Fridolin just can't get enough from jumping on the number line

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Fridolin just can't get enough from jumping on the number line

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Source: Bundeswettbewerb Mathematik 2025, Round 1 - Problem 1

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#1 h Mar 17, 2025, 12:20 PM

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Fridolin the frog jumps on the number line: He starts at $0$ , then jumps in some order on each of the numbers $1,2,\dots,9$ exactly once and finally returns with his last jump to $0$ . Can the total distance he travelled with these $10$ jumps be a) $20$ , b) $25$ ?

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#2 h Mar 26, 2025, 5:39 PM

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Let $a_1, a_2, \dots, a_9$ be the order in which the frog visited the nine points (from $1$ to $9$ ). Then the total distance traveled is:
$S = a_1 + \left|a_1 - a_2\right| + \left|a_2 - a_3\right| + \dots + \left|a_8 - a_9\right| + a_9.$ Since each $a_i$ appears twice in the expression for $S$ , we deduce that $S$ is even. Therefore, it is not possible for $S$ to be equal to $25$ , solving part $b)$ .

To show that it is possible for $S$ to be equal to $20$ , it is enough to present the following jump sequence:
$0 \to 2 \to 1 \to 9 \to 8 \to 7 \to 6 \to 5 \to 4 \to 3 \to 0.$

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#3 h Apr 6, 2025, 11:10 AM

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Let $\{a_1;a_2;...;a_9\}={1;2;...;9}$ . $a_1$ is this frog's first step, $a_2$ is its second step etc. Total distance is $A=|a_1-0|+|a_2-a_1|+|a_3-a_2|+...+|a_9-a_8|+|0-a_9|$
Since $|a|+|b|\equiv |a+b|~(mod~2)$ , we have $A\equiv a_1+|a_2-a_1+a_3-a_2+...+a_9-a_8|+a_9=a_1+a_9+|a_1-a_9|\equiv 0~(mod~2)$
So, $25$ is not possible

For $20$ , consider this $\{a_1;a_2;a_3...;a_9\}={9;8;7;6;5;3;4;2;1}$ $A=9+1+1+1+1+2+1+2+1+1=20$

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