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Arithmetic Game for Beginners [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to prove an arithmetic convolution identity?Exponential Diophantine Equations for BeginnersFind the last member remained in gameCalculating unique “damage-per-second” for a gameArithmetic SeriesInterval arithmetic for open intervalsChopping arithmeticBasic arithmeticIs there a general algorithm for these “simple” arithmetic problems?!Puzzle involving arithmetic series - will this game always end
$begingroup$
I came up with this game yesterday, intended for first year college students in a very elementary math course, here it goes:
Suppose we have the numbers 1 2=3. If we place the $+$ sign between 1 and 2 we get 1+2=3, which is correct. Suppose now we have 1 2 3=4, I can make -1+2+3=4, which is also correct. I can also make, for 5, 6 and 7
$-1cdot2+3+4=5$
$1-2+3cdot4-5=6$
$1+2-3-4+5+6=7$
So the challenge is, to reach the most natural numbers from $$1quad2quad3quad4quadldotsquad (n-1)=n$$ by using addition subtraction, product and division in between the numbers without rearranging them.
At first this game seemed interesting, and eventually begin to wonder if this was always possible, and for instance, noticed (though not formally proved) that for the numbers $n=3+4k$ it is always possible. Feel free to post any interesting property you find.
Thanks
number-theory arithmetic
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
$endgroup$
closed as too broad by hardmath, Cesareo, Shailesh, Parcly Taxel, dantopa Mar 28 at 2:22
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I came up with this game yesterday, intended for first year college students in a very elementary math course, here it goes:
Suppose we have the numbers 1 2=3. If we place the $+$ sign between 1 and 2 we get 1+2=3, which is correct. Suppose now we have 1 2 3=4, I can make -1+2+3=4, which is also correct. I can also make, for 5, 6 and 7
$-1cdot2+3+4=5$
$1-2+3cdot4-5=6$
$1+2-3-4+5+6=7$
So the challenge is, to reach the most natural numbers from $$1quad2quad3quad4quadldotsquad (n-1)=n$$ by using addition subtraction, product and division in between the numbers without rearranging them.
At first this game seemed interesting, and eventually begin to wonder if this was always possible, and for instance, noticed (though not formally proved) that for the numbers $n=3+4k$ it is always possible. Feel free to post any interesting property you find.
Thanks
number-theory arithmetic
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
$endgroup$
closed as too broad by hardmath, Cesareo, Shailesh, Parcly Taxel, dantopa Mar 28 at 2:22
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
"Feel free to post any interesting property you find" is not a question. If you review How to Ask, you'll see that Math.SE seeks Q&A content that is less chatty and more definitive than many forums collect. For example you might ask what the smallest $ngt 3$ is for which the expression is not possible, or (if you really treat this as a game) a problem about a winning strategy.
$endgroup$
– hardmath
Mar 27 at 16:25
$begingroup$
You're right, is not a question... but I'm interested in any interesting property somebody may find.
$endgroup$
– Cristian Baeza
Mar 27 at 17:10
add a comment |
$begingroup$
I came up with this game yesterday, intended for first year college students in a very elementary math course, here it goes:
Suppose we have the numbers 1 2=3. If we place the $+$ sign between 1 and 2 we get 1+2=3, which is correct. Suppose now we have 1 2 3=4, I can make -1+2+3=4, which is also correct. I can also make, for 5, 6 and 7
$-1cdot2+3+4=5$
$1-2+3cdot4-5=6$
$1+2-3-4+5+6=7$
So the challenge is, to reach the most natural numbers from $$1quad2quad3quad4quadldotsquad (n-1)=n$$ by using addition subtraction, product and division in between the numbers without rearranging them.
At first this game seemed interesting, and eventually begin to wonder if this was always possible, and for instance, noticed (though not formally proved) that for the numbers $n=3+4k$ it is always possible. Feel free to post any interesting property you find.
Thanks
number-theory arithmetic
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
$endgroup$
I came up with this game yesterday, intended for first year college students in a very elementary math course, here it goes:
Suppose we have the numbers 1 2=3. If we place the $+$ sign between 1 and 2 we get 1+2=3, which is correct. Suppose now we have 1 2 3=4, I can make -1+2+3=4, which is also correct. I can also make, for 5, 6 and 7
$-1cdot2+3+4=5$
$1-2+3cdot4-5=6$
$1+2-3-4+5+6=7$
So the challenge is, to reach the most natural numbers from $$1quad2quad3quad4quadldotsquad (n-1)=n$$ by using addition subtraction, product and division in between the numbers without rearranging them.
