Bernardino identifies unaltered dollar words
Python, 39 38 bytes
lambda s:sum(ord(i)-96for i in s)==100
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-1 byte thanks to @JonathanAllan
05AB1E, 8 bytes
Code:
Ç96-O4bQ
Uses the CP-1252 encoding. Try it online!
Explanation:
Ç # Convert the string into a list of character codes
96- # Subtract 96 of each element
O # Take the sum
4b # Push 100 (4 in binary)
Q # Check if equal
Perl 6, 21 bytes
{100==[+] .ords X%32}
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Alternate:
{Ⅽ==[+] .ords X%32}
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Note that the Ⅽ
is ROMAN NUMERAL ONE HUNDRED
U+216D with a unival of … 100
Which takes 3 bytes to encode.
Expanded:
{ # bare block lambda with implicit parameter $_
100 # is 100
== # equal to
[+] # the sum of the following
.ords # the ordinals of the input (implicit method call on $_)
X[%] # crossed using the modulus operator
32 # with 32 to get 65..90 or 97..122 to become 1..26
}