Bernardino identifies unaltered dollar words

Question

Definition

A dollar word is a word where when each of its letters is given a cent value, from a = 1 to z = 26, and the letters are summed, the result is 100. Here is an example on CodeReview, and here is a list of dollar words I found online.

Input

Input will be alphabetical from a-z, in your one language's text datatypes (arrays are allowed). You do not need to account for any other input - there will be no spaces, apostrophes, or hyphens. You can take as lowercase, uppercase, or a combination. Trailing newlines are allowed.

Output

Output a truthy value if the input is a dollar word, and a falsey value if it is not.

Test Cases

Truthy:

buzzy
boycott
identifies
adiabatically
ttttt

Falsey:

zzz
zzzzzzz
abcdefghiljjjzz
tttt
basic

This is code-golf, so the shortest answer in bytes wins! Standard loopholes and rules apply. Tie goes to first poster.

Answer 1

GS2, 6 bytes

▲1Θd←q

Input must be in uppercase.

Try it online!

How it works

  Θ       Combine the previous two tokens into a block and map it over the input.
▲             Push 64.
 1            Subtract 64 from the character on the stack.
   d      Take the sum of the resulting character array.
    ←     Push 100.
     q    Compare the two items on the stack for equality.

Answer 2

Python, 39 38 bytes

lambda s:sum(ord(i)-96for i in s)==100

Try it online!

---

-1 byte thanks to @JonathanAllan

Answer 3

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

Answer 4

Perl 6, 21 bytes

{100==[+] .ords X%32}

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Alternate:

{Ⅽ==[+] .ords X%32}

Try it

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
}

Answer 5

MATL, 8 bytes

96-s100=

Uses lowercase input.

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Explanation

The code is as readable as it gets:

96-   % Implicitly input string. Subtract 96 from each char interpreted as ASCII code
s     % Sum of array
100=  % Does it equal 100? Implicitly display

Answer 6

JavaScript (ES6), 46 bytes

Returns 0 or 1.

let f =

w=>[...w].map(c=>p-=parseInt(c,36)-9,p=100)|!p

console.log(f('buzzy'))
console.log(f('qwerty'))

Answer 7

C, 45 43 bytes

Thanks to @Neil for saving two bytes and making the solution case-insensitive!

n;f(char*s){for(n=100;*s;)n-=*s++&31;n=!n;}

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Answer 8

Haskell, 32 bytes

f s=sum[1|c<-s,_<-['a'..c]]==100

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The idea is to make a list of characters from a to the given character for each character in the list, and check that the total length is 100.

Other attempts:

f s=sum[1|c<-s,_<-['a'..c]]==100

f s=sum[fromEnum c-96|c<-s]==100
f s=100==length((\c->['a'..c])=<<s)
(==100).length.(>>= \c->['a'..c])
(==100).length.(=<<)(\c->['a'..c])
(==100).length.(enumFromTo 'a'=<<)
f s=100==length(do c<-s;['a'..c])

Too bad enumFromTo is so long.

Answer 9

Jelly, 9 7?* 8 bytes

ɠO%32S⁼³

Full program, outputting 1 if the input is a dollar word, or 0 if not.

Try it online!

How?

ɠO%32S⁼³ - Main link
ɠ        - read a line of input from STDIN
 O       - cast to ordinals
  %32    - mod 32 (vectorises) (-3*32=96 from lowercase; -2*32=64 from uppercase)
     S   - sum
       ³ - literal: 100
      ⁼  - equal?

---

* Could it be 7 bytes?

The only reason this took input with ɠ was to keep ³ as the literal 100 rather than the 3^rd command line input (1^st program input).

A way to avoid that would be, as pointed out by Dennis, to create 100 using the raw literal form ȷ2 which is 10². This leads to another 8 byte O%32S=ȷ2, but this is now an unnamed monadic function (as well as operating as a full program with a 3^rd argument).

Since, in golf, one may create variables or helper functions which restrict the program in which they may reside (one cannot reuse the name in-scope without stopping the function from being reusable), maybe restricting the program to only taking input from STDIN may also be acceptable, in which case the 7 byte O%32S=³ would be acceptable here as an unnamed function.

Answer 10

Haskell, 32 bytes

f w=sum[fromEnum c-96|c<-w]==100

This works for lowercase input. For uppercase, s/96/64/. Mixed-case support would add a bunch of bytes.

Answer 11

Mathematica, 23 bytes

100==Tr@LetterNumber@#&

Pure function taking a string (or an array of letters) as input, case-insensitive, and returning True or False. Here Tr just adds the letter-numbers together; everything else is self-explanatory.

Answer 12

R, 55 54 bytes

function(x)sum(match(el(strsplit(x,"")),letters))==100

-1 byte thanks to BLT

• returns a function that does the required computation, which returns TRUE and FALSE as one would expect.

