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Olympiad Algebra Practice Question
A Math Olympiad question regarding GeometryHelp with inequality pleaseMath Olympiad Algebra QuestionAlgebra books for olympiad preparationFind all possible sum of digits of perfect squares$10times 10$ checkerboard with relatively prime elements9-10th grade olympiad problemMotivation for this solution to a British olympiad problemReal analysis question from olympiad practice book
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I just needed some help for this problem they gave us at our mathematics class, its’s an Olympiad type question, I suspect the answer is 2019, I tried with smaller cases and tried to use a recursive relation, or some type of induction. I also tried factoring by noticing that $a+b+ab=(a+1)(b+1)-1$ but couldn’t manage to solve it. Hoping someone could help. Here is the problem:
Martin wrote the following list of numbers on a whiteboard:
$$1, frac12, frac13, frac14, frac15,...,frac12019$$
Martin asked for a volunteer. Vincent offered to play, Martin explained the rules to Vincent. Vincent has to choose two of the numbers on the board, lets say $a$ and $b$. He has to wipe off these numbers and write the number $a+b+ab$ on the board.
For example, if Vincent would have chosen $frac12$ and $frac13$, he would wipe these numbers off and write $frac12+frac13+frac12·frac13=1$ on the board. Then the board would look like this:
$$1, 1, frac14, frac15,...,frac12019$$
Martin asked Vincent to do this over and over again choosing any two numbers each time, until only one number was left. When Vincent was done, Martin asked him to open an envelope where he had written a prediction, Vincent opened it and was surprised to find that the number on the whiteboard and on the envelope were the same.
What was Martin’s prediction? Justify your answer.
contest-math
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
$endgroup$
add a comment |
$begingroup$
I just needed some help for this problem they gave us at our mathematics class, its’s an Olympiad type question, I suspect the answer is 2019, I tried with smaller cases and tried to use a recursive relation, or some type of induction. I also tried factoring by noticing that $a+b+ab=(a+1)(b+1)-1$ but couldn’t manage to solve it. Hoping someone could help. Here is the problem:
Martin wrote the following list of numbers on a whiteboard:
$$1, frac12, frac13, frac14, frac15,...,frac12019$$
Martin asked for a volunteer. Vincent offered to play, Martin explained the rules to Vincent. Vincent has to choose two of the numbers on the board, lets say $a$ and $b$. He has to wipe off these numbers and write the number $a+b+ab$ on the board.
For example, if Vincent would have chosen $frac12$ and $frac13$, he would wipe these numbers off and write $frac12+frac13+frac12·frac13=1$ on the board. Then the board would look like this:
$$1, 1, frac14, frac15,...,frac12019$$
Martin asked Vincent to do this over and over again choosing any two numbers each time, until only one number was left. When Vincent was done, Martin asked him to open an envelope where he had written a prediction, Vincent opened it and was surprised to find that the number on the whiteboard and on the envelope were the same.
What was Martin’s prediction? Justify your answer.
contest-math
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
$endgroup$
1
$begingroup$
Well, to see that the result (if it exists) can only be $2019$, work inductively. Do the game $G_n$ with $1, frac 12, cdots ,frac 1n$. Easy to see that $G_2$ ends in $2$. Assume that $G_n$ ends in $n$. Then, to do $G_n+1$ you first play the game $G_n$, leaving you with $n$ and $frac 1n+1$. But $n+frac 1n+1+frac nn+1=n+1$ as desired. You still need to show that the order in which you do things doesn't matter, however.
$endgroup$
– lulu
8 hours ago
$begingroup$
Yeah, that’s the thing, I don’t think is induction because at my level we are not supposed to know that or solve these problems with those methods, I think its something simpler. But thanks for the response.
