Some amount of billiard balls were arranged in an equilateral triangle. And 4 balls were extra. When the same set of billiard balls were arranged into triangle in which each side has one more ball than in the first arrangement there were 3 balls shortage. How many balls were at the set?

25 The number of balls to make an equilateral triangle with the sides of length n is expressed by the formula n (n+1) / 2 So to express the number of balls you have is b = n (n+1) / 2 + 4 With the larger arrangement you have b = (n+1) (n+2) / 2 - 3 Both of those qualities are equal to each other, so set an equation where they're equal. Then solve for n n (n+1) / 2 + 4 = (n+1) (n+2) / 2 - 3 Distribute the n term on the left (n^2 + n) / 2 + 4 = (n+1) (n+2) / 2 - 3 Distribute the / 2 on the left. 0.5n^2 + 0.5n + 4 = (n+1) (n+2) / 2 - 3 Multiply (n+1) (n+2) on right, then distribute the / 2 0.5n^2 + 0.5n + 4 = 0.5n^2 + 1.5n + 1 - 3 Subtract 0.5n^2 + 0.5n from both sides 4 = n + 1 - 3 Add 2 to each side 6 = n So the original triangle had sides with a length of 6, for a total number of balls of 6 (6+1) / 2 = 21 with 4 extra giving 25 balls. Let's check with the next larger triangle 7 (7+1) / 2 = 28 with 3 balls shortage. Which 25 balls would make happen. So the number of balls in the set is 25