• takes input in as lowercase; would only be a switch from letters to LETTERS for all uppercase

Answer 13

Java 8, 36 bytes

s->s.chars().map(c->c%32).sum()==100

Try it online!

Note: case independent.

Answer 14

Alice, 23 bytes

/o!
\i@/e)q&w[?'`-+k3-n

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Input should be lower case. Prints 1 for dollar words and 0 otherwise.

Explanation

Time to show off Alice's tape and some advanced control flow. Despite being fairly good at working with integers and strings individually, Alice has no built-ins to a) determine a string's length, b) convert between characters and their code points. The reason for this is that all of Alice's commands either map integers to integers or strings to strings. But both of those would require mapping strings to integers or vice versa, so they don't fit into either of Alice's modes.

However, in addition to it's stack, Alice also has a tape and Cardinal and Ordinal mode interpret the data on the tape in different ways

• In Cardinal mode, it's a regular tape familiar from other languages like Brainfuck. You can store one integer in each cell and you can move a tape head around. The tape is infinitely long and initially holds a -1 in every cell. The cells are also indexed and the tape head starts at index 0.
• Ordinal mode has its own tape head (also starting at index 0) and it interprets the tape as a list of strings. Strings are terminated by non-character cells (i.e. any values which are not a valid Unicode code point), in particular -1. So for Ordinal mode, the tape is initially filled with empty strings.

This tape can be used for both of the above operations: to get a string length, we write it to the tape in Ordinal mode, seek the terminating -1 in Cardinal mode and retrieve the position of the tape head. To convert characters to their code points, we simply read them off the tape in Cardinal mode.

The other two important features used in this solution are the return stack and an iterator. Alice has a return stack which is usually filled when using the jump command j, and which you can pop an address from to jump back with k. However, it's also possible to push the current address to the return stack without jumping anywhere with w. If we combine w with the repeat command &, we can push the current address to the return stack n times. Now each time we reach k, one copy is popped off the return stack and we perform another iteration from w (starting at the cell after it, because the IP moves before executing another command). When the return stack becomes empty, k does nothing at all and the IP simply passes through. Hence &w...k pops an integer n and then executes ... n+1 times, which gives us a very concise way to express a simple for loop.

On to the code itself...

/     Reflect to SE. Switch to Ordinal.
i     Read the input word as a string.
      Bounce off bottom boundary, move NE.
!     Store the input word on the tape.
      Bounce off top boundary, move SE.
/     Reflect to E. Switch to Cardinal.
e     Push -1.
)     Seek right on the tape for a -1, which finds the -1 terminating
      the input word.
q     Push the tape head's position, which gives us the string length N.
&w    Repeat this loop n+1 times (see above for an explanation)...
  [     Move the tape head left by one cell.
  ?     Retrieve the code point of the character in that cell.
  '`    Push 96.
  -     Subtract it from the code point to convert the letters to 1...26.
  +     Add the result to a running total. This total is initialised to 
        zero, because in Cardinal mode, the stack is implicitly filled with
        an infinite amount of zeros at the bottom.
k    End of loop.
     Note that the above loop ran once more than we have characters in the
     string. This is actually really convenient, because it means that we've
     added a "-1 character" to the running total. After subtracting 96 to
     convert it to its "letter value" this gives 97. So dollar words will
     actually result in 100 - 97 = 3, which we can check against for one
     byte less than for equality with 100.
3-   Subtract 3 to give 0 for dollar words.
n    Logical NOT. Turns 0 (dollar words) into 1 and everything else into 0.
     The IP wraps around to the beginning of the first line.
\    Reflect to NE. Switch to Ordinal.
o    Implicitly convert the result to a string and print it.
     Bounce off top boundary, move SE.
@    Terminate the program.

Answer 15

Ruby, 25 bytes

->s{s.sum-s.size*64==100}

Works for uppercase.

I see a couple of more complex Ruby entries, but it really is this simple. s.sum adds the ASCII codes of the input string, and from this we subtract 64 times the length of the string.

Example of use

f=->s{s.sum-s.size*64==100}

puts f["ADIABATICALLY"]
puts f["ZZZ"]

Answer 16

Perl 5, 30 bytes

-1 byte thanks to @Neil (31& instead of -96+).

29 bytes of code + -p flag.

$@+=31&ord for/./g;$_=$@==100

Try it online!

Answer 17

05AB1E, 9 bytes

5bIvAyk>-

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Explanation

As 1 is the only truthy value in 05AB1E we can save a byte using subtraction over comparing to 100.

5b         # convert 5 to binary (results in 101)
  Iv       # for each letter in input word
    Ayk    # get the index of the letter in the alphabet
       >   # increment
        -  # subtract from total

Answer 18

PowerShell, 36 30 bytes

$args|%{$s+=$_[0]-96};$s-eq100

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Inputs as an array, but I'm wondering if there is a better way to handle characters.

