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A class with $2N$ students took a quiz, on which the possible scores were $0,1,\dots,10$. Each of these scores occurred at least once, and the average score was exactly $7.4$. Show that the class can be divided into two groups of $N$ students in such a way that the average score for each group was exactly $7.4$

I do not know, I am Clueless

Initially I assumed that there are $a_0$ $0's$,$a_1$ $1's$ $ \dots $ $a_{10}$ $10's$. Where $a_0+ a_1 \dots a_{10}=N$ each $a_i>0$ Further we have Average 7.4 but I do not know what to do next

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closed as off-topic by Andrés E. Caicedo, Namaste, Isaac Browne, Shailesh, user223391 Jul 5 '18 at 1:03

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  • $\begingroup$ What have you done? How did you start? Where did you get stuck? $\endgroup$ – saulspatz Jul 2 '18 at 16:46
  • $\begingroup$ Initially I assumed that there are $a_0$ $0's$,$a_1$ $1's$ $ \dots $ $a_{10}$ $10's$. Where $a_0+ a_1 \dots a_{10}=N$ each $a_i>0$ Further we have Average 7.4 but I do not know what to do next $\endgroup$ – user567182 Jul 2 '18 at 16:50
  • $\begingroup$ What can you say about the number of students? Is it possible that there were 12 students, for example? Let me just add that the idea of trying to figure out how many students got each grade seems way too complicated, at least to begin with. Start with the easy stuff. $\endgroup$ – saulspatz Jul 2 '18 at 16:54
  • $\begingroup$ No, 12 is not a possibility. If it is then total score is supposed to be $ 7.4 \times 12=98.8$ which is not an integer, Our total score is an integer because its summation of $0,1,2 \dots 10$ $\endgroup$ – user567182 Jul 2 '18 at 16:57
  • $\begingroup$ I think this was a recent Putnam exam problem. Don't have the time to find the reference. $\endgroup$ – Ethan Bolker Jul 3 '18 at 20:50
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Let $a_1,\dots,a_{2N}$ be the scores in nondecreasing order, and define the sums $$s_i = \sum_{j=i+1}^{i+N} a_i \text{ for } i=0,\dots,N$$.

Then $$s_0 \leq \cdots \leq s_{N}$$

and

$$s_0 + s_{N} = \sum_{j=1}^{2N} a_i = 7.4(2N)$$,

so $$s_0 \leq 7.4N \leq s_N$$.

Let $i$ be the largest index for which $$s_i \leq 7.4N$$;

note that we cannot have $i = N$, as otherwise $s_0 = s_N = 7.4N$ and hence $a_1 = \cdots = a_{2N} = 7.4$, contradiction.

Then $$7.4N - s_i < s_{i+1} - s_i = a_{i+N+1} - a_i$$ and so $$ a_i < s_i + a_{i+N+1} - 7.4N \leq a_{i+N+1}; $$

since all possible scores occur, this means that we can find $N$ scores with sum $7.4N$ by taking $$a_i, \dots, a_{i+N+1}$$ and omitting one occurrence of the value $$s_i + a_{i+N+1} - 7.4N$$.

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  • $\begingroup$ You can put the punctuation marks inside the $ signs to avoid these weird lines with just a comma or a period. $\endgroup$ – saulspatz Jul 2 '18 at 17:08

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