Mathematical guessing

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A total of 2,879 votes have been distributed among the seven candidates in a one-party primary election. It is known that none of the candidates has obtained the same number of votes as another and that if the number of votes obtained by any of the candidates is divided by the number of votes obtained by any other candidate who has obtained fewer votes, the result is always a whole number. How many votes has each candidate obtained?




I was thinking of solving it by means of congruences and the Chinese Theorem of the Rest, but I don't know very well how to solve this riddle, which has been proposed to me by an acquaintance.










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  • 4




    Hint: order the candidates vote count as $v_1<v_2<cdots <v_7$. Then $v_1$ divides each of the $v_i$, so $v_1$ must divide the total. Reason from there.
    – lulu
    Nov 16 at 19:16











  • I don't understand very well the hint @lulu
    – Carlos
    Nov 16 at 19:20






  • 2




    Let $v_i=n_iv_1$. Then $2879=v_1+v_2+cdots+v_7=v_1+n_2v_1+cdots +n_7v_1=v_1times (1+n_1+n_2+cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number.
    – lulu
    Nov 16 at 19:22










  • But how can I factor in a number I don't know? @lulu
    – Carlos
    Nov 16 at 19:25











  • Just factor $2879$. Nobody is asking you to "factor in a number you don't know."
    – lulu
    Nov 16 at 19:28















up vote
6
down vote

favorite













A total of 2,879 votes have been distributed among the seven candidates in a one-party primary election. It is known that none of the candidates has obtained the same number of votes as another and that if the number of votes obtained by any of the candidates is divided by the number of votes obtained by any other candidate who has obtained fewer votes, the result is always a whole number. How many votes has each candidate obtained?




I was thinking of solving it by means of congruences and the Chinese Theorem of the Rest, but I don't know very well how to solve this riddle, which has been proposed to me by an acquaintance.










share|cite|improve this question

















  • 4




    Hint: order the candidates vote count as $v_1<v_2<cdots <v_7$. Then $v_1$ divides each of the $v_i$, so $v_1$ must divide the total. Reason from there.
    – lulu
    Nov 16 at 19:16











  • I don't understand very well the hint @lulu
    – Carlos
    Nov 16 at 19:20






  • 2




    Let $v_i=n_iv_1$. Then $2879=v_1+v_2+cdots+v_7=v_1+n_2v_1+cdots +n_7v_1=v_1times (1+n_1+n_2+cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number.
    – lulu
    Nov 16 at 19:22










  • But how can I factor in a number I don't know? @lulu
    – Carlos
    Nov 16 at 19:25











  • Just factor $2879$. Nobody is asking you to "factor in a number you don't know."
    – lulu
    Nov 16 at 19:28













up vote
6
down vote

favorite









up vote
6
down vote

favorite












A total of 2,879 votes have been distributed among the seven candidates in a one-party primary election. It is known that none of the candidates has obtained the same number of votes as another and that if the number of votes obtained by any of the candidates is divided by the number of votes obtained by any other candidate who has obtained fewer votes, the result is always a whole number. How many votes has each candidate obtained?




I was thinking of solving it by means of congruences and the Chinese Theorem of the Rest, but I don't know very well how to solve this riddle, which has been proposed to me by an acquaintance.










share|cite|improve this question














A total of 2,879 votes have been distributed among the seven candidates in a one-party primary election. It is known that none of the candidates has obtained the same number of votes as another and that if the number of votes obtained by any of the candidates is divided by the number of votes obtained by any other candidate who has obtained fewer votes, the result is always a whole number. How many votes has each candidate obtained?




I was thinking of solving it by means of congruences and the Chinese Theorem of the Rest, but I don't know very well how to solve this riddle, which has been proposed to me by an acquaintance.







puzzle






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asked Nov 16 at 19:13









Carlos

354




354







  • 4




    Hint: order the candidates vote count as $v_1<v_2<cdots <v_7$. Then $v_1$ divides each of the $v_i$, so $v_1$ must divide the total. Reason from there.
    – lulu
    Nov 16 at 19:16











  • I don't understand very well the hint @lulu
    – Carlos
    Nov 16 at 19:20






  • 2




    Let $v_i=n_iv_1$. Then $2879=v_1+v_2+cdots+v_7=v_1+n_2v_1+cdots +n_7v_1=v_1times (1+n_1+n_2+cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number.
    – lulu
    Nov 16 at 19:22










  • But how can I factor in a number I don't know? @lulu
    – Carlos
    Nov 16 at 19:25











  • Just factor $2879$. Nobody is asking you to "factor in a number you don't know."
    – lulu
    Nov 16 at 19:28