EDIT Missed an easy space but @AdmBorkBork kindly let me know :P also, there was in fact a better way to handle the characters!

Answer 19

Dyalog APL, 17 15 bytes

100=+/17-⍨⎕AV⍳⍞

Uses the Dyalog Classic character set.

              ⍞  ⍝ string input
          ⎕AV⍳   ⍝ index in the character map
      17-⍨       ⍝ subtract 17 from each ('a' = 18)
    +/           ⍝ sum
100=             ⍝ equal to 100?

Answer 20

Alice, 28 18 bytes

Thanks to @MartinEnder for golfing 10 bytes

=I.!'`-+?hn
>3-nO@

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This submission uses a different method than @MartinEnder's answer.

This submission outputs 0x00 for falsy and 0x01 for truthy.

So here is a version that outputs 0 or 1 instead: Try it!

Explanation

The explanation below is for the "visible" version. Both are very similar, except in the first program, the last o doesn't convert the 0 or 1 into a string (because we are in cardinal mode), but instead takes the number and output the character at that code point.

=                 Does nothing, but will be useful later on
I                 Read a character and push its code point onto the stack
                  If there is no more input, -1 is pushed instead
.                 Duplicate it
!                 Store it on the tape
#                 Skip the next command
o                 Gets skipped
'`                Push 96
-                 Subtract it from the character
+                 And add it to the total
?                 Load the number on the tape
h                 Increment it
n                 And negate it
                  For all characters that are read, ?hn results in 0,
                  but if -1 is pushed, then the result becomes 1

After this the IP wraps around to the left edge at the =. If the top value of the stack is 0, the IP continues on with its path, increasing the total sum of all the characters, once it is done with the input (the top of the stack will be 1), then the IP turns right (90 degrees clockwise).

One thing is important to note, the loop on the first line will iterate once after the input has ended. This will subtract 97 (96 from the '` and -1 from the lack of input) from the total.

>                Set the direction of the IP to East
3-               Subtract 3 from it (yields 0 if sum is 100, something else otherwise)
n                Negate it; Zero becomes 1, non-zero numbers become 0
/                Mirror; the IP gets redirected South-East
                 The IP reflects off the bottom and goes North-East
                 Now the program is in Ordinal mode, where numbers are automatically converted into strings when being used
o                Output the top of the stack as a string
                 IP reflects off the top and heads South-East
@                End the program

Answer 21

Taxi, 1259 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to Auctioneer School.Go to Auctioneer School:s 1 r 1 l 1 l.Pickup a passenger going to Chop Suey.0 is waiting at Starchild Numerology.Go to Starchild Numerology:s 1 l.Pickup a passenger going to Addition Alley.Go to Chop Suey:e 1 l 2 r 3 r 3 r.[a]Switch to plan "b" if no one is waiting.Pickup a passenger going to Charboil Grill.Go to Charboil Grill:n 1 l 3 l 3 l.Pickup a passenger going to What's The Difference.Go to Go More:e.64 is waiting at Starchild Numerology.Go to Starchild Numerology:e 2 r.Pickup a passenger going to What's The Difference.Go to What's The Difference:e 1 l 2 r 1 l.Pickup a passenger going to Addition Alley.Go to Addition Alley:e 2 r.Pickup a passenger going to Addition Alley.Go to Chop Suey:n 1 r 2 r.Switch to plan "a".[b]Go to Addition Alley:n 1 l 2 l.Pickup a passenger going to Equal's Corner.100 is waiting at Starchild Numerology.Go to Starchild Numerology:n 1 l 1 l 3 l 2 r.Pickup a passenger going to Equal's Corner.Go to Equal's Corner:w 1 l.Switch to plan "c" if no one is waiting."TRUE" is waiting at Writer's Depot.[c]"FALSE" is waiting at Writer's Depot.Go to Writer's Depot:n 1 l 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 r 2 r 1 l.

With line breaks, it looks like this:

Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to Auctioneer School.
Go to Auctioneer School:s 1 r 1 l 1 l.
Pickup a passenger going to Chop Suey.
0 is waiting at Starchild Numerology.
Go to Starchild Numerology:s 1 l.
Pickup a passenger going to Addition Alley.
Go to Chop Suey:e 1 l 2 r 3 r 3 r.
[a]
Switch to plan "b" if no one is waiting.
Pickup a passenger going to Charboil Grill.
Go to Charboil Grill:n 1 l 3 l 3 l.
Pickup a passenger going to What's The Difference.
Go to Go More:e.
64 is waiting at Starchild Numerology.
Go to Starchild Numerology:e 2 r.
Pickup a passenger going to What's The Difference.
Go to What's The Difference:e 1 l 2 r 1 l.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:e 2 r.
Pickup a passenger going to Addition Alley.
Go to Chop Suey:n 1 r 2 r.
Switch to plan "a".
[b]
Go to Addition Alley:n 1 l 2 l.
Pickup a passenger going to Equal's Corner.
100 is waiting at Starchild Numerology.
Go to Starchild Numerology:n 1 l 1 l 3 l 2 r.
Pickup a passenger going to Equal's Corner.
Go to Equal's Corner:w 1 l.
Switch to plan "c" if no one is waiting.
TRUE is waiting at Writer's Depot.
[c]
FALSE is waiting at Writer's Depot.
Go to Writer's Depot:n 1 l 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 r 2 r 1 l.