$endgroup$
– Eurekaepicstyle
8 hours ago
2
$begingroup$
The result will be $(1+1/1)(1+1/2)cdots(1+1/2019)-1=2019.$ If we define the uperation $*$ as $a*b=(1+a)(1+b)-1$ then $*$ is commutative and associative.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
$begingroup$
I just needed some help for this problem they gave us at our mathematics class, its’s an Olympiad type question, I suspect the answer is 2019, I tried with smaller cases and tried to use a recursive relation, or some type of induction. I also tried factoring by noticing that $a+b+ab=(a+1)(b+1)-1$ but couldn’t manage to solve it. Hoping someone could help. Here is the problem:
Martin wrote the following list of numbers on a whiteboard:
$$1, frac12, frac13, frac14, frac15,...,frac12019$$
Martin asked for a volunteer. Vincent offered to play, Martin explained the rules to Vincent. Vincent has to choose two of the numbers on the board, lets say $a$ and $b$. He has to wipe off these numbers and write the number $a+b+ab$ on the board.
For example, if Vincent would have chosen $frac12$ and $frac13$, he would wipe these numbers off and write $frac12+frac13+frac12·frac13=1$ on the board. Then the board would look like this:
$$1, 1, frac14, frac15,...,frac12019$$
Martin asked Vincent to do this over and over again choosing any two numbers each time, until only one number was left. When Vincent was done, Martin asked him to open an envelope where he had written a prediction, Vincent opened it and was surprised to find that the number on the whiteboard and on the envelope were the same.
What was Martin’s prediction? Justify your answer.
contest-math
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
$endgroup$
I just needed some help for this problem they gave us at our mathematics class, its’s an Olympiad type question, I suspect the answer is 2019, I tried with smaller cases and tried to use a recursive relation, or some type of induction. I also tried factoring by noticing that $a+b+ab=(a+1)(b+1)-1$ but couldn’t manage to solve it. Hoping someone could help. Here is the problem:
Martin wrote the following list of numbers on a whiteboard:
$$1, frac12, frac13, frac14, frac15,...,frac12019$$
Martin asked for a volunteer. Vincent offered to play, Martin explained the rules to Vincent. Vincent has to choose two of the numbers on the board, lets say $a$ and $b$. He has to wipe off these numbers and write the number $a+b+ab$ on the board.
For example, if Vincent would have chosen $frac12$ and $frac13$, he would wipe these numbers off and write $frac12+frac13+frac12·frac13=1$ on the board. Then the board would look like this:
$$1, 1, frac14, frac15,...,frac12019$$
Martin asked Vincent to do this over and over again choosing any two numbers each time, until only one number was left. When Vincent was done, Martin asked him to open an envelope where he had written a prediction, Vincent opened it and was surprised to find that the number on the whiteboard and on the envelope were the same.
What was Martin’s prediction? Justify your answer.
contest-math
contest-math
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
asked 8 hours ago
EurekaepicstyleEurekaepicstyle
363 bronze badges
363 bronze badges
1
$begingroup$
Well, to see that the result (if it exists) can only be $2019$, work inductively. Do the game $G_n$ with $1, frac 12, cdots ,frac 1n$. Easy to see that $G_2$ ends in $2$. Assume that $G_n$ ends in $n$. Then, to do $G_n+1$ you first play the game $G_n$, leaving you with $n$ and $frac 1n+1$. But $n+frac 1n+1+frac nn+1=n+1$ as desired. You still need to show that the order in which you do things doesn't matter, however.
$endgroup$
– lulu
8 hours ago
$begingroup$
Yeah, that’s the thing, I don’t think is induction because at my level we are not supposed to know that or solve these problems with those methods, I think its something simpler. But thanks for the response.
$endgroup$
– Eurekaepicstyle
8 hours ago
2
$begingroup$
The result will be $(1+1/1)(1+1/2)cdots(1+1/2019)-1=2019.$ If we define the uperation $*$ as $a*b=(1+a)(1+b)-1$ then $*$ is commutative and associative.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
1
$begingroup$
Well, to see that the result (if it exists) can only be $2019$, work inductively. Do the game $G_n$ with $1, frac 12, cdots ,frac 1n$. Easy to see that $G_2$ ends in $2$. Assume that $G_n$ ends in $n$. Then, to do $G_n+1$ you first play the game $G_n$, leaving you with $n$ and $frac 1n+1$. But $n+frac 1n+1+frac nn+1=n+1$ as desired. You still need to show that the order in which you do things doesn't matter, however.
