 at 13:17
$begingroup$
This is basically the Frobenius Coin Problem, which asks for the largest amount (the "Frobenius number") that cannot be obtained from $n$ coin types. For $n=2$ coins of value $x$ and $y$, the number is $xy-x-y$; in this problem, $x=7$ and $y=4$, so the largest non-attainable amount is $17$. How many non-attainable amounts there are remains to be determined, but we know the answer isn't "infinitely many", and we know the number of cases to check is limited. (Related: McNugget Numbers.)
$endgroup$
– Blue
Jan 22 at 13:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
All integers which are multiples of $4$ are possible scores, using only KMDs.
All integers which are $1$ less than a multiple of $4$ require at least one JBL, and all these are possible using one JBL and an appropriate number of KMDs except for $3$.
All integers which are $2$ less than a multiple of $4$ require at least two JBLs to make, and all these are possible using exactly two JBLs and an appropriate number of KMDs except for $2$, $6$ and $10$.
All integers which are $3$ less than a multiple of $4$ require at least three JBLS, and all can be made using exactly three except $1,5,9,13$ and $17$. So there are nine impossible scores in total.
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
add a comment |
$begingroup$
There is no way a score of $1,2,3,5,6,9,10$ can be made from just $4$s and $7$s, so straight away we have $7$ scores that cannot be made, which rules out answer A. Scores of $11=4+7$ and $12=3times4$ can be made.
Once we have reached $20=5times4$ we can:
a) replace the five $4$s with three sevens to reach $21=3times7$.
b) now we can replace a $7$ with two $4$s to reach $22 = 2times7 + 2times4$.
c) and finally we can replace another $7$ with two $4$s to reach $23 = 7 +4times4$.
Then we have $24=6times4$, and we can repeat the steps a), b) and c) to reach $25,26,27,dots$. So any score from $20$ upwards can be made. This rules out answer E.
To decide between answers B, C and D, you just have to work out how many scores between $13$ and $19$ can be made with just $4$s and $7$s.
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
$endgroup$
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
add a comment |
Your Answer
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
All integers which are multiples of $4$ are possible scores, using only KMDs.
All integers which are $1$ less than a multiple of $4$ require at least one JBL, and all these are possible using one JBL and an appropriate number of KMDs except for $3$.
All integers which are $2$ less than a multiple of $4$ require at least two JBLs to make, and all these are possible using exactly two JBLs and an appropriate number of KMDs except for $2$, $6$ and $10$.
All integers which are $3$ less than a multiple of $4$ require at least three JBLS, and all can be made using exactly three except $1,5,9,13$ and $17$. So there are nine impossible scores in total.
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
add a comment |
$begingroup$
All integers which are multiples of $4$ are possible scores, using only KMDs.
All integers which are $1$ less than a multiple of $4$ require at least one JBL, and all these are possible using one JBL and an appropriate number of KMDs except for $3$.
All integers which are $2$ less than a multiple of $4$ require at least two JBLs to make, and all these are possible using exactly two JBLs and an appropriate number of KMDs except for $2$, $6$ and $10$.
All integers which are $3$ less than a multiple of $4$ require at least three JBLS, and all can be made using exactly three except $1,5,9,13$ and $17$. So there are nine impossible scores in total.
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
add a comment |
$begingroup$
All integers which are multiples of $4$ are possible scores, using only KMDs.
All integers which are $1$ less than a multiple of $4$ require at least one JBL, and all these are possible using one JBL and an appropriate number of KMDs except for $3$.
All integers which are $2$ less than a multiple of $4$ require at least two JBLs to make, and all these are possible using exactly two JBLs and an appropriate number of KMDs except for $2$, $6$ and $10$.
All integers which are $3$ less than a multiple of $4$ require at least three JBLS, and all can be made using exactly three except $1,5,9,13$ and $17$. So there are nine impossible scores in total.
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
$endgroup$
All integers which are multiples of $4$ are possible scores, using only KMDs.
All integers which are $1$ less than a multiple of $4$ require at least one JBL, and all these are possible using one JBL and an appropriate number of KMDs except for $3$.
All integers which are $2$ less than a multiple of $4$ require at least two JBLs to make, and all these are possible using exactly two JBLs and an appropriate number of KMDs except for $2$, $6$ and $10$.
All integers which are $3$ less than a multiple of $4$ require at least three JBLS, and all can be made using exactly three except $1,5,9,13$ and $17$. So there are nine impossible scores in total.
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
answered Jan 22 at 13:38
Especially LimeEspecially Lime
22.3k22858
22.3k22858
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
add a comment |
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
$begingroup$
Thank you I got the idea. It is right
$endgroup$
– nar
Jan 25 at 4:10
add a comment |
$begingroup$
There is no way a score of $1,2,3,5,6,9,10$ can be made from just $4$s and $7$s, so straight away we have $7$ scores that cannot be made, which rules out answer A. Scores of $11=4+7$ and $12=3times4$ can be made.