It accepts upper or lowercase because the Auctioneer School converts it all to uppercase.
Chop Suey breaks it into individual characters.
Charboil Grill converts characters to their ASCII code.
We pickup one character a time, convert it to ASCII, subtract 65, and add it to the running total.
Once there aren't any more characters, compare the total to 100.

Returns TRUE for dollar words and FALSE for everything else.

Answer 22

IA-32 machine code, 21 bytes

Hexdump:

33 c0 6a 64 5a 8a 01 41 24 1f 75 05 83 ea 01 d6
c3 2b d0 eb f0

Assembly code:

    xor eax, eax;   initialize eax to 0
    push 100;       initialize edx
    pop edx;            to 100
myloop:
    mov al, [ecx];  load a byte
    inc ecx;        go to next byte
    and al, 31;     convert from letter to number
    jnz cont;       not done? continue

    ;               done:
    sub edx, 1;     check whether edx got to 0; result is in CF
    __emit(0xd6);   aka SALC - set al to CF
    ret
cont:
    sub edx, eax
    jmp myloop

Counts from 100 to 0. If arrived to 0, returns true (0xff); otherwise false (0x00).

Answer 23

Python, 38 bytes

lambda s:sum(map(ord,s))==4-96*~len(s)

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Same length as ovs's solution. Rather than subtracting 96 from each ord value, this checks if the ord total equals 100+96*len(s). This is expressed one byte shorter as 4-96*~len(s), which equals 4-96*(-len(s)-1).

Answer 24

Japt, 13 12 10 bytes

L¥U¬x_c %H

Explanation:

L¥ U¬x _c %H
L¥(U¬x(_c %H))
L¥(          )   // 100==
   U¬            //   Input split into a char array
     x(     )    //   The sum of:
       _         //     At each char:
        c        //       Get the char-code and
          %H     //       Mod 32

Test it online!

12-bytes:

L¥U¬mc m%H x

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Another 12-byte solution using a different technique

L¥U¬x@;CaX Ä

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Answer 25

Ruby, 35 30 bytes

->w{196==eval(w.bytes*'-96+')}

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Answer 26

Retina, 47 23 bytes

\w
!$&
}T`l`_l
^!{100}$

Try it online! Note: Header lowercases input and splits it into words; results appear on separate lines. Edit: Saved far too many bytes thanks to @MartinEnder.

Answer 27

Octave, 18 bytes

@(x)sum(x-96)==100

Subtracts 96 from the input string x (lower case), to get the numeric values of the letters. Takes the sum and compares it to 100. Returns a logical 1 for truthy cases, and a logical 0 for false cases.

I could save one byte if it was OK to give false for "dollar words" and true for "non-dollar words".

Answer 28

///, 564 210 189 185 bytes

/~/1\/\///4/11~3/41//6/33//2/66//5/22~a/~b/1~c/4~d/3~e/31~f/34~g/6~h/16~i/46~j/36~k/316~l/346~m/2~n/12~o/42~p/32~q/312~r/342~s/62~t/162~u/462~v/362~w/3162~x/3462~y/22~z/122~5555/0///1/0

Try it online!

Prints a 1 if it is a "dollar word", otherwise prints a "0"

Input is the following: (Scroll all the way to the right)

/~/1\/\///4/11~3/41//6/33//2/66//5/22~a/~b/1~c/4~d/3~e/31~f/34~g/6~h/16~i/46~j/36~k/316~l/346~m/2~n/12~o/42~p/32~q/312~r/342~s/62~t/162~u/462~v/362~w/3162~x/3462~y/22~z/122~5555/0//INPUT WORD HERE/1/0

Works by replacing each letter with its value in unary, then replacing a unary 100 with a 0. It then substitutes whatever the word's value is with a 1. If the word's value is 0, then it will print a 1 because at the end of the code, it is replacing a 0. If the word's value is anything else, it will just print that 0.

The golf works by using common occurrences in the code as replacements.

Answer 29

Ruby (2.4+), 38 bytes

Takes input in lowercase. Requires Ruby 2.4's Array#sum so it won't run on TIO.

->a{a.chars.map{|c|c.ord-96}.sum==100}

Answer 30

Aceto, 49 bytes

TL;DR: 3rd-order Hilbert curve. Uses character input, looping until a newline is given. Sums up the values (96 (5*3+9*9) minus them), and in the end adds 100 (5*5*4) and checks whether it's zero. Uses the fairly new catch marks.

