Question

$111{,}111{,}111$ 的平方的各位数字之和是多少？

What is the sum of the digits of the square of $111{,}111{,}111?$

$18$

$27$

$45$

$63$

$81$

这个各位都是一的九位数的平方是回文数 $\begin{aligned} &111{,}111{,}111^2 \\ &= 12{,}345{,}678{,}987{,}654{,}321. \end{aligned}$

它的各位数字依次为 $1, 2, \ldots, 9, 8, \ldots, 1$ ，所以数字和为 $\begin{aligned} &2(1 + 2 + \cdots + 8) + 9 \\ &= 2 \cdot 36 + 9 = 81. \end{aligned}$

所以正确答案是 E。

The square of the nine-digit repunit is the palindrome $\begin{aligned} &111{,}111{,}111^2 \\ &= 12{,}345{,}678{,}987{,}654{,}321. \end{aligned}$

Its digits are $1, 2, \ldots, 9, 8, \ldots, 1,$ so the sum is $\begin{aligned} &2(1 + 2 + \cdots + 8) + 9 \\ &= 2 \cdot 36 + 9 = 81. \end{aligned}$

Thus, the correct answer is E.