Once we have reached $20=5times4$ we can:
a) replace the five $4$s with three sevens to reach $21=3times7$.
b) now we can replace a $7$ with two $4$s to reach $22 = 2times7 + 2times4$.
c) and finally we can replace another $7$ with two $4$s to reach $23 = 7 +4times4$.
Then we have $24=6times4$, and we can repeat the steps a), b) and c) to reach $25,26,27,dots$. So any score from $20$ upwards can be made. This rules out answer E.
To decide between answers B, C and D, you just have to work out how many scores between $13$ and $19$ can be made with just $4$s and $7$s.
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
$endgroup$
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
add a comment |
$begingroup$
There is no way a score of $1,2,3,5,6,9,10$ can be made from just $4$s and $7$s, so straight away we have $7$ scores that cannot be made, which rules out answer A. Scores of $11=4+7$ and $12=3times4$ can be made.
Once we have reached $20=5times4$ we can:
a) replace the five $4$s with three sevens to reach $21=3times7$.
b) now we can replace a $7$ with two $4$s to reach $22 = 2times7 + 2times4$.
c) and finally we can replace another $7$ with two $4$s to reach $23 = 7 +4times4$.
Then we have $24=6times4$, and we can repeat the steps a), b) and c) to reach $25,26,27,dots$. So any score from $20$ upwards can be made. This rules out answer E.
To decide between answers B, C and D, you just have to work out how many scores between $13$ and $19$ can be made with just $4$s and $7$s.
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
$endgroup$
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
add a comment |
$begingroup$
There is no way a score of $1,2,3,5,6,9,10$ can be made from just $4$s and $7$s, so straight away we have $7$ scores that cannot be made, which rules out answer A. Scores of $11=4+7$ and $12=3times4$ can be made.
Once we have reached $20=5times4$ we can:
a) replace the five $4$s with three sevens to reach $21=3times7$.
b) now we can replace a $7$ with two $4$s to reach $22 = 2times7 + 2times4$.
c) and finally we can replace another $7$ with two $4$s to reach $23 = 7 +4times4$.
Then we have $24=6times4$, and we can repeat the steps a), b) and c) to reach $25,26,27,dots$. So any score from $20$ upwards can be made. This rules out answer E.
To decide between answers B, C and D, you just have to work out how many scores between $13$ and $19$ can be made with just $4$s and $7$s.
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
$endgroup$
There is no way a score of $1,2,3,5,6,9,10$ can be made from just $4$s and $7$s, so straight away we have $7$ scores that cannot be made, which rules out answer A. Scores of $11=4+7$ and $12=3times4$ can be made.
Once we have reached $20=5times4$ we can:
a) replace the five $4$s with three sevens to reach $21=3times7$.
b) now we can replace a $7$ with two $4$s to reach $22 = 2times7 + 2times4$.
c) and finally we can replace another $7$ with two $4$s to reach $23 = 7 +4times4$.
Then we have $24=6times4$, and we can repeat the steps a), b) and c) to reach $25,26,27,dots$. So any score from $20$ upwards can be made. This rules out answer E.
To decide between answers B, C and D, you just have to work out how many scores between $13$ and $19$ can be made with just $4$s and $7$s.
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
answered Jan 22 at 13:05
gandalf61gandalf61
8,946725
8,946725
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
add a comment |
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
$begingroup$
Between 13 and 19, there were 2 numbers can't reach by 4s and 7s, 17 and 13. So the answer is 9 numbers. Thank you I got the idea.
$endgroup$
– nar
Jan 25 at 4:09
add a comment |
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$begingroup$
That is just, if I understand correctly, a problem with a very poor statement. The question is: what integers can be written as sums of fours and sevens?
$endgroup$
– Mindlack
Jan 22 at 12:15
2
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Well, essentially we're considering numbers that are linear combinations of 7 and 4, so $$ text{score} = 7n + 4 m $$ where $n$ and $m$ are natural numbers. Note that $ngeq0$ and $mgeq 0$, otherwise this wouldn't be a very interesting question (since $gcd(7,4)=1$)
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– Matti P.
Jan 22 at 12:16
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Without knowing the inner workings of the game (how do you get the scores?) it is impossible to know, but if each sequence of those two scores is possible then we get all the (non. neg) linear combinations.
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– ploosu2
Jan 22 at 13:15
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This is basically the Frobenius Coin Problem, which asks for the largest amount (the "Frobenius number") that cannot be obtained from $n$ coin types. For $n=2$ coins of value $x$ and $y$, the number is $xy-x-y$; in this problem, $x=7$ and $y=4$, so the largest non-attainable amount is $17$. How many non-attainable amounts there are remains to be determined, but we know the answer isn't "infinitely many", and we know the number of cases to check is limited. (Related: McNugget Numbers.)
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– Blue
Jan 22 at 13:17