}){(
(o-+
d{))
|(&
=,d@*45(
{d""*+5(
3*)+=0
599*p

Outputs True if the given word is a dollar word, False otherwise.

Explanation:

We first construct a 96 by calculating 5*3+9*9: 53*99*+. Next, we move a stack to the right and push an empty string: )"". Because we're about to enter a repeating part, we set the catching mark using @. This is where we will return if an error occurs.

We duplicate the empty string, read a single character (,), duplicate that character, and move it to the original stack ({). Still on the right stack, we compare the top two values (the empty string and our character) and use the conditional mirror to jump to the right edge if they are equal: =|.

Assuming they're not, we now move on the left stack, push the character once more one stack to the left, duplicate the 96, go to the left stack, push the character back on the middle stack and go there again: ({d(}).

Now we convert the character to the value of its unicode codepoint and subtract it from 96, resulting in it's negated dollar value: o-. We do it this way, because we'd have to swap the values otherwise. We push the value to the left stack, go there, and add the two values there (initially 0 and our negative character value): {(+. To prepare for reading another character, we go to the middle stack again ), and then raise an error: &.

Once a newline is read and the character is therefore empty and the equality truthful, we got mirrored somewhere on the top-left quadrant; the exact position doesn't matter in our case. The next operation that does something is moving back on the stack of our final value: ((. On it, we should now have the sum of all character values, negative. That means, if it's a dollar word, it should be -100. To test whether it is, we add 100 (554**+), push 0, test for equality and print the result: =p.

Answer 31

Bash + GNU utils, 47

od -An -td1 -vw1|sed 's/^/a+=-96+/;$a!a-100'|bc

Try it online.

Answer 32

R, 36 39 bytes

Using a different method than the other R answer

function(x)sum(utf8ToInt(x)-96)==100

This is an unnamed function for lowercase characters. utf8ToInt(x) converts the string into a vector of ascii values. 96 is taken away from all items and sum checked against 100.

In use

> f=
+ function(x)sum(utf8ToInt(x)-96)==100
> f('buzzy')
[1] TRUE
> f('boycott')
[1] TRUE
> f('identifies')
[1] TRUE
> f('adiabatically')
[1] TRUE
> f('ttttt')
[1] TRUE
> f('zzz')
[1] FALSE
> f('zzzzzzz')
[1] FALSE
> f('abcdefghiljjjzz')
[1] FALSE
> f('tttt')
[1] FALSE
> f('basic')
[1] FALSE

Answer 33

CJam, 9 bytes

q'`f-:+E=

Input should be in lowercase with a trailing newline. This is a tad cheaty as there's no reason to append a newline to the input, but it follows the rules as written.

Try it online!

How it works

q         e# Read all input from STDIN and push the resulting string on the stack.
 '`       e# Push the backtick character.
   f-     e# Subtract the backtick character from each input character. The
          e# difference of two characters is the difference of their code points,
          e# so this maps a..z to 1..26 and the newline to -86.
     :+   e# Take the sum.
       E  e# Push 14.
        = e# Test the sum and 14 for equality.

Answer 34

C++ (gcc), 45 bytes

Unnamed generic lambda, accepting char[], std::string or any other container of char and returning via reference parameter.

• 0 means dollar word
• everything else is no dollar word

Even works for capital letters ;>

[](auto&s,int&r){r=100;for(auto x:s)r-=x&31;}

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Answer 35

PHP, 45 Bytes

combination 3 Bytes saved by @user63956

input as argument list

for(;$s+=ord($argv[++$i])%32?:die($s==100););

input as string using the -R option

for(;$s+=ord($argn[$i++])%32?:die($s==100););

PHP, 48 Bytes

combination

for(;$c=$argv[++$i];)$s+=ord($c)%32;echo$s==100;

PHP, 50 Bytes

Running PHP from the commandline without a file $argv[0] is "-"

prints 1 for true and nothing for false

lowercase

echo array_sum(array_map(ord,$argv))-96*$argc==49;

uppercase

echo array_sum(array_map(ord,$argv))-64*$argc==81;

Answer 36

Java 7, 77 74 bytes

boolean c(String s){int r=0;for(int c:s.getBytes())r+=c&31;return r==100;}

Case-insensitive thanks to @Neil by changing c-96 to c&31.
-3 bytes thanks to @SuperChafouin.

