The nine cells of the triangle contain nine zeros. Try to change the numbers, with any number of moves, so that they become nine consecutive positive integers, the smallest of which is a. The arrangement of these nine numbers in the triangle is irrelevant. A move is the increase or decrease by 1 of the two numbers occupying two adjacent cells, where adjacent cells share a side. For which values of a is the problem solvable?
A pair of adjacent cells always consists of one orange and one black triangle. Initially, the sum r of the numbers on the orange cells is equal to the sum b of the numbers on the black cells, namely, 0. A move does not change this equilibrium r = b. At the end, the nine numbers of the triangle should be a , a + 1, …, and a + 8. Thus, because b is the sum of six of the numbers, b ≥ a + (a + 1) + (a + 2) + (a + 3) + (a + 4) + (a + 5) = 6a + 15, and because r is the sum of three of the numbers, r ≤ (a + 6) + (a + 7) + (a + 8) = 3a + 21. Because b = r, 6a + 15 ≤ 3a + 21, or a ≤ 2. Furthermore, r = b means that r + b = 2r, so r + b must be an even number. For a = 1, the sum of the nine numbers 1 + 2 + … + 9 = 45, which is odd. Therefore, if there is a solution to the problem at all, it is only for a = 2.
Instead of increasing the numbers on the same pair of cells by 1 n times, one can directly increase both numbers by n. This does not change the problem in any way. The order of the individual moves also plays no role. The image shows in three steps that there is indeed a solution for a = 2.
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This puzzle originally appeared in Spektrum der Wissenschaft and was reproduced with permission. It was translated from the original German version with the assistance of artificial intelligence and reviewed by our editors.
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