  • 4




    Hint: order the candidates vote count as $v_1<v_2<cdots <v_7$. Then $v_1$ divides each of the $v_i$, so $v_1$ must divide the total. Reason from there.
    – lulu
    Nov 16 at 19:16











  • I don't understand very well the hint @lulu
    – Carlos
    Nov 16 at 19:20






  • 2




    Let $v_i=n_iv_1$. Then $2879=v_1+v_2+cdots+v_7=v_1+n_2v_1+cdots +n_7v_1=v_1times (1+n_1+n_2+cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number.
    – lulu
    Nov 16 at 19:22










  • But how can I factor in a number I don't know? @lulu
    – Carlos
    Nov 16 at 19:25











  • Just factor $2879$. Nobody is asking you to "factor in a number you don't know."
    – lulu
    Nov 16 at 19:28








4




4




Hint: order the candidates vote count as $v_1<v_2<cdots <v_7$. Then $v_1$ divides each of the $v_i$, so $v_1$ must divide the total. Reason from there.
– lulu
Nov 16 at 19:16





Hint: order the candidates vote count as $v_1<v_2<cdots <v_7$. Then $v_1$ divides each of the $v_i$, so $v_1$ must divide the total. Reason from there.
– lulu
Nov 16 at 19:16













I don't understand very well the hint @lulu
– Carlos
Nov 16 at 19:20




I don't understand very well the hint @lulu
– Carlos
Nov 16 at 19:20




2




2




Let $v_i=n_iv_1$. Then $2879=v_1+v_2+cdots+v_7=v_1+n_2v_1+cdots +n_7v_1=v_1times (1+n_1+n_2+cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number.
– lulu
Nov 16 at 19:22




Let $v_i=n_iv_1$. Then $2879=v_1+v_2+cdots+v_7=v_1+n_2v_1+cdots +n_7v_1=v_1times (1+n_1+n_2+cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number.
– lulu
Nov 16 at 19:22












But how can I factor in a number I don't know? @lulu
– Carlos
Nov 16 at 19:25





But how can I factor in a number I don't know? @lulu
– Carlos
Nov 16 at 19:25













Just factor $2879$. Nobody is asking you to "factor in a number you don't know."
– lulu
Nov 16 at 19:28





Just factor $2879$. Nobody is asking you to "factor in a number you don't know."
– lulu
Nov 16 at 19:28











1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










Call the number of votes the candidate that gotten the fewest votes for $a_1$.



The candidate that got the second lowest number of votes must then have gotten a multiplum of $a_1$, let's define $a_2$ so that $a_2a_1$.



We can continue and define $a_3, ldots, a_7$ similarly.



As $a_i$ for $igeq 2$ is the factor between the number of votes for candidate $i-1$ and candidate $i$, they can't be $1$.



If we add the number of votes we get
$$
2879 = a_1+a_2a_1+cdots+a_7a_6a_5a_4a_3a_2a_1
= a_1(1+a_2(1+a_3(1+cdots)))
$$



So we can conclude that $a_1 | 2879$, but that number is prime, so the only factors are $1$ and $2879$. $a_1=2879$ is obviously absurd, so $a_1 = 1$.



Putting that into the equation above we get $2879=1+a_2(1+cdots)$, or $2878=a_2(1+cdots)$. Factoring $2878$ (and ruling out the absurd possibility) we conclude $a_2=2$.



Try to go on from there.



Addition: If anyone tries to complete the calculations, the $a_i$'s are not the number of votes, and thus don't have to be different (hint: they won't be).






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  • Ooops... let me check again. Thanks.
    – David G. Stork
    Nov 16 at 20:55










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Call the number of votes the candidate that gotten the fewest votes for $a_1$.



The candidate that got the second lowest number of votes must then have gotten a multiplum of $a_1$, let's define $a_2$ so that $a_2a_1$.



We can continue and define $a_3, ldots, a_7$ similarly.



As $a_i$ for $igeq 2$ is the factor between the number of votes for candidate $i-1$ and candidate $i$, they can't be $1$.



If we add the number of votes we get
$$
2879 = a_1+a_2a_1+cdots+a_7a_6a_5a_4a_3a_2a_1
= a_1(1+a_2(1+a_3(1+cdots)))
$$



So we can conclude that $a_1 | 2879$, but that number is prime, so the only factors are $1$ and $2879$. $a_1=2879$ is obviously absurd, so $a_1 = 1$.



Putting that into the equation above we get $2879=1+a_2(1+cdots)$, or $2878=a_2(1+cdots)$. Factoring $2878$ (and ruling out the absurd possibility) we conclude $a_2=2$.



