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.

  • 4
    $\begingroup$ 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. $\endgroup$ – lulu Nov 16 '18 at 19:16
  • $\begingroup$ I don't understand very well the hint @lulu $\endgroup$ – Carlos Nov 16 '18 at 19:20
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    $\begingroup$ Let $v_i=n_iv_1$. Then $2879=v_1+v_2+\cdots+v_7=v_1+n_2v_1+\cdots +n_7v_1=v_1\times (1+n_1+n_2+\cdots +n_7)$ thus $v_1$ must divide $2879$. Now factor that number. $\endgroup$ – lulu Nov 16 '18 at 19:22
  • $\begingroup$ But how can I factor in a number I don't know? @lulu $\endgroup$ – Carlos Nov 16 '18 at 19:25
  • $\begingroup$ Just factor $2879$. Nobody is asking you to "factor in a number you don't know." $\endgroup$ – lulu Nov 16 '18 at 19:28

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 $i\geq 2$ 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).

  • $\begingroup$ Ooops... let me check again. Thanks. $\endgroup$ – David G. Stork Nov 16 '18 at 20:55

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