Explanation:

Try it here.

boolean c(String s){          // Method with String parameter and boolean return-type
  int r=0;                    //  Resulting sum-integer
  for(int c:s.getBytes())     //  For each character in the String (as integer value)
    r+=c&31;                  //   Sum its alphabetic 1-indexed index (case-insensitive)
                              //  End of loop (implicit / single-line body)
  return r==100;              //  return if the sum equals 100
}                             // End of method

Answer 37

C#, 72 64 bytes

---

Golfed

(string w)=>{int s=0;foreach(var c in w)s+=c-96;return s==100;};

---

Ungolfed

( string w ) => {
   int s = 0;

   foreach( var c in w )
      s += c - 96;

   return s == 100;
};

---

Ungolfed readable

( string w ) => {
   // Initialize the var to store the sum of the chars
   //    of the given word
   int s = 0;

   // Cycle through each char
   foreach( var c in w )

      // Add it to the sum minus 96
      //    'a' == 97
      s += c - 96;

   // Return if the sum is equal, or not, to 100
   return s == 100;
};

---

Full code

using System;
using System.Collections.Generic;

namespace Namespace {
   class Program {
      static void Main( String[] args ) {
         Func<String, Boolean> f = ( string w ) => {
            int s = 0;

            foreach( var c in w )
               s += c - 96;

            return s == 100;
         };
         List<String>
            testCases = new List<String>() {
               "buzzy",
               "boycott",
               "identifies",
               "adiabatically",
               "ttttt",
               "zzz",
               "zzzzzzz",
               "abcdefghiljjjzz",
               "tttt",
               "basic",
            };

         foreach( String testCase in testCases ) {
            Console.WriteLine( $" Input: {testCase}\nOutput:{f( testCase )}\n" );
         }

         Console.ReadLine();
      }
   }
}

---

Releases

• v1.1 - -8 bytes - Swapped the for loop with a foreach one
• v1.0 - 72 bytes - Initial solution.

---

Notes

• $

Answer 38

braingasm, 20 bytes

Assumes input without trailing newlines, e.g echo -n "unaltered" | braingasm dollarwords.bg

,[96-[->+<],]>100-z:

Could easily been done in plain brainfuck, if I had the patience for it.

,                       Read a byte to the current cell.
 [                      While there's something in the current cell:
  96-                     subtract 96 from it,
     [->+<]               move the remainder to the next cell,
           ,              read another byte.
            ]           
             >100-      Move to next cell and subtract 100 from it,
                  z:    print 1 if remainder is zero, 0 if not.

Answer 39

Brain-Flak, 92 bytes

({{}[(((((()()()){}){}){}){}){}]})((((((()()()){}){}){}()){}){})({}[{}]<(())>){((<{}{}>))}{}

Try it online!

Answer 40

Brachylog, 12 9 bytes

-3 bytes thanks to Fatalize

ạ-₉₆ᵐ+100

Takes input as a string of lowercase letters. Try it online!

Explanation

ạ          Convert input string to a list of ASCII codes
    ᵐ      To that list, map this predicate:
 -₉₆        Subtract 96 from each charcode
     +     Sum the resulting list of numbers 1-26
      100  Succeed if the sum is 100, fail otherwise

Answer 41

Stax, 9 bytes

ü☺ïΦΘ╬╟₧╘

_{I love the ☺ in the code! Potential Stax/Emojicode polyglot in the future...?}

Run and debug it at staxlang.xyz!

Unpacked (10 bytes):

{64-m|+AJ=

Explanation:

{   m         Form a block and map it over input.
 64-            Push 64 and subtract it from the character code.
     |+       Sum the resulting array.
       AJ     Push 10 and square it.
         =    Check for equality!

I'm quite new to Stax (and golfing languages); this can almost certainly be shortened. I think the { can probably be dropped with proper placement of the m...

Another approach (imperative, 10 bytes):

ú☼7Tí:£_▓δ

12 unpacked:

c|+AJ-s%64*=
c               Copy the top item on the stack.
 |+             Sum it. For a string, this sums code points.
   AJ           Push 10 and square, yielding 100.
     -          Subtract this 100 from the earlier sum.
      s         Swap the top two items, putting the input back on top.
       %        Push the input's length.
        64*     Push 64 and multiply.
           =    Check for equality!

Input for both is capital letters. For lowercase, replace 64 with 96.

Answer 42

Rust, 158 bytes

fn main(){let(i,mut c)=(&mut"".into(),0);std::io::stdin().read_line(i);let l:Vec<u8>=i.trim().bytes().collect();for b in&l{c+=*b as u8-96}print!("{}",c==100)}

Ungolfed

fn main() {
    let (input_string, mut count) = (&mut "".into(), 0);
    std::io::stdin().read_line(input_string);
    let bytes: Vec<u8> = input_string.trim().bytes().collect();
    for byte in &bytes {
        count += *byte as u8 - 96;
    }
    print!("{}", count == 100);
}

Prints true if the input is a dollar word, and false if it is not. Only works on lower case letters.

Answer 43

R, 45 bytes

sum(as.numeric(charToRaw(scan(,"")))-96)==100

Try it online!

Returns TRUE or FALSE.

As stated in the question, expects user input in a-z, converts the characters to numbers, brings them to the 1–26 range, and checks if the sum is equal to 100.