$endgroup$
– lulu
8 hours ago
$begingroup$
Yeah, that’s the thing, I don’t think is induction because at my level we are not supposed to know that or solve these problems with those methods, I think its something simpler. But thanks for the response.
$endgroup$
– Eurekaepicstyle
8 hours ago
2
$begingroup$
The result will be $(1+1/1)(1+1/2)cdots(1+1/2019)-1=2019.$ If we define the uperation $*$ as $a*b=(1+a)(1+b)-1$ then $*$ is commutative and associative.
$endgroup$
– Thomas Andrews
8 hours ago
1
1
$begingroup$
Well, to see that the result (if it exists) can only be $2019$, work inductively. Do the game $G_n$ with $1, frac 12, cdots ,frac 1n$. Easy to see that $G_2$ ends in $2$. Assume that $G_n$ ends in $n$. Then, to do $G_n+1$ you first play the game $G_n$, leaving you with $n$ and $frac 1n+1$. But $n+frac 1n+1+frac nn+1=n+1$ as desired. You still need to show that the order in which you do things doesn't matter, however.
$endgroup$
– lulu
8 hours ago
$begingroup$
Well, to see that the result (if it exists) can only be $2019$, work inductively. Do the game $G_n$ with $1, frac 12, cdots ,frac 1n$. Easy to see that $G_2$ ends in $2$. Assume that $G_n$ ends in $n$. Then, to do $G_n+1$ you first play the game $G_n$, leaving you with $n$ and $frac 1n+1$. But $n+frac 1n+1+frac nn+1=n+1$ as desired. You still need to show that the order in which you do things doesn't matter, however.
$endgroup$
– lulu
8 hours ago
$begingroup$
Yeah, that’s the thing, I don’t think is induction because at my level we are not supposed to know that or solve these problems with those methods, I think its something simpler. But thanks for the response.
$endgroup$
– Eurekaepicstyle
8 hours ago
$begingroup$
Yeah, that’s the thing, I don’t think is induction because at my level we are not supposed to know that or solve these problems with those methods, I think its something simpler. But thanks for the response.
$endgroup$
– Eurekaepicstyle
8 hours ago
2
2
$begingroup$
The result will be $(1+1/1)(1+1/2)cdots(1+1/2019)-1=2019.$ If we define the uperation $*$ as $a*b=(1+a)(1+b)-1$ then $*$ is commutative and associative.
$endgroup$
– Thomas Andrews
8 hours ago
$begingroup$
The result will be $(1+1/1)(1+1/2)cdots(1+1/2019)-1=2019.$ If we define the uperation $*$ as $a*b=(1+a)(1+b)-1$ then $*$ is commutative and associative.
$endgroup$
– Thomas Andrews
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let the terms on the board in any given move be $a_1, a_2, a_3, a_4 ... a_n$. We will prove that $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ never changes, and therefore is an invariant.
First, consider any move, with $a_i$ and $a_j$. Notice that $(1+a_i)(1+a_j) = 1+a_i+a_j+a_ia_j$, and therefore the value of $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ constant.
Let the value that Martin gets at the end be $n$. Notice that $(1+frac11)(1+frac12)...(1+frac12019)-1 = frac21frac32....frac20202019 - 1 = (1 + n) - 1 implies n = 2019$, so we are done.
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
$endgroup$
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
Let the terms on the board in any given move be $a_1, a_2, a_3, a_4 ... a_n$. We will prove that $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ never changes, and therefore is an invariant.
First, consider any move, with $a_i$ and $a_j$. Notice that $(1+a_i)(1+a_j) = 1+a_i+a_j+a_ia_j$, and therefore the value of $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ constant.
Let the value that Martin gets at the end be $n$. Notice that $(1+frac11)(1+frac12)...(1+frac12019)-1 = frac21frac32....frac20202019 - 1 = (1 + n) - 1 implies n = 2019$, so we are done.
