XXIII OM - I - Problem 6

Determine for which digits $ a $ the decimal representation of a number $ \frac{n(n+1)}{2} $ ($ n\in \mathbb{N} $) consists entirely of the digit $ a $.
Of course, $ a $ cannot be zero. If the decimal representation of the number
$ \displaystyle t_n = \frac{1}{2} n (n + 1) $ consists of $ k $ ones, where $ k \geq 2 $, then $ 9 t_n = 10^k - 1 $. From this, after simple transformations, we obtain $ (3n + 1) \cdot (3n + 2) = 2 \cdot 10^k = 2^{k+1} \cdot 5^k $. Since the consecutive natural numbers $ 3n + 1 $ and $ 3n + 2 $ are relatively prime and $ 2^{k+1} \leq 2^{2k} = 4^k < 5^k $, it follows that $ 3n + 1 = 2^{k+1} $ and $ 3n + 2 = 5^k $. Subtracting these equations side by side, we get $ 1 = 5^k - 2^{k+1} > 4^k - 2^{k+1} = 2^k (2^k - 2) \geq 8 $, since $ k \geq 2 $. The obtained contradiction proves that $ a $ cannot be one.
Notice that $ 8 t_n + 1 = (2n + 1)^2 $. If the decimal representation of the number $ t_n $ ends with the digits $ 2 $, $ 4 $, $ 7 $, or $ 9 $, then the decimal representation of the number $ 8 t_n + 1 $ would end with the digits $ 7 $ or $ 3 $. However, the last digit of the square of any natural number can only be $ 0, 1, 4, 5, 6 $, or $ 9 $. Therefore, $ a \ne 2, 4, 7, 9 $.
If the decimal representation of the number $ t_n $ ends with the digits $ 33 $ or $ 88 $, then the decimal representation of the number $ 8 t_n + 1 $ would end with the digits $ 65 $ or $ 05 $. However, the decimal representation of the square of any natural number $ m $ does not end with the digits $ 65 $ or $ 05 $. Such a number $ m $ would be odd and divisible by $ 5 $, i.e., $ m = 10k + 5 $. Therefore, the number $ m^2 = 100k^2 + 100k + 25 $ would give a remainder of $ 25 $ when divided by $ 100 $, meaning the decimal representation of the number $ m^2 $ would end with the digits $ 25 $. Therefore, $ a \ne 3, 8 $.
However, $ a $ can be equal to $ 5 $ or $ 6 $. We have $ t_{10} = 55 $, $ t_{11} = 66 $, $ r_{36} = 666 $.
Note. It has been proven that among the numbers $ t_n $ for $ n > 3 $, only the numbers $ t_{10} $, $ t_{11} $, and $ t_{36} $ have a decimal representation composed of the same digits. See David W. Ballew and Ronald C. Weger in Notices Amer. Math. Soc. 19 (1972), p. A - 511, abstract 72 T - A 152.