Money!

Logic Level 4

There are $108$ points on a circle with the names $A(0), A(1), \ldots , A(107)$ . For future reference, let points $A(n)$ and $A(n+108)$ refer to the same point for any integer $n$ .

Now Hugo enters the game with his big pockets filled with money. He places money with a positive integer value on each point, so that the sum of the value of the money placed on the points $A(n), A(n+1), A(n+2), \dots , A(n+18), A(n+19)$ equals $1000$ for every integer $n$ with $0 \le n \le 107$ .

Now by coincidence, the value of the money placed on the points $A(1), A(19), A(50)$ equals $1, 19, 50$ .

Can you compute now the value of the money Hugo placed on field $A(100)$ ? Type its value in the answer field.

The answer is 130.

2 solutions

Marilyn Bretscher
Sep 29, 2015

This is a very clever and entertaining problem, as one would expect from an ETH student.

Applying the given sum formula $...=1000$ to $n$ and $n+1$ , we can conclude that $A(n)=A(n+20)$ for all $n$ . Together with the given information that $A(n)=A(n+108)$ , if follows that $A(n)=A(n+4)$ . Now the sum formula simplifies to $A(1)+A(2)+A(3)+A(4)=200$ since we are going through the period 5 times. Finally $A(100)=A(4)=200-A(1)-A(50)-A(19)=\boxed{130}$

Lam Nguyen
Oct 26, 2015

On a side note, this, and a lot of other high level logic problems could be (and more likely) to be assigned as combinatorics instead.

Money!

The answer is 130.

2 solutions

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