Even if one converts it to a function, it still beats Giuseppe’s answer in terms of byte count (it would be 48):

function(x)sum(as.numeric(charToRaw(x))-96)==100

Answer 44

J 14 bytes or 10 bytes if rules allow!!!

100=+/_96+a.i.

        100=+/_96+a.i. 'identifies'
    1
        100=+/_96+a.i. 'ttttt'
    1
        100=+/_96+a.i. 'wrong'
    0

    Because:
                        a.i. gives position in code table 0-255
                    _96+ subtracts 96 ('a') from each value
                +/   adds them all together
            100= compares with 100  

IDEA: Can a "truthy" value be 100, and a "falsey value be <> 100?

If so...

+/_96+a.i.

    +/_96+a.i. 'boycott'
    100
    +/_96+a.i. 'boycot'
    80

:O)

Answer 45

jq, 28 characters

(25 characters code + 3 characters command line option)

[explode[]|.-96]|add==100

Sample run:

bash-4.3$ jq -R '[explode[]|.-96]|add==100' <<< 'bernardino'
true

On-line test

Answer 46

Groovy, 32 characters

{s->s*.value.sum{it[0]-96}==100}

Sample run:

groovy:000> ({s->s*.value.sum{it[0]-96}==100}("bernardino"))
===> true

Answer 47

Ruby 2.4, 30 + 1 = 31 bytes

One extra byte for the n flag: $ ruby -n dollar_words.rb

p chop.bytes.sum{|c|c-96}==100

Prints true or false for each given line of input:

$ ruby -n dollar_words.rb 
zzzzz
false
unaltered
true
dollar
false

Answer 48

QBIC, 34 bytes

;[_lA||p=p+asc(_sA,a,1|)%32]?p=100

Explanation:

;               Get A$ from the cmd-line
[_lA||          FOR a = 0; a <= a$.Length(); a++
p=p+asc(        Increment P with the ASCII value of
_sA,a,1|)%32]     A$.Substring(a,1) mod 32 (thank you @JonathanAllan, that's a neat trick!)
?p=100          When done, print -1 if p is 100, 0 otherwise

Answer 49

Pyth, 16

!-100+Fm+1xGdcQ1

Try online.

Answer 50

Röda, 25 bytes

{[ord(_)-96]|sum|[_=100]}

Try it online!

Answer 51

Batch, 189 bytes

@set/p,=
@set z=1
@for %%a in (a b c d e f g h i j k l m n o p q r s t u v w z y)do @set/a%%a=z,z+=1
@set .=100
:l
@set/a.-=%,:~,1%
@set ,=%,:~1%
@if not "%,%"=="" goto l
@exit/b%.%

Takes lowercase input on STDIN and ouputs via ERRORLEVEL (0 is truthy). Batch has no character code operator so I set the variables a-z to their value in a loop, then I individually evaluate all the characters in the input word and subtract them all from 100.

By using non-alphabetic variable names the code is readily extended to handle uppercase input.

Answer 52

Pyth, 9 bytes

q100smhxG

Explanation:

q100smhxGdQ autofill variables
       xGd  [index of d in "abcdefghijklmnopqrstuvwxyz"
      h      + 1
     m    Q  for d in eval(input())]
    s       sum(^)
q100        == 100

Test suite.

Answer 53

Excel, 52 + 2 bytes

{=SUM(IFERROR(CODE(MID(A1,ROW(A1:A100),1))-64,0))=100}

Input is in cell A1 in all uppercase.
Must be entered as an array formula with Ctrl+Shift+Enter (Adds the curly brackets { }).
Returns TRUE for dollar words and FALSE for others.

Answer 54

Rip, 27 bytes

10gDiW[8DmsagDi]P9iDmsI[d]O

Works only for uppercase input, which it reads from stdin. Please do not end the input with a newline. :)

Explanation:

1                            0I[Push a 1 for later]
 0                           0I[Sum is 0 now]
  gDiW[                      0I[Getchar, duplicate, increment, while: while not EOF]
       8Dms                  0I[8 8 dup mul subtract: subtract 65]
           a                 0I[Add to the current total]
            gDi]             0I[Getchar, dup, increment, end while: while not EOF]
                P            0I[Pop the last -1 (=EOF) read]
                 9iDms       0I[Subtract 100 (9 incr dup mul subtract)]
                      I[d]   0I[If not zero, decrement the first 1 (see first line) to a 0]
                          O  0I[Output the result (1 if ==100, 0 else) as a number]

Answer 55

C# / Linq, 42 bytes

Is Linq cheating?

(string w)=>{return w.Sum(x=>x-96)==100;};

Answer 56

Braingolf, 13 bytes [non.competing]

{#`-}&+#de1:0

Try it online!

