The Tenacious Thai Calendar In the Thai calendar the year 2017 corresponds to 2560. The Thai calendar is always 543 years ahead of the Gregorian calendar.
Observant coders will note that 2560 is equal to \$2^9 \times 5\$, in other words it has 10 prime factors. This will not happen again for another 896 years! We call a year tenacious if it has exactly ten prime factors.
Write a program which outputs a truthy value if the current year using the Thai calendar, based on the system clock, is tenacious, and a falsey value otherwise.
Test cases:

If the program is run during 2017, true
If the program is run during any of the years 2018 to 2912, false
If the program is run during 2913, true (\$2913+543 = 2^7 \times 3^3\$)
Bash + coreutils, 35 bytes

factor $[`date +%Y`+543]|awk NF==11

Output is either a non-empty string (truthy) or an empty string (falsy).
Try it online!
Alternate version: 37 bytes.
date -d 543year +%Y|factor|awk NF==11

Not as golfy, but I like this one.
Try it online!
How it works
The arithmetic expansion
$[`date +%Y`+543]
executes date +%Y to get the current (full) year and adds 543 to the year.
Factor takes the sum as an argument and prints it prime factorization: first the number to be factored, then a list of individual prime factors.
Finally, awk filters the input, printing only lines with exactly 11 fields (the number plus 10 prime factors).