At first this game seemed interesting, and eventually begin to wonder if this was always possible, and for instance, noticed (though not formally proved) that for the numbers $n=3+4k$ it is always possible. Feel free to post any interesting property you find.
Thanks
number-theory arithmetic
number-theory arithmetic
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
asked Mar 27 at 16:16
Cristian BaezaCristian Baeza
433213
433213
closed as too broad by hardmath, Cesareo, Shailesh, Parcly Taxel, dantopa Mar 28 at 2:22
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by hardmath, Cesareo, Shailesh, Parcly Taxel, dantopa Mar 28 at 2:22
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
"Feel free to post any interesting property you find" is not a question. If you review How to Ask, you'll see that Math.SE seeks Q&A content that is less chatty and more definitive than many forums collect. For example you might ask what the smallest $ngt 3$ is for which the expression is not possible, or (if you really treat this as a game) a problem about a winning strategy.
$endgroup$
– hardmath
Mar 27 at 16:25
$begingroup$
You're right, is not a question... but I'm interested in any interesting property somebody may find.
$endgroup$
– Cristian Baeza
Mar 27 at 17:10
add a comment |
3
$begingroup$
"Feel free to post any interesting property you find" is not a question. If you review How to Ask, you'll see that Math.SE seeks Q&A content that is less chatty and more definitive than many forums collect. For example you might ask what the smallest $ngt 3$ is for which the expression is not possible, or (if you really treat this as a game) a problem about a winning strategy.
$endgroup$
– hardmath
Mar 27 at 16:25
$begingroup$
You're right, is not a question... but I'm interested in any interesting property somebody may find.
$endgroup$
– Cristian Baeza
Mar 27 at 17:10
3
3
$begingroup$
"Feel free to post any interesting property you find" is not a question. If you review How to Ask, you'll see that Math.SE seeks Q&A content that is less chatty and more definitive than many forums collect. For example you might ask what the smallest $ngt 3$ is for which the expression is not possible, or (if you really treat this as a game) a problem about a winning strategy.
$endgroup$
– hardmath
Mar 27 at 16:25
$begingroup$
"Feel free to post any interesting property you find" is not a question. If you review How to Ask, you'll see that Math.SE seeks Q&A content that is less chatty and more definitive than many forums collect. For example you might ask what the smallest $ngt 3$ is for which the expression is not possible, or (if you really treat this as a game) a problem about a winning strategy.
$endgroup$
– hardmath
Mar 27 at 16:25
$begingroup$
You're right, is not a question... but I'm interested in any interesting property somebody may find.
$endgroup$
– Cristian Baeza
Mar 27 at 17:10
$begingroup$
You're right, is not a question... but I'm interested in any interesting property somebody may find.
$endgroup$
– Cristian Baeza
Mar 27 at 17:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Seems like there are many ways to do this. Thought of removing $*$ but then parity/modulo 2 problem prevents certain cases from working. Below is one possibility (middle rows are zeroes):
$4k+1$:
$$
beginalign
&-1*2+3\
&4-5-6+7,8-9-10+11,dots,+(4k-4)-(4k-3)-(4k-2)+(4k-1)\
&+(4k)
endalign
$$
$4k+2$:
$$
beginalign
&1*2 +3 -4\
&5-6-7+8,9-10-11+12,dots,+(4k-3)-(4k-2)-(4k-1)+(4k)\
&+(4k+1)
endalign
$$
$4k+3$:
$$
beginalign
&1\
&2-3-4+5,6-7-8+9,dots,+(4k-2)-(4k-1)-(4k)+(4k+1)\
&+(4k+2)
endalign
$$
$4k+4$:
$$
beginalign
&-1+2\
&3-4-5+6,7-8-9+10,dots,(4k-1)-(4k)-(4k+1)+(4k+2)\
&+(4k+3)
endalign
$$
