Mixing
AMC 12 2010B Problem Help #18
$begingroup$
Can someone explain this solution?
A frog makes 3 jumps, each exactly 1 meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than 1 meter from its starting position?
Solution 1
We will let the moves be complex numbers a,b , and c, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set
$$ {|a+b+c|,|a+b-c|,|a-b+c|,|a-b-c|} $$
has magnitude less than or equal to 1. Hence, the probability is 1/4.
contest-math
edited Jan 24 at 5:35
Presh
661718
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
$endgroup$
|
show 2 more comments
$begingroup$
Can someone explain this solution?
A frog makes 3 jumps, each exactly 1 meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than 1 meter from its starting position?
Solution 1
We will let the moves be complex numbers a,b , and c, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set
$$ {|a+b+c|,|a+b-c|,|a-b+c|,|a-b-c|} $$
has magnitude less than or equal to 1. Hence, the probability is 1/4.
contest-math
edited Jan 24 at 5:35
Presh
661718
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
$endgroup$
$begingroup$
Do we know how many jumps it makes?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:30
$begingroup$
I remember this question, it makes two jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:31
$begingroup$
but in the answer it says a,b,c. Sure it isn't three?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:34
$begingroup$
Nevermind, I was right the first time, it's three jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:35
$begingroup$
oops. so sorry, yes it is three
$endgroup$
– Yadnarav3
Feb 4 '14 at 0:36
|
show 2 more comments
$begingroup$
Can someone explain this solution?
A frog makes 3 jumps, each exactly 1 meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than 1 meter from its starting position?
Solution 1
We will let the moves be complex numbers a,b , and c, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set
$$ {|a+b+c|,|a+b-c|,|a-b+c|,|a-b-c|} $$
has magnitude less than or equal to 1. Hence, the probability is 1/4.
contest-math
edited Jan 24 at 5:35
Presh
661718
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
$endgroup$
Can someone explain this solution?
A frog makes 3 jumps, each exactly 1 meter long. The directions of the jumps are chosen independently at random. What is the probability that the frog's final position is no more than 1 meter from its starting position?
Solution 1
We will let the moves be complex numbers a,b , and c, each of magnitude one. The frog starts on the origin. It is relatively easy to show that exactly one element in the set
$$ {|a+b+c|,|a+b-c|,|a-b+c|,|a-b-c|} $$
has magnitude less than or equal to 1. Hence, the probability is 1/4.
contest-math
contest-math
edited Jan 24 at 5:35
Presh
661718
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
edited Jan 24 at 5:35
Presh
661718
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
edited Jan 24 at 5:35
Presh
661718
edited Jan 24 at 5:35
Presh
661718
edited Jan 24 at 5:35
Presh
661718
661718
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
asked Feb 4 '14 at 0:24
Yadnarav3Yadnarav3
523214
523214
$begingroup$
Do we know how many jumps it makes?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:30
$begingroup$
I remember this question, it makes two jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:31
$begingroup$
but in the answer it says a,b,c. Sure it isn't three?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:34
$begingroup$
Nevermind, I was right the first time, it's three jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:35
$begingroup$
oops. so sorry, yes it is three
$endgroup$
– Yadnarav3
Feb 4 '14 at 0:36
|
show 2 more comments
$begingroup$
Do we know how many jumps it makes?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:30
$begingroup$
I remember this question, it makes two jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:31
$begingroup$
but in the answer it says a,b,c. Sure it isn't three?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:34
$begingroup$
Nevermind, I was right the first time, it's three jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:35
$begingroup$
oops. so sorry, yes it is three
$endgroup$
– Yadnarav3
Feb 4 '14 at 0:36
$begingroup$
Do we know how many jumps it makes?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:30
$begingroup$
Do we know how many jumps it makes?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:30
$begingroup$
I remember this question, it makes two jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:31
