1000 nonnegative numbers are placed in 1000 boxes.
The sum of every 4 consecutive numbers is 12.
Only first ten and five last are shown below, revealing only 3 numbers.
$0$$$$1$$$ .., .. $2$$$
It is not too difficult to deduct the last number.
This is your task .
(In reply to
Solution by tomarken)
By definition of consecutive numbers, the numbers must follow each other continuously and be increasing from smallest to largest. The four "consecutive" numbers 0,1,2,9 may fulfill the requirement if there were but four boxes. Yet it is given that there are a 1000 nonnegative numbers placed in 1000 boxes It is given that the second box contains the number 0. It is an easily deducible fact that the second box's number (0) can not be consecutive to the number in the first box. No number less than 0 is nonnegative. It therefore can be concluded that not every sequence of four boxes must contain consecutive numbers. It is possible in the 1000 boxes that there is no sequence of four consecutive numbers. It is but given that if there are four consecutive numbers, the total of those numbers sums to 12. The last box, like the first box is not required by the givens to be a part of a string of four consecutive numbers.
To deduce the entire string is just a sequence of four numbers repeated over and over is invalid. The first box can not be part of any consecutive sequence of four, but must be alone in a sequence of one, with the assumption that a set of one number can be considered to be a set of consecutive number(s). Thus, the first box may contain any nonnegative number. The sum of the next three numbers are thus not required to be (12n) By logic it should be deduced that the last box may also be any nonnegative number, thus making the assertion that the number can be deduced false.

Posted by Dej Mar
on 20210615 00:35:59 