Fewest number of steps to reach 200 using special calculator
$begingroup$
This is a problem from the AMC 8 (math contest):
A certain calculator has only two keys $[+1]$ and $[times 2]$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$9$” and you pressed $[+1]$, it would display “$10$”. If you then pressed $[times 2]$, it would display “$20$”. Starting with the display “$1$”, what is the fewest number of keystrokes you would need to reach “$200$”?
Intuitively I worked back from $200$, dividing by $2$ until I reached an odd number, subtracting $1$ when I did, etc..to reach the correct answer of $9$ steps.
However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically.
The best I can come up with is that beyond the first step from $1$ to $2$, multiplication by $2$ is always going to yield a larger step than addition by $1$ and therefore I should take as many $[times 2]$ steps as I can. This doesn't feel rigorous enough, though.
EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.
combinatorics algebra-precalculus contest-math
$endgroup$
add a comment |
$begingroup$
This is a problem from the AMC 8 (math contest):
A certain calculator has only two keys $[+1]$ and $[times 2]$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$9$” and you pressed $[+1]$, it would display “$10$”. If you then pressed $[times 2]$, it would display “$20$”. Starting with the display “$1$”, what is the fewest number of keystrokes you would need to reach “$200$”?
Intuitively I worked back from $200$, dividing by $2$ until I reached an odd number, subtracting $1$ when I did, etc..to reach the correct answer of $9$ steps.
However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically.
The best I can come up with is that beyond the first step from $1$ to $2$, multiplication by $2$ is always going to yield a larger step than addition by $1$ and therefore I should take as many $[times 2]$ steps as I can. This doesn't feel rigorous enough, though.
EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.
combinatorics algebra-precalculus contest-math
$endgroup$
$begingroup$
Do you want the fewest steps to get to exactly $200$ or at least $200$?
$endgroup$
– John Douma
3 hours ago
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@JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice.
$endgroup$
– TonyK
2 hours ago
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@TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise.
$endgroup$
– John Douma
2 hours ago
1
$begingroup$
@JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$.
$endgroup$
– jeremy radcliff
2 hours ago
add a comment |
$begingroup$
This is a problem from the AMC 8 (math contest):
A certain calculator has only two keys $[+1]$ and $[times 2]$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$9$” and you pressed $[+1]$, it would display “$10$”. If you then pressed $[times 2]$, it would display “$20$”. Starting with the display “$1$”, what is the fewest number of keystrokes you would need to reach “$200$”?
Intuitively I worked back from $200$, dividing by $2$ until I reached an odd number, subtracting $1$ when I did, etc..to reach the correct answer of $9$ steps.
However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically.
The best I can come up with is that beyond the first step from $1$ to $2$, multiplication by $2$ is always going to yield a larger step than addition by $1$ and therefore I should take as many $[times 2]$ steps as I can. This doesn't feel rigorous enough, though.
EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.
combinatorics algebra-precalculus contest-math
$endgroup$
This is a problem from the AMC 8 (math contest):
A certain calculator has only two keys $[+1]$ and $[times 2]$. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “$9$” and you pressed $[+1]$, it would display “$10$”. If you then pressed $[times 2]$, it would display “$20$”. Starting with the display “$1$”, what is the fewest number of keystrokes you would need to reach “$200$”?
Intuitively I worked back from $200$, dividing by $2$ until I reached an odd number, subtracting $1$ when I did, etc..to reach the correct answer of $9$ steps.
However, I can't figure out how to convince myself beyond any doubt that it is the optimal solution. In other words, I can't prove it mathematically.
The best I can come up with is that beyond the first step from $1$ to $2$, multiplication by $2$ is always going to yield a larger step than addition by $1$ and therefore I should take as many $[times 2]$ steps as I can. This doesn't feel rigorous enough, though.
EDIT: Just to be clear, I am asking for a proof or at least more rigorous explanation of why this is the optimal solution.
combinatorics algebra-precalculus contest-math
combinatorics algebra-precalculus contest-math
asked 3 hours ago
jeremy radcliffjeremy radcliff
2,15612240
2,15612240
$begingroup$
Do you want the fewest steps to get to exactly $200$ or at least $200$?
$endgroup$
– John Douma
3 hours ago
$begingroup$
@JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise.