$begingroup$
I remember this question, it makes two jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:31
$begingroup$
but in the answer it says a,b,c. Sure it isn't three?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:34
$begingroup$
but in the answer it says a,b,c. Sure it isn't three?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:34
$begingroup$
Nevermind, I was right the first time, it's three jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:35
$begingroup$
Nevermind, I was right the first time, it's three jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:35
$begingroup$
oops. so sorry, yes it is three
$endgroup$
– Yadnarav3
Feb 4 '14 at 0:36
$begingroup$
oops. so sorry, yes it is three
$endgroup$
– Yadnarav3
Feb 4 '14 at 0:36
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Funny that I remember this problem. Anyway this is how I solved it:
The frog starts at $(0, 0)$ and makes a jump, WLOG let it land at $(1, 0)$. From here, the frog can make two jumps and land anywhere on a circle of radius $2$ centered at $(1, 0)$, each point being equally likely. This has an area of $4 pi$, and this circle is large enough so that it contains the entirety of the desired circle of radius $1$ centered at $(0, 0)$ that we want the frog to end up in. Thus, the probability is $frac{pi}{4pi} = frac{1}{4}$
As for the solution you linked, I don't know particularly what's troubling you about it so I'll explain as much as I can:
We choose at random three steps $a, b, c$. If exactly one element has absolute value (talking vector length here with complex numbers) less than or equal to $1$, then, out of $4$ possibilities, one will be inside the inner circle. So the probability is $frac{1}{4}$.
How would you prove that exactly one is? Try considering whether it's possible to have more than one step which satisfies the conditions, and furthermore consider whether if none of them are. (This is pretty much the essence of the solution, so I don't want to just spoon feed it to you, especially since this is contest math)
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
$endgroup$
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f662709%2famc-12-2010b-problem-help-18%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Funny that I remember this problem. Anyway this is how I solved it:
The frog starts at $(0, 0)$ and makes a jump, WLOG let it land at $(1, 0)$. From here, the frog can make two jumps and land anywhere on a circle of radius $2$ centered at $(1, 0)$, each point being equally likely. This has an area of $4 pi$, and this circle is large enough so that it contains the entirety of the desired circle of radius $1$ centered at $(0, 0)$ that we want the frog to end up in. Thus, the probability is $frac{pi}{4pi} = frac{1}{4}$
As for the solution you linked, I don't know particularly what's troubling you about it so I'll explain as much as I can:
We choose at random three steps $a, b, c$. If exactly one element has absolute value (talking vector length here with complex numbers) less than or equal to $1$, then, out of $4$ possibilities, one will be inside the inner circle. So the probability is $frac{1}{4}$.
How would you prove that exactly one is? Try considering whether it's possible to have more than one step which satisfies the conditions, and furthermore consider whether if none of them are. (This is pretty much the essence of the solution, so I don't want to just spoon feed it to you, especially since this is contest math)
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
$endgroup$
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
add a comment |
$begingroup$
Funny that I remember this problem. Anyway this is how I solved it:
The frog starts at $(0, 0)$ and makes a jump, WLOG let it land at $(1, 0)$. From here, the frog can make two jumps and land anywhere on a circle of radius $2$ centered at $(1, 0)$, each point being equally likely. This has an area of $4 pi$, and this circle is large enough so that it contains the entirety of the desired circle of radius $1$ centered at $(0, 0)$ that we want the frog to end up in. Thus, the probability is $frac{pi}{4pi} = frac{1}{4}$
As for the solution you linked, I don't know particularly what's troubling you about it so I'll explain as much as I can:
We choose at random three steps $a, b, c$. If exactly one element has absolute value (talking vector length here with complex numbers) less than or equal to $1$, then, out of $4$ possibilities, one will be inside the inner circle. So the probability is $frac{1}{4}$.
