Find numbers of the form $\frac{n(n+1)}{2}$ such that the digits of the number are all same Find all $a \in \{1, 2, \ldots, 8, 9\}$ such that $\exists n \in \mathbb{N}$ and the digits of $\dfrac{n(n+1)}{2}$ are all $a$.


In other words, we have to find the values of $a$ such that $\dfrac{n(n+1)}{2} = a \cdot (1111\ldots1111)$ where $a$ is a Numerical Digit of Base $10$, i.e. $a \in \{1, 2, \ldots, 8, 9\}$.

I made the following progress:

Let's take $(1111\ldots1111) = R$ for convenience.
So we get the following equation: $n^2 + n - 2aR = 0$.
Taking the roots of $n$ by using the Quadratic Formula, we get $n = \dfrac{-1\pm\sqrt{1+8aR}}{2}$ .

As $n \in \mathbb{N}$, $-1\pm\sqrt{1+8aR}$ must be even, so $\sqrt{1+8aR}$ must be odd and hence $1+8aR$ must also be odd.
We also know that $1+8aR$ must be a perfect square.

So, our problem reduces down to $1+8aR = f^2$ where $f \in \mathbb{N}$.

I couldn't proceed ahead from here.



Series of natural numbers which has all same digits also talks about pretty much the same problem. However, no conclusive solution was reached in that thread. I want to do the above question with an analytical approach. Hence, I beleive that my question is not a duplicate.

[Source: As far as I know, this problem is from IMOTC India]
Let $T_n$ be the $n^{th}$ triangular number. 

Modulo $10$, $T_n$ is $0,1,3,5,6$ or $8$.

Clearly $8$ is not a triangle number. 

Modulo $100$, $T_n$ is never $88$. 

Therefore $8$ does not work, and the only solutions are
$1,3,5,6$ (as shown in the comments)
Let $T_n$ be the $n^{th}$ triangular number. 

Modulo $10$, $T_n$ is $0,1,3,5,6$ or $8$.

Clearly $8$ is not a triangle number. 

Modulo $100$, $T_n$ is never $88$. 

Therefore $8$ does not work, and the only solutions are
$1,3,5,6$ (as shown in the comments)