How to find and show the content of the file with the highest name value?

Question

linux

How to find and show the contents of the file with the highest value name ?
Condition :
The directory /opt/1/ contains files with names like

1436811088_1
1436811088_15
1436813877_7

...
There are 100,000 such files. The files are named in ascending order of date. Thus, when comparing the names of two files in a string, the name of the file created later will have a greater value.
Specify the bash command, which can display the contents of the last file in the /opt/1/ directory.
I sorted and found the largest using
ls -1v | sort | tail -n 1
but somehow I don’t understand how to do it in 1 command in such a way that it would display the contents of the resulting output file.

Answer 1

cat `ls -1v | sort | tail -n 1`
right?

Answer 2

ls -1c | head

-c with -lt: sort by, and show, ctime (time of last modification of file status information); with -l: show ctime and sort by name; otherwise: sort by ctime, newest first
--sort=WORD
sort by WORD instead of name: none (-U), size (-S), time (-t), version (-v), extension (-X)