How would you prove that exactly one is? Try considering whether it's possible to have more than one step which satisfies the conditions, and furthermore consider whether if none of them are. (This is pretty much the essence of the solution, so I don't want to just spoon feed it to you, especially since this is contest math)
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
$endgroup$
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
add a comment |
$begingroup$
Funny that I remember this problem. Anyway this is how I solved it:
The frog starts at $(0, 0)$ and makes a jump, WLOG let it land at $(1, 0)$. From here, the frog can make two jumps and land anywhere on a circle of radius $2$ centered at $(1, 0)$, each point being equally likely. This has an area of $4 pi$, and this circle is large enough so that it contains the entirety of the desired circle of radius $1$ centered at $(0, 0)$ that we want the frog to end up in. Thus, the probability is $frac{pi}{4pi} = frac{1}{4}$
As for the solution you linked, I don't know particularly what's troubling you about it so I'll explain as much as I can:
We choose at random three steps $a, b, c$. If exactly one element has absolute value (talking vector length here with complex numbers) less than or equal to $1$, then, out of $4$ possibilities, one will be inside the inner circle. So the probability is $frac{1}{4}$.
How would you prove that exactly one is? Try considering whether it's possible to have more than one step which satisfies the conditions, and furthermore consider whether if none of them are. (This is pretty much the essence of the solution, so I don't want to just spoon feed it to you, especially since this is contest math)
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
$endgroup$
Funny that I remember this problem. Anyway this is how I solved it:
The frog starts at $(0, 0)$ and makes a jump, WLOG let it land at $(1, 0)$. From here, the frog can make two jumps and land anywhere on a circle of radius $2$ centered at $(1, 0)$, each point being equally likely. This has an area of $4 pi$, and this circle is large enough so that it contains the entirety of the desired circle of radius $1$ centered at $(0, 0)$ that we want the frog to end up in. Thus, the probability is $frac{pi}{4pi} = frac{1}{4}$
As for the solution you linked, I don't know particularly what's troubling you about it so I'll explain as much as I can:
We choose at random three steps $a, b, c$. If exactly one element has absolute value (talking vector length here with complex numbers) less than or equal to $1$, then, out of $4$ possibilities, one will be inside the inner circle. So the probability is $frac{1}{4}$.
How would you prove that exactly one is? Try considering whether it's possible to have more than one step which satisfies the conditions, and furthermore consider whether if none of them are. (This is pretty much the essence of the solution, so I don't want to just spoon feed it to you, especially since this is contest math)
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
edited Feb 4 '14 at 0:58
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
answered Feb 4 '14 at 0:34
MCTMCT
14.5k42668
14.5k42668
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
add a comment |
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
but I wanna be spoon fed this time :[ ] feed me
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:07
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
The proof of why exactly one element has magnitude less than or equal to one is what stumps me.
$endgroup$
– Yadnarav3
Feb 4 '14 at 1:08
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Well, what you have in the set is all of the possible permutations of negative and positive signs (considering absolute value) of the three steps. Assuming it exists, consider the permutation that leads to an absolute value of less than or equal to 1. What would happen if you changed the sign of one of the steps?
$endgroup$
– MCT
Feb 4 '14 at 1:28
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
Is there a nice proof that the sum of two unit vectors is uniformly distributed in the disc of radius 2? There is clear radial symmetry, but I'm not sure that the distribution (conditional on lying on a particular radial line) is uniform. In particular, if $theta$ is uniform $[0,2pi]$, I think the radial distribution is $2cos theta.$
$endgroup$
– Aaron
Feb 4 '14 at 4:11
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
$begingroup$
@Aaron The sum of two uniform unit vectors is indeed not uniform in the disc of radius 2. The radius $R$ of the sum has as its CDF $mathbb{P}(R leq r) = 1-frac{1}{pi}arccosleft(frac{r^2-2}{2}right)$ for $0 leq r leq 2$.
$endgroup$
– kccu
Jan 24 at 4:44
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f662709%2famc-12-2010b-problem-help-18%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Do we know how many jumps it makes?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:30
$begingroup$
I remember this question, it makes two jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:31
$begingroup$
but in the answer it says a,b,c. Sure it isn't three?
$endgroup$
– Jorge Fernández Hidalgo
Feb 4 '14 at 0:34
$begingroup$
Nevermind, I was right the first time, it's three jumps.
$endgroup$
– MCT
Feb 4 '14 at 0:35
$begingroup$
oops. so sorry, yes it is three
$endgroup$
– Yadnarav3
Feb 4 '14 at 0:36