Go ahead: if you use a string like "10" in a place where PHP expects an integer, it will convert it into an integer "54"+"10" equals 64, not "5410" - that's "54"."10".
There's a section in the manual on this called "Type Juggling".
If you want to be certain that the value of $var is treated as an integer, you can insist on it by referring to it as (integer)$var. Often this is more for documentary purposes than any requirement by the language.