$endgroup$
– John Douma
2 hours ago
1
$begingroup$
@JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$.
$endgroup$
– jeremy radcliff
2 hours ago
add a comment |
$begingroup$
Do you want the fewest steps to get to exactly $200$ or at least $200$?
$endgroup$
– John Douma
3 hours ago
$begingroup$
@JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise.
$endgroup$
– John Douma
2 hours ago
1
$begingroup$
@JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$.
$endgroup$
– jeremy radcliff
2 hours ago
$begingroup$
Do you want the fewest steps to get to exactly $200$ or at least $200$?
$endgroup$
– John Douma
3 hours ago
$begingroup$
Do you want the fewest steps to get to exactly $200$ or at least $200$?
$endgroup$
– John Douma
3 hours ago
$begingroup$
@JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice.
$endgroup$
– TonyK
2 hours ago
$begingroup$
@TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise.
$endgroup$
– John Douma
2 hours ago
$begingroup$
@TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise.
$endgroup$
– John Douma
2 hours ago
1
1
$begingroup$
@JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$.
$endgroup$
– jeremy radcliff
2 hours ago
$begingroup$
@JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$.
$endgroup$
– jeremy radcliff
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Look at what the operations $+$ and $times$ do to the binary expansion of a number:
$times$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;- if the final digit is $0$, then $+$ increases the number of $1$'s by one, but doesn't change the length;
- if the final digit is $1$, then $+$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.
Therefore:
- you can increase the length by one, but this won't increase the number of $1$'s;
- you can increase the number of $1$'s by one, but this won't increase the length.
The binary expansion of $200$ is $200_{10}=11001000_2$. This has three $1$'s, and a length of eight. Starting from $1$, we must increase the length by seven, and increase the number of $1$'s by two. So this requires at least nine steps.
$endgroup$
add a comment |
$begingroup$
Setting the display to binary base, $[times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.
Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.
$endgroup$
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
add a comment |
$begingroup$
I'll make a try
Since $200=2^7+2^6+2^3$ you will need at least $8$ steps to reach $200$ (since we start from $1$ and we get a number of the form $2^a+...+2^l$) so it remains to show that $8$ steps are not enough.
Maybe you could try to show that if there was a solution with $8$ steps then it would contain only one $+1$ which contradicts the fact that in $200=2^7+2^6+2^3$ we have two $+$
$endgroup$
add a comment |
$begingroup$
You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).
Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Look at what the operations $+$ and $times$ do to the binary expansion of a number:
$times$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;- if the final digit is $0$, then $+$ increases the number of $1$'s by one, but doesn't change the length;
- if the final digit is $1$, then $+$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.
Therefore:
- you can increase the length by one, but this won't increase the number of $1$'s;
- you can increase the number of $1$'s by one, but this won't increase the length.
The binary expansion of $200$ is $200_{10}=11001000_2$. This has three $1$'s, and a length of eight. Starting from $1$, we must increase the length by seven, and increase the number of $1$'s by two. So this requires at least nine steps.
$endgroup$
add a comment |
$begingroup$
Look at what the operations $+$ and $times$ do to the binary expansion of a number:
$times$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;- if the final digit is $0$, then $+$ increases the number of $1$'s by one, but doesn't change the length;
- if the final digit is $1$, then $+$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.
Therefore:
- you can increase the length by one, but this won't increase the number of $1$'s;
- you can increase the number of $1$'s by one, but this won't increase the length.
The binary expansion of $200$ is $200_{10}=11001000_2$. This has three $1$'s, and a length of eight. Starting from $1$, we must increase the length by seven, and increase the number of $1$'s by two. So this requires at least nine steps.
$endgroup$
add a comment |
$begingroup$
Look at what the operations $+$ and $times$ do to the binary expansion of a number:
$times$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;- if the final digit is $0$, then $+$ increases the number of $1$'s by one, but doesn't change the length;
- if the final digit is $1$, then $+$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.
Therefore:
- you can increase the length by one, but this won't increase the number of $1$'s;
- you can increase the number of $1$'s by one, but this won't increase the length.
The binary expansion of $200$ is $200_{10}=11001000_2$. This has three $1$'s, and a length of eight. Starting from $1$, we must increase the length by seven, and increase the number of $1$'s by two. So this requires at least nine steps.
