Guy has 17 cards. Each of them has an integer written on it (the numbers are not necessarily positive, and not necessarily different from each other). Guy noticed that for each card, the square of the number written on it equals the sum of the numbers on the 16 other cards. What are the numbers on Guy's cards? Find all of the options.
Problem
Source: Israel National Olympiad 2019 Q5
Tags: number theory
07.08.2019 23:45
08.08.2019 00:03
Mixer_V wrote:
08.08.2019 06:25
I did some simulation. For number of cards as odd integers, the solution is simple. either all 0 or $n-1$. For even numbers the solutions are more complicated
08.08.2019 07:32
After play around the code, I notice (not yet to prove) that besides all 0 and all 16, the other solutions (if exist) are of the form $(a,a,a,..b,b,b,b)$, where $a + b +1 =0$. So for 17 cards, we have two extra solutions: 5 pieces of $-6$ + 12 pieces $5$ 5 pieces of $-2$ + 12 pieces $1$
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08.08.2019 08:01
Assume $1 \leq i \leq 17$. Let $x_i$ be the integer written on the card $i$. Futhermore let $(1) \;\; S_k = x_1^k + x_2^k + ... + x_{17}^k, k \in \{1,2\}$. The given condition yields $x_i^2 = S_1 - x_i$, i.e. $(2) \;\; x_i^2 + x_i = S_1$. The fact that $x_i=a$ is a solution of equation (2) iff $x_i = -1 - a$ is a solution of equation (2) implies $x_i=x_1$ or $x_i=-1-x_1$. Let $n$ be the number of $i$´s for which $x_i = -1 - x_1$. According to equation (2) $\sum_{i=1}^{17} (x_i^2 + x_i) = \sum_{i=1}^{17} S_1$ $S_2 + S_1 = 17S_1$ $S_2 = 16S_1$ $(17 - n)x_1^2 + n(-1 - x_1)^2= 16[(17 - n)x_1 + n(-1 - x_1)]$ $n(2x_1 + 1) = x_1(16 - x_1)$, yielding $(3) \;\; n = \frac{1}{4} \Big( -2x_1 + 33 - \frac{33}{2x_1+1} \Big)$. Hence $2x_1 + 1 \mid 3 \cdot 11$, which give us $(x_1,n) = (-17,17), (-6,12), (-2,12), (-1,17), (0,0), (1,5), (5,5), (16,0)$. Now $n \neq 17$ since $n < 17$, which give us the following four solutions: $\bullet$ All 17 cards have the same number $m$, where $m \in \{0,16\}$. $\bullet$ 5 cards have the number -2 and 12 cards have the number 1. $\bullet$ 5 cards have the number -6 and 12 cards have the number 5.
08.08.2019 09:03
Solar Plexsus wrote: Assume $1 \leq i \leq 17$. Let $x_i$ be the integer written on the card $i$. Futhermore let $(1) \;\; S_k = x_1^k + x_2^k + ... + x_{17}^k, k \in \{1,2\}$. The given condition yields $x_i^2 = S_1 - x_i$, i.e. $(2) \;\; x_i^2 + x_i = S_1$. The fact that $x_i=a$ is a solution of equation (2) iff $x_i = -1 - a$ is a solution of equation (2) implies $x_i=x_1$ or $x_i=1-x_1$. Let $n$ be the number of $i$´s for which $x_i = 1 - x_1$. According to equation (2) $\sum_{i=1}^{17} (x_i^2 + x_i) = \sum_{i=1}^{17} S_1$ $S_2 + S_1 = 17S_1$ $S_2 = 16S_1$ $(17 - n)x_1^2 + n(1 - x_1)^2= 16[(17 - n)x_1 + n(1 - x_1)]$ which is equivalent to $(3) \;\; 15n(2x_1 - 1) = -17x_1(x_1 - 16)$. Assume $17 \nmid 2x_1 - 1$. Then $17|n$ by equation (3), which give us $n=0$ (since $n<17$). Hence $x_i=x_1$ for all $1 \leq i \leq 17$, where $x_1=0$ or $x_1=16$ by equation (3). Next assume $17 \mid 2x_1 - 1$. The fact that $GCD(x_1,2x_1-1)=1$ combined with equation (3) implies $\frac{34(x_1 - 16)}{2x_1 - 1} = 17 - \frac{17 \cdot 31}{2x_1 - 1} \in \mathbb{Z}$, which (since $17 \mid 2x_1-1$) means $(4) \;\; 2x_1 - 1 = -17t, t \in \{-31,-1,1,31\}$. Then by equation (4) $2x_1 - 1 \equiv -17 \cdot (\pm 1) = \mp 17 \equiv \mp 2 \pmod{5}$, yielding $(5) \;\; x_1 \equiv 2,4 \pmod{5}$. But $x_1 \equiv 0,1 \pmod{5}$ by equation (3), contradicting congruence (5). Conclusion: The integers written on all 17 cards are the same. The number is either 0 or 16. but 5 pieces of $-6$ + 12 pieces of $5$ 5 pieces of $-2$ + 12 pieces of $1$ are indeed two valid solutions. You can verify them by hand.
08.08.2019 09:39
08.08.2019 12:02
I have edited my solution. My first solution was wrong because I obtain $x_i = 1 - x_1$ instead of $x_i = -1 - x_1$ (which is correct)