'100' is a literal. When you take a reference to a literal, it's still a literal. Literals cannot be altered.

Think of it this way: You can take a reference to a number of "things". You can take a reference to an @array, to a %hash, to a $scalar, to a &sub, and even to a literal. If you modify the value of the target of your hash reference, you change the referent hash's value(s). If you modify the value of the target of a scalar ref, you modify the referent scalar container's value. But these are called variables for a reason; they hold variable data. Literals aren't, by their very definition, variable.

An example:

my $string = 'This';
my $number = 100;
my $scalar_ref = \$string; # Take a reference to $string.
$$scalar_ref = 'That'; # Modify the value of the scalar that
                       # $scalar_ref refers to.
print "$$scalar_ref = $string\n"; # That = That because we 
                                  # changed the value of the 
                                  # thing that $scalar_ref
                                  # points to, which happens to 
                                  # also be what $string holds.
$scalar_ref = \$number;  # Change what $scalar_ref points to
                         # doesn't change the referent values.
print "$$scalar_ref isn't $string, it's $number\n";
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You can modify where a reference points, and can modify the values held in the variable containers referred to by the reference all day long if you want. But if your reference doesn't point to a variable container, but instead points to a literal value, all you can do is modify where it points, not what the underlying value is.

$var = 100;
$var = 60;
print $var;
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