$endgroup$
Look at what the operations $+$ and $times$ do to the binary expansion of a number:
$times$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;- if the final digit is $0$, then $+$ increases the number of $1$'s by one, but doesn't change the length;
- if the final digit is $1$, then $+$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.
Therefore:
- you can increase the length by one, but this won't increase the number of $1$'s;
- you can increase the number of $1$'s by one, but this won't increase the length.
The binary expansion of $200$ is $200_{10}=11001000_2$. This has three $1$'s, and a length of eight. Starting from $1$, we must increase the length by seven, and increase the number of $1$'s by two. So this requires at least nine steps.
edited 1 hour ago
answered 2 hours ago
TonyKTonyK
43.2k356135
43.2k356135
add a comment |
add a comment |
$begingroup$
Setting the display to binary base, $[times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.
Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.
$endgroup$
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
add a comment |
$begingroup$
Setting the display to binary base, $[times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.
Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.
$endgroup$
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
add a comment |
$begingroup$
Setting the display to binary base, $[times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.
Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.
$endgroup$
Setting the display to binary base, $[times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.
Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.
answered 2 hours ago
Yves DaoustYves Daoust
130k676229
130k676229
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
add a comment |
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
you just exactly reproduces the binary 200, why should we think it is not optimal?
$endgroup$
– dEmigOd
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
$begingroup$
@dEmigOd: I didn't prove that inserting the bits one by one with $times2$ or $times2+1$ is optimal.
$endgroup$
– Yves Daoust
2 hours ago
add a comment |
$begingroup$
I'll make a try
Since $200=2^7+2^6+2^3$ you will need at least $8$ steps to reach $200$ (since we start from $1$ and we get a number of the form $2^a+...+2^l$) so it remains to show that $8$ steps are not enough.
Maybe you could try to show that if there was a solution with $8$ steps then it would contain only one $+1$ which contradicts the fact that in $200=2^7+2^6+2^3$ we have two $+$
$endgroup$
add a comment |
$begingroup$
I'll make a try
Since $200=2^7+2^6+2^3$ you will need at least $8$ steps to reach $200$ (since we start from $1$ and we get a number of the form $2^a+...+2^l$) so it remains to show that $8$ steps are not enough.
Maybe you could try to show that if there was a solution with $8$ steps then it would contain only one $+1$ which contradicts the fact that in $200=2^7+2^6+2^3$ we have two $+$
$endgroup$
add a comment |
$begingroup$
I'll make a try
Since $200=2^7+2^6+2^3$ you will need at least $8$ steps to reach $200$ (since we start from $1$ and we get a number of the form $2^a+...+2^l$) so it remains to show that $8$ steps are not enough.
Maybe you could try to show that if there was a solution with $8$ steps then it would contain only one $+1$ which contradicts the fact that in $200=2^7+2^6+2^3$ we have two $+$
$endgroup$
I'll make a try
Since $200=2^7+2^6+2^3$ you will need at least $8$ steps to reach $200$ (since we start from $1$ and we get a number of the form $2^a+...+2^l$) so it remains to show that $8$ steps are not enough.
Maybe you could try to show that if there was a solution with $8$ steps then it would contain only one $+1$ which contradicts the fact that in $200=2^7+2^6+2^3$ we have two $+$
answered 2 hours ago
giannispapavgiannispapav
1,806325
1,806325
add a comment |
add a comment |
$begingroup$
You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).
Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.
$endgroup$
add a comment |
$begingroup$
You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).
Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.
$endgroup$
add a comment |
$begingroup$
You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).
Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.
$endgroup$
You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).
Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.
edited 1 hour ago
answered 2 hours ago
Barry CipraBarry Cipra
60.2k654126
60.2k654126
add a comment |
add a comment |
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Do you want the fewest steps to get to exactly $200$ or at least $200$?
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– John Douma
3 hours ago
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@JohnDouma: Exactly $200$, obviously. Otherwise eight steps would suffice.
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– TonyK
2 hours ago
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@TonyK That is not obvious. In fact, the last paragraph before the EDIT implies otherwise.
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– John Douma
2 hours ago
1
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@JohnDouma It's exactly $200$, but I disagree with your comment that the last paragraph implies otherwise. I can try to take as many X2 steps as I can and still intend to not get past $200$.
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– jeremy radcliff
2 hours ago