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A puzzle by Paul Vaderlind:
Is it possible to arrange 25 whole numbers (not necessarily all different) so that the sum of any three successive terms is even but the sum of all 25 is odd?
I found this puzzle on the Futility Closet website.
The puzzle has been solved already, but one concrete example involves the Fibonacci numbers, usually defined to begin with
1, 1, 2, 3, 5, 8,...
with of course each term after the second being the sum of the previous two. Exactly every third Fibonacci number is even, so any sum of three consecutive Fibonacci numbers is the sum of two odds and one even.
The sum of the first $N$ Fibonacci numbers is equal to the $(N+2)$-th number minus 1, and the 27th Fibonacci number, with 27 a multiple of 3, happens to be even. Hence the sum of the first 25 Fibonacci numbers is (even minus 1 = odd). The actual value of this sum is 196,417.
There are only two possible solutions (disregarding meaningless number swaps).
110 110 110 110 110 110 110 110 1 or 101 101 ... 101 1
Explanation
We are just using 1s and zeros for simplicity. The first two numbers determine the entire rest of the sequence and it's always going to be periodic with 3. Each period has two 1s and one 0. That makes the possible starters 110, 101, or 011. You can pick the parity of the entire sequence by choosing the proper starting sequence. Since sum from 1-24 must is even, the 25th value must be a one, so the only possible starters are 110 and 101. These are the only two possible solutions (Disregarding swapping 1s and 0s for any other odd or even number).
What I mean by "meaningless number swaps"
For this puzzles it only matters if each individual number is even or odd. Nothing changes at all, if you swap any odd number with any other odd number or if you swap any even number with any other even number. That means you create a new sequence by simply swapping any 1 with with ...-7,-5,-3,-1,3,5,7,... Same for swapping 0 with an even number.
1 1 2 1 1 2
In this group, the sum of any 3 successive terms is even and the total is even.
Repeat this 4 times, you get 24 numbers. Still, the sum of any 3 successive terms is even and the total is even.
Add a 1 on the end. The sum of the last 3 numbers is still even. As the sum of the 24 first numbers is even, the total of all numbers is odd.
This is just one of numerous possibilities.
1,2,3,1,2,3 ,1,2,3 ,1,2,3 ,1,2,3 ,1,2,3 ,1,2,3 ,1,2,3,1
1011011011011011011011011and1101101101101101101101101.
Explanation and method of solution.
Without loss of generality, the only numbers in the sequence are 0 and 1 (you can add any even number to any element of a solution to get another equivalent solution). The requirement for the sum of each three consecutive numbers to be even means that the entire sequence is determined by the first two elements since the third element is fixed by the first two element, and then the fourth by the second and third and so on. That means, just considering the 3-sums being even requirement there are 4 possible solutions (grouping them in threes to make them easier to read):
000 000 000 000 000 000 000 000 0
011 011 011 011 011 011 011 011 0
101 101 101 101 101 101 101 101 1
110 110 110 110 110 110 110 110 1
The last two have odd sum.
<br> or with two spaces at the end of a line.
$\endgroup$
110 110 110 110 110 110 110 110 1
101 101 101 101 101 101 101 101 1
Explanation
The sum of N whole numbers being odd or even only depends on each of the numbers being odd or even. So we use 0 and 1, but:
any 0 can be replaced with any even number any 1 can be replace with any odd number.
Since we need any 3 successive numbers to be even, we need a repeating pattern of 3 numbers.
The possible values of a 3 digit-pattern using only 0 and 1 are:
000even
001odd
010odd
011even
100odd
101even
110even
111odd
That leaves us with000,011,101and110:
000 000-> the sum of any 3 consecutive digits (always000) is even.
011 011-> the 3 consecutive digits can be either011,110or101, all 3 sums are even.
101 101-> it's the same 3 groups just offset by 1
110 110-> ditto
Another way of getting this list is considering that to get an even sum of 3 digits, we need either:
Only even digits (0 -> 000)Two odd digits (1, the sum of which will be even) and one even digit (0) -> 011,101or110.
Now, by repeating 8 times either of these 4 patterns, we get a series of 24 numbers where the sum of any 3 consecutive numbers is even. Clearly the sum of all 24 numbers is even as well.
We still need to add one digit, and for the sum of all numbers to be odd. This condition means the last digit must be 1.
Since we need to continue repeating our pattern, it means the pattern must start with 1, which only leaves us101and110.
Working:
Use only the numbers 0 and 1. Any multiple of 2 added to any number has no effect on the even-ness or odd-ness in any of the sums it is involved in.
If you specify the first two numbers in a 0-1 sequence, the sum of the first three numbers uniquely determines the next member, and so on.
If you start with 0,0 the rest of the members are 0, so the overall sum cannot be odd.
If you start with 1,1.. it will continue ..0 1,1,0,1,1. It has a repeating element of 1,1,0. The sum of this is element is even. The series will have eight repeats of this element, so the last number must be 1 for the overall sum to be odd.
There are two solutions composed of 0 and 1 as a 25 number sequence from a repeating copy of 1,1,0 ending in 1.
Solution:
1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1 or 1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1
1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 is one solution. The sum of any three consecutive terms is always 2, and the sum of the series is 17. The zeroes and ones can represent any even or odd numbers respectively, and so there are infinitely many solutions to the problem.
