Repdigit Triangular Numbers
Theorem
The only repdigit numbers which are also triangular are:
- $55, 66, 666$
This sequence is A045914 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
Proof
\(\ds 55\) | \(=\) | \(\ds \dfrac {10 \times 11} 2\) | ||||||||||||
\(\ds 66\) | \(=\) | \(\ds \dfrac {11 \times 12} 2\) | ||||||||||||
\(\ds 666\) | \(=\) | \(\ds \dfrac {36 \times 37} 2\) |
Let $\dfrac {d \paren {10^j - 1}} 9$ be a $j$-digit repdigit number.
Suppose it is triangular.
For $d = 1$, write:
\(\ds \frac {k \paren {k + 1} } 2\) | \(=\) | \(\ds \frac {10^j - 1} 9\) | ||||||||||||
\(\ds 9 k \paren {k + 1}\) | \(=\) | \(\ds 2 \times 10^j - 2\) | ||||||||||||
\(\ds \paren {3 k + 1} \paren {3 k + 2}\) | \(=\) | \(\ds 2^{j + 1} \times 5^j\) |
Since $3 k + 1 \perp 3 k + 2$, one of them is equal to $2^{j + 1}$ while the other is equal to $5^j$.
This gives $2^{j + 1} \pm 1 = 5^j$.
Since $j \ge 1$, taking modulo $4$:
- $0 \pm 1 \equiv 1^j \equiv 1 \pmod 4$
so we have:
- $2^{j + 1} + 1 = 5^j$
By 1 plus Power of 2 is not Perfect Power except 9, the above equation has no solutions for $j > 1$.
For $j = 1$ we have equality.
Hence $1$ is the only triangular repunit.
$\Box$
By Odd Square is Eight Triangles Plus One:
- $x = 1 + \dfrac {8 d \paren {10^j - 1}} 9$ is a perfect square.
For $d = 2, 4, 7, 9$, $x$ ends in $2$ or $3$, which is not a perfect square by Square Modulo 5/Corollary.
$\Box$
For $d = 3, 8$, $x$ ends in $5$.
Since it is a perfect square, $x$ must end in $25$.
This is satisfied only when $d = 3, j = 1$.
For $d = 3, j > 1$ and $d = 8, j = 1$, $x$ ends in $65$.
For $d = 8, j > 1$, $x$ ends in $05$.
This corresponds to $\dfrac {x - 1} 8 = 3$.
$\Box$
For $d = 5$, $x$ is in the form $44 \dots 41$.
We show that $x$ is square only when $x = 441$.
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This corresponds to $\dfrac {x - 1} 8 = 55$.
$\Box$
For $d = 6$, $x$ is either:
- $49$ for $j = 1$
- $53 \dots 329$ for $j > 1$, with $j - 1$ $3$'s
we show that $x$ is square only when $x = 49, 529, 5329$.
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These correspond to $\dfrac {x - 1} 8 = 6$, $66$ and $666$.
$\Box$
We have hence demonstrated that the only triangular repdigit numbers are:
- $1, 3, 6, 55, 66, 666$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $55$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $55$