/proc/meminfo reports memory in KiloBytes and so needs a multiplier of
1024 instead of 1000.
The kernel reports in terms of pages and the proc filesystem is left
shifting by 2 for 4KB pages to get KB. Since this is a binary shift,
Bytes will need to shift by 10 and so get multiplied by 1024.
From the kernel code. PAGE_SHIFT = 12 for 4KB pages
"MemTotal: %8lu kB\n", K(i.totalram)
Thanks to @subhachandrachandra!
Cameron Sparr