Requires all lowercase input, outputs 1 for dollar word, 0 otherwise

Explanation

{#`-}&+#de1:0  Implicit input of string as char values to stack
{...}          Foreach item on stack..
 #`            ..Push char value of ` (96)
   -           ..Subtract from input char value
     &+        Sum entire stack
       #d      Push char value of d (100)
         e     If last 2 items are equal..
          1    ..Push 1
           :   else
            0  ..Push 0
               Implicit output of last item on stack

Answer 57

Proton, 32 bytes

map(ord+((-)&96))+sum+((==)&100)

Try it online!

Waiting for pull from TIO.

Answer 58

Perl 5, 28 + 1 (-p) = 29 bytes

s/./96-ord$&/ge;$_=-100^eval

Try it online!

Returns 0 for truthy, anything else for falsey. Assumes lowercase input.

How?

Replace each character with the additive inverse of its position in the alphabet. Then, evaluate that equation and XOR it with -100 to determine if this is a dollar word.

Answer 59

Common Lisp, 56 bytes

(=(loop as x across(read-line)sum(-(char-code x)96))100)

Try it online!

Answer 60

SmileBASIC, 48 bytes

INPUT W$WHILE""<W$D=D+ASC(POP(W$))-64WEND?D==100

The classic string eating loop.

Answer 61

Pyth, 15 bytes

Vz=+ZhxGN)qZ^T2

Try it online!

The other Pyth answers use Q=eval(input()), which means the input must be in quotes in order to be valid. This program does not have that stipulation.

Pyth (indented) | Python 3 (translation)

             | G="abcdefghijklmnopqrstuvwxyz"
             | T=10
             | Z=0
             | z=input()
Vz           | for N in z:
    =+ZhxGN) |     Z+=G.find(N)+1
qZ^T2        | print(Z==T**2)

Answer 62

Befunge-98, 14 13 bytes

+~'@ #.-"d"$#

Try it online!

Takes input as uppercase letters. Outputs 0 for truthy, any other number for falsey.

How It Works:

 ~          Get input and continue right.
+ '@   -    Subtract 64 from the letter and add it to the total
 ~          On EOF go left instead
       -"d" Subtract 100 from the total
   @  .     Print the value and end the program

Answer 63

Pip, 11 bytes

A_-96MSa==h

Takes a string of lowercase letters as a command-line argument. Try it online!

How it works

             a is 1st command-line arg; h is 100 (implicit)
A_           An anonymous function that takes the ASCII value of its argument
  -96        and subtracts 96
     MSa     Map that to the characters of a and sum the result
        ==h  Test if that's equal to 100
             The result (0 or 1) is autoprinted

We use == (exact equality) instead of the usual = (numeric equality) because it has lower precedence than MS.

Answer 64

Add++, 18 bytes

D,f,@~,€O96€_s100=

Try it online!

Answer 65

Julia 0.6, 27 bytes

s->sum(c->Int(c)-96,s)==100

Try it online!

Answer 66

Tcl, 70 bytes

puts [expr 100==[join [lmap c [split $argv ""] {scan $c %c}] -96+]-96]

Try it online!

---

Tcl, 71 bytes

proc D s {expr 100==[join [lmap c [split $s ""] {scan $c %c}] -96+]-96}

Try it online!

---

Tcl, 74 bytes

puts [expr 100==[join [lmap c [split $argv ""] {expr [scan $c %c]-96}] +]]

Try it online!

Saved a byte using a command line argument instead of a function.

---

Tcl, 75 bytes

proc D s {expr 100==[join [lmap c [split $s ""] {expr [scan $c %c]-96}] +]}

Try it online!

Answer 67

Kotlin, 23 bytes

{it.sumBy{it-'`'}==100}

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Answer 68

Ahead, 21 bytes

>jvi96-+v
^~>'d=O~<~@

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Answer 69

Attache, 20 bytes

{100=Sum[_-96]}@Ords

Try it online!

Explanation

{100=Sum[_-96]}@Ords
               @Ords    convert input to ordinates
{             }         anonymous lambda, input: _
         _-96           subtract 96 from each char
     Sum[    ]          take the sum
 100=                   is it equal to 100?

Alternatives

20 bytes: {100=Sum[Ords@_-96]}

21 bytes: 100&`=@Sum@`-&96@Ords

22 bytes: {100=_}@Sum@`-&96@Ords

22 bytes: {100=Sum@_}@`-&96@Ords

24 bytes: {100=Sum!-~_}##STN=>List

25 bytes: {100-#_=Sum@_}##STN=>List

25 bytes: {100=Sum[_+1]}##STN=>List

26 bytes: {100=Sum[_+1]}##STN=>Chars

26 bytes: 100&`=@Sum##Succ@STN=>List

27 bytes: {100=_}@Sum##Succ@STN=>List

Answer 70

F# (Mono), 38 bytes

fun s->Seq.sumBy(fun c->int c-96)s=100

Try it online!