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
$endgroup$
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
add a comment |
$begingroup$
Let the terms on the board in any given move be $a_1, a_2, a_3, a_4 ... a_n$. We will prove that $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ never changes, and therefore is an invariant.
First, consider any move, with $a_i$ and $a_j$. Notice that $(1+a_i)(1+a_j) = 1+a_i+a_j+a_ia_j$, and therefore the value of $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ constant.
Let the value that Martin gets at the end be $n$. Notice that $(1+frac11)(1+frac12)...(1+frac12019)-1 = frac21frac32....frac20202019 - 1 = (1 + n) - 1 implies n = 2019$, so we are done.
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
$endgroup$
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
add a comment |
$begingroup$
Let the terms on the board in any given move be $a_1, a_2, a_3, a_4 ... a_n$. We will prove that $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ never changes, and therefore is an invariant.
First, consider any move, with $a_i$ and $a_j$. Notice that $(1+a_i)(1+a_j) = 1+a_i+a_j+a_ia_j$, and therefore the value of $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ constant.
Let the value that Martin gets at the end be $n$. Notice that $(1+frac11)(1+frac12)...(1+frac12019)-1 = frac21frac32....frac20202019 - 1 = (1 + n) - 1 implies n = 2019$, so we are done.
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
$endgroup$
Let the terms on the board in any given move be $a_1, a_2, a_3, a_4 ... a_n$. We will prove that $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ never changes, and therefore is an invariant.
First, consider any move, with $a_i$ and $a_j$. Notice that $(1+a_i)(1+a_j) = 1+a_i+a_j+a_ia_j$, and therefore the value of $(1+a_1)(1+a_2)(1+a_3)....(1+a_n)-1$ constant.
Let the value that Martin gets at the end be $n$. Notice that $(1+frac11)(1+frac12)...(1+frac12019)-1 = frac21frac32....frac20202019 - 1 = (1 + n) - 1 implies n = 2019$, so we are done.
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
answered 7 hours ago
ETS1331ETS1331
6174 silver badges17 bronze badges
6174 silver badges17 bronze badges
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
add a comment |
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
Excellent answer, thanks a lot. Nice to see that it had something to do with factoring. I think Invariance is a really powerful method. I saw 3blue1browns’s video of the IMO problem and was amazed.
$endgroup$
– Eurekaepicstyle
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
That "and therefore" in the middle (lines 3-4) bewilders me. Care about fleshing that out a bit?
$endgroup$
– DonAntonio
7 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
$begingroup$
Without loss of generality, assume that $a_i$ and $a_j$ are in fact $a_1$ and $a_2$ (to make writing this up easier). Then, notice how when they erase $a_1$ and $a_2$, the new term, $a_n = a_1a_2 + a_1 + a_2$, and therefore $a_n + 1 = a_1a_2 + a_1 + a_2 + 1 = (a_1+1)(a_2+1)$ so the two produts are equal.
$endgroup$
– ETS1331
4 hours ago
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1
$begingroup$
Well, to see that the result (if it exists) can only be $2019$, work inductively. Do the game $G_n$ with $1, frac 12, cdots ,frac 1n$. Easy to see that $G_2$ ends in $2$. Assume that $G_n$ ends in $n$. Then, to do $G_n+1$ you first play the game $G_n$, leaving you with $n$ and $frac 1n+1$. But $n+frac 1n+1+frac nn+1=n+1$ as desired. You still need to show that the order in which you do things doesn't matter, however.
$endgroup$
– lulu
8 hours ago
$begingroup$
Yeah, that’s the thing, I don’t think is induction because at my level we are not supposed to know that or solve these problems with those methods, I think its something simpler. But thanks for the response.
$endgroup$
– Eurekaepicstyle
8 hours ago
2
$begingroup$
The result will be $(1+1/1)(1+1/2)cdots(1+1/2019)-1=2019.$ If we define the uperation $*$ as $a*b=(1+a)(1+b)-1$ then $*$ is commutative and associative.
$endgroup$
– Thomas Andrews
8 hours ago