Triangle party

Let the consecutive integers be a, (a+1) and (a+2).

Therefore we have,

n(n+1)/2 = a(a+1)(a+2)

upon simplification.. it leads to,

n^2 + n - 2a(a+1)(a+2) = 0

The solutions to this equation is given by,

n=[-1 + {1 + 8a(a+1)(a+2)}^1/2] / 2 and n= [-1 - {1 + 8a(a+1)(a+2)}^1/2] / 2

Clearly, the latter one gives a negative value and so we ignore it.

Now, since n is an integer, the discriminant must be a perfect square.

So we must have,

8a(a+1)(a+2) + 1 = x^2

But we know that " product of 4 consecutive numbers exceeded by 1 is a perfect square. thus 8a(a+1)(a+2) + 1 will be a perfect square if 8a(a+1)(a+2) is the product of 4 consecutive numbers.

Since we are looking for the minimum n, we must also look for the minimum a, so in this case,8 must form the fourth consecutive number.

Thus we take a = 4. plugging it in the equation and simplification leads us to our desired solution.

Listing the first few products of 3 consecutive integers greater than 1:

• $2 \times 3 \times 4 = 24$
• $3 \times 4 \times 5 = 60$
• $4 \times 5 \times 6 = 120$

Listing the first few triangle numbers:

• $T_{1} = 1$
• $T_{2} = 3$
• $T_{3} = 6$
• $T_{4} = 10$
• $T_{5} = 15$
• $T_{6} = 21$
• $T_{7} = 28$
• $T_{8} = 36$
• $T_{9} = 45$
• $T_{10} = 55$
• $T_{11} = 66$
• $T_{12} = 78$
• $T_{13} = 91$
• $T_{14} = 105$
• $T_{15} = 120$

Right away we find a $T_{n}$ that is the product of 3 consecutive integers greater than 1:

$T_{15} = 4 \times 5 \times 6 = 120$

Therefore, n = 15.

We list out the first few products of $3$ consecutive integers greater than $1$ :

• $2 \times 3 \times 4 = 24,$
• $3 \times 4 \times 5 = 60,$
• $4 \times 5 \times 6 = 120,$
• $5 \times 6 \times 7 = 210,\ldots$

The numbers $24$ and $60$ are not triangular numbers (which you can verify by finding the first few triangular numbers), but $120$ is a triangular number, since we have $120 = \dfrac{15(16)}{2}.$

Therefore, the smallest possible $n$ is $n = \boxed{15}.$

We can do this question via trial and error as it is asking for the smallest possible integer. First of all we could try 2x3x4=24, not a prime. Continue until you try 4x5x6=120 which is the 15th triangle number.

i couldn't think of any other method but trial and error. however, we can have restrictions to make our trials easier. knowing that product of 3 successive integers is an even number, for this problem, n(n+1) will be divisible by 4. so, we can try with all values of n(n+1) in which, either n or n+1 are multiples of 4. moreover, for n>4 we can continue eliminating all options which have either n or n+1 as a prime because we need the number to be a product of three consecutive numbers. by doing this, we can end up at the number 15 as the answer.

First let's start plugging in values of x for Tx. The series goes 1, 3, 6, 10, 15, etc., with each term being increased by the next highest integer (in other words, 1 + 2 + 3 + 4 + 5...). We then start listing numbers that can be written as the product of 3 consecutive numbers: 24, 60, 120, etc. Continuing the list of Tx, we see that 120 is the answer.