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Seems like there are many ways to do this. Thought of removing $*$ but then parity/modulo 2 problem prevents certain cases from working. Below is one possibility (middle rows are zeroes):
$4k+1$:
$$
beginalign
&-1*2+3\
&4-5-6+7,8-9-10+11,dots,+(4k-4)-(4k-3)-(4k-2)+(4k-1)\
&+(4k)
endalign
$$
$4k+2$:
$$
beginalign
&1*2 +3 -4\
&5-6-7+8,9-10-11+12,dots,+(4k-3)-(4k-2)-(4k-1)+(4k)\
&+(4k+1)
endalign
$$
$4k+3$:
$$
beginalign
&1\
&2-3-4+5,6-7-8+9,dots,+(4k-2)-(4k-1)-(4k)+(4k+1)\
&+(4k+2)
endalign
$$
$4k+4$:
$$
beginalign
&-1+2\
&3-4-5+6,7-8-9+10,dots,(4k-1)-(4k)-(4k+1)+(4k+2)\
&+(4k+3)
endalign
$$
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
add a comment |
$begingroup$
Seems like there are many ways to do this. Thought of removing $*$ but then parity/modulo 2 problem prevents certain cases from working. Below is one possibility (middle rows are zeroes):
$4k+1$:
$$
beginalign
&-1*2+3\
&4-5-6+7,8-9-10+11,dots,+(4k-4)-(4k-3)-(4k-2)+(4k-1)\
&+(4k)
endalign
$$
$4k+2$:
$$
beginalign
&1*2 +3 -4\
&5-6-7+8,9-10-11+12,dots,+(4k-3)-(4k-2)-(4k-1)+(4k)\
&+(4k+1)
endalign
$$
$4k+3$:
$$
beginalign
&1\
&2-3-4+5,6-7-8+9,dots,+(4k-2)-(4k-1)-(4k)+(4k+1)\
&+(4k+2)
endalign
$$
$4k+4$:
$$
beginalign
&-1+2\
&3-4-5+6,7-8-9+10,dots,(4k-1)-(4k)-(4k+1)+(4k+2)\
&+(4k+3)
endalign
$$
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
add a comment |
$begingroup$
Seems like there are many ways to do this. Thought of removing $*$ but then parity/modulo 2 problem prevents certain cases from working. Below is one possibility (middle rows are zeroes):
$4k+1$:
$$
beginalign
&-1*2+3\
&4-5-6+7,8-9-10+11,dots,+(4k-4)-(4k-3)-(4k-2)+(4k-1)\
&+(4k)
endalign
$$
$4k+2$:
$$
beginalign
&1*2 +3 -4\
&5-6-7+8,9-10-11+12,dots,+(4k-3)-(4k-2)-(4k-1)+(4k)\
&+(4k+1)
endalign
$$
$4k+3$:
$$
beginalign
&1\
&2-3-4+5,6-7-8+9,dots,+(4k-2)-(4k-1)-(4k)+(4k+1)\
&+(4k+2)
endalign
$$
$4k+4$:
$$
beginalign
&-1+2\
&3-4-5+6,7-8-9+10,dots,(4k-1)-(4k)-(4k+1)+(4k+2)\
&+(4k+3)
endalign
$$
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
$endgroup$
Seems like there are many ways to do this. Thought of removing $*$ but then parity/modulo 2 problem prevents certain cases from working. Below is one possibility (middle rows are zeroes):
$4k+1$:
$$
beginalign
&-1*2+3\
&4-5-6+7,8-9-10+11,dots,+(4k-4)-(4k-3)-(4k-2)+(4k-1)\
&+(4k)
endalign
$$
$4k+2$:
$$
beginalign
&1*2 +3 -4\
&5-6-7+8,9-10-11+12,dots,+(4k-3)-(4k-2)-(4k-1)+(4k)\
&+(4k+1)
endalign
$$
$4k+3$:
$$
beginalign
&1\
&2-3-4+5,6-7-8+9,dots,+(4k-2)-(4k-1)-(4k)+(4k+1)\
&+(4k+2)
endalign
$$
$4k+4$:
$$
beginalign
&-1+2\
&3-4-5+6,7-8-9+10,dots,(4k-1)-(4k)-(4k+1)+(4k+2)\
&+(4k+3)
endalign
$$
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
answered Mar 27 at 16:58
Yong Hao NgYong Hao Ng
3,7091222
3,7091222
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
add a comment |
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
$begingroup$
Unfortunately the rules of the "game" are not clear. In particular with a mixture of operations the role of parentheses, not mention in your Answer (or in the Question), becomes paramount.
$endgroup$
– hardmath
Mar 27 at 19:45
add a comment |
3
$begingroup$
"Feel free to post any interesting property you find" is not a question. If you review How to Ask, you'll see that Math.SE seeks Q&A content that is less chatty and more definitive than many forums collect. For example you might ask what the smallest $ngt 3$ is for which the expression is not possible, or (if you really treat this as a game) a problem about a winning strategy.
$endgroup$
– hardmath
Mar 27 at 16:25
$begingroup$
You're right, is not a question... but I'm interested in any interesting property somebody may find.
$endgroup$
– Cristian Baeza
Mar 27 at 17:10