Try to go on from there.



Addition: If anyone tries to complete the calculations, the $a_i$'s are not the number of votes, and thus don't have to be different (hint: they won't be).






share|cite|improve this answer






















  • Ooops... let me check again. Thanks.
    – David G. Stork
    Nov 16 at 20:55














up vote
5
down vote



accepted










Call the number of votes the candidate that gotten the fewest votes for $a_1$.



The candidate that got the second lowest number of votes must then have gotten a multiplum of $a_1$, let's define $a_2$ so that $a_2a_1$.



We can continue and define $a_3, ldots, a_7$ similarly.



As $a_i$ for $igeq 2$ is the factor between the number of votes for candidate $i-1$ and candidate $i$, they can't be $1$.



If we add the number of votes we get
$$
2879 = a_1+a_2a_1+cdots+a_7a_6a_5a_4a_3a_2a_1
= a_1(1+a_2(1+a_3(1+cdots)))
$$



So we can conclude that $a_1 | 2879$, but that number is prime, so the only factors are $1$ and $2879$. $a_1=2879$ is obviously absurd, so $a_1 = 1$.



Putting that into the equation above we get $2879=1+a_2(1+cdots)$, or $2878=a_2(1+cdots)$. Factoring $2878$ (and ruling out the absurd possibility) we conclude $a_2=2$.



Try to go on from there.



Addition: If anyone tries to complete the calculations, the $a_i$'s are not the number of votes, and thus don't have to be different (hint: they won't be).






share|cite|improve this answer






















  • Ooops... let me check again. Thanks.
    – David G. Stork
    Nov 16 at 20:55












up vote
5
down vote



accepted







up vote
5
down vote



accepted






Call the number of votes the candidate that gotten the fewest votes for $a_1$.



The candidate that got the second lowest number of votes must then have gotten a multiplum of $a_1$, let's define $a_2$ so that $a_2a_1$.



We can continue and define $a_3, ldots, a_7$ similarly.



As $a_i$ for $igeq 2$ is the factor between the number of votes for candidate $i-1$ and candidate $i$, they can't be $1$.



If we add the number of votes we get
$$
2879 = a_1+a_2a_1+cdots+a_7a_6a_5a_4a_3a_2a_1
= a_1(1+a_2(1+a_3(1+cdots)))
$$



So we can conclude that $a_1 | 2879$, but that number is prime, so the only factors are $1$ and $2879$. $a_1=2879$ is obviously absurd, so $a_1 = 1$.



Putting that into the equation above we get $2879=1+a_2(1+cdots)$, or $2878=a_2(1+cdots)$. Factoring $2878$ (and ruling out the absurd possibility) we conclude $a_2=2$.



Try to go on from there.



Addition: If anyone tries to complete the calculations, the $a_i$'s are not the number of votes, and thus don't have to be different (hint: they won't be).






share|cite|improve this answer














Call the number of votes the candidate that gotten the fewest votes for $a_1$.



The candidate that got the second lowest number of votes must then have gotten a multiplum of $a_1$, let's define $a_2$ so that $a_2a_1$.



We can continue and define $a_3, ldots, a_7$ similarly.



As $a_i$ for $igeq 2$ is the factor between the number of votes for candidate $i-1$ and candidate $i$, they can't be $1$.



If we add the number of votes we get
$$
2879 = a_1+a_2a_1+cdots+a_7a_6a_5a_4a_3a_2a_1
= a_1(1+a_2(1+a_3(1+cdots)))
$$



So we can conclude that $a_1 | 2879$, but that number is prime, so the only factors are $1$ and $2879$. $a_1=2879$ is obviously absurd, so $a_1 = 1$.



Putting that into the equation above we get $2879=1+a_2(1+cdots)$, or $2878=a_2(1+cdots)$. Factoring $2878$ (and ruling out the absurd possibility) we conclude $a_2=2$.



Try to go on from there.



Addition: If anyone tries to complete the calculations, the $a_i$'s are not the number of votes, and thus don't have to be different (hint: they won't be).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 16 at 21:57

























answered Nov 16 at 19:35









Henrik

5,93282030




5,93282030











  • Ooops... let me check again. Thanks.
    – David G. Stork
    Nov 16 at 20:55
















  • Ooops... let me check again. Thanks.
    – David G. Stork
    Nov 16 at 20:55















Ooops... let me check again. Thanks.
– David G. Stork
Nov 16 at 20:55




Ooops... let me check again. Thanks.
– David G. Stork
Nov 16 at 20:55

















 

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