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First a part of the set of same balls was arranged into an equilateral triangle, 19 balls were not used, but when the sides of this triangle were needed to be one-uped, it was a 5 ball insufficiency. How many balls were in the set?

I feel like it's a contradiction in the problem.

Now I see that I misunderstood the problem. I thought it was the circuit of the triangle.

MelanieJohansson
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  • What exactly is the question? Also do you mean "19 balls were used" instead of "19 balls were not used"? – KBusc Oct 20 '14 at 13:25
  • I would interpret this as: With $N$ balls, I can make an equilateral triangle with 19 left over. But if I were to try to make the triangle one row deeper, I would be short by 5. – Semiclassical Oct 20 '14 at 13:30
  • Thank you for making an edit. this makes more sense now – KBusc Oct 20 '14 at 13:31

3 Answers3

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Hint:

  • this is about triangle numbers

  • the partial last row would have $19+5$ balls if complete, so the complete last row has one fewer

  • find the number of balls in the complete triangle and add those left over

Henry
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Since after the first triangle you had 19 balls left over and after the next one you were 5 balls short you are looking for 2 conescutive triangle numbers whos differnce is $24(19+5)$ using the sequence from the OEIS https://oeis.org/A000217 we can see that 276 and 300 are the triangle numbers that fit this criteria. So a triangle with 276 balls has 19 left over, that means that the set started with 295 balls. And when you try and make the next biggest triangle you will be 5 balls short

KBusc
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Suppose you have $N$ balls and the sides of your triangle have $n$ balls. Then the number of balls needed for the triangle is: $$n + (n-1) + (n-2) + \ldots + 3 + 2 + 1 = \textrm{number of balls in triangle}.$$ Now we have two unknowns, $N$ and $n$, which means we need two equations. For the first triangle 19 balls were not used, so: $$n + (n-1) + (n-2) + \ldots + 3 + 2 + 1 = N-19.$$ With the triangle sized up we are 5 balls short, so we have: $$(n+1) + \Big[n + (n-1) + (n-2) + \ldots + 3 + 2 + 1\Big] = N+5.$$ Now we can combine the latter two equations to find: $$(n+1) + (N-19) = N+5.$$ This equation is easy to solve and will give you $n=23$, which implies that $N=295$.

You can also derive this easier: increasing the size of the triangle can be achieved by laying down one extra row of balls on one side of the triangle. If you previously had 19 balls to spare and now you are 5 short, that means you were trying to lay down 24 extra balls. Hence, previously your triangle had sides of 23 balls. Then: $$N=19+\sum_{k=1}^{23}k = 295.$$

SPK.z
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