I have devised a rule that assigns a positive integer to each English word. Here is a list of words and the integers the rule assigns them.

$$ \begin{array}{l|l} \text{Word}&\text{Integer}\\\hline \text{atmosphere}&3\\ \text{day}&2\\ \text{eclipse}&7\\ \text{geology}&4\\ \text{heavy}&3\\ \text{inflation}&1\\ \text{neanderthal}&3\\ \text{parsimoniously}&4\\ \text{span}&1\\ \text{zealous}&8\\ &&\\ &&\\ \end{array} \hspace{1cm} \begin{array}{l|l} \text{Word}&\text{Integer}\\\hline \text{a}&1\\ \text{beret}&1\\ \text{cheaters}&6\\ \text{coins}&2\\ \text{dad}&5\\ \text{fad}&2\\ \text{hectares}&6\\ \text{highways}&1\\ \text{I}&5\\ \text{laziness}&2\\ \text{lid}&1\\ \text{transportation}&2\\ \end{array}\hspace{1cm} \begin{array}{l|l} \text{Word}&\text{Integer}\\\hline \text{ab}&3\\ \text{ad}&1\\ \text{bag}&1\\ \text{banana}&5\\ \text{bann}&3\\ \text{hear}&4\\ \text{her}&3\\ \text{males}&1\\ \text{prevent}&1\\ \text{snap}&1\\ \text{thorny}&1\\ &\\ \end{array} $$

The rule can be described in a single sentence.

What is my rule?

Edit: I have chosen some additional words and computed the numbers the rule assigns them. Also, find some hints below.

Hint 1:

It is unlikely you will find a word in the dictionary that is assigned a value larger than $10$. However, the rule is defined for arbitrary strings of English letters, and it is possible write down strings with arbitrarily large output. To get an output of $11$ requires a string of at least 16 characters, and the 16 character string with output $11$ that comes first alphabetically is $\text{'dzzzzzzzzzzzzzzz'}$.

Hint 2:

It might be useful to start by figuring out what the words assigned $1$ have in common.

Hint 3:

The US coins are: penny (1¢), nickel (5¢), dime (10¢), quarter (25¢), half dollar (50¢), and dollar (100¢). This is relevant.

Hint 4:

The rule assigns anagrams the same number.

Hint 5:

The rule does not depend on any of the following:
Case, font, keyboard layout, letter frequency, word meaning

Hint 6:

The longest string that the rule assigns $1$ has one hundred letters.

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    $\begingroup$ The meaning of the words does not matter. The output is always strictly positive, never zero. $\endgroup$ – Julian Rosen May 29 '15 at 12:59
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    $\begingroup$ I got nowhere using scrabble scoring $\endgroup$ – Raystafarian May 29 '15 at 17:09
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    $\begingroup$ @TroyAndAbed No, each word determines one (and only one) value $\endgroup$ – Julian Rosen Jun 2 '15 at 15:16
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    $\begingroup$ The "I" (worth 5) is capitalized and in a serif font. Would it be 5 if it was lowercase? $\endgroup$ – Raystafarian Jun 2 '15 at 18:07
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    $\begingroup$ @Raystafarian Yes, I should have clarified: the rule does not depend on case or font $\endgroup$ – Julian Rosen Jun 3 '15 at 1:02

I have found the rule.

Give each letter a value. A = 1, B = 2, C = 3, and so on. Then add all the values. I have taken "snap" as an example. "snap" = 19 + 14 + 1 + 16 = 50. Then find what is the least number of coins needed to make the total, or in this case, 50 cents. You only need one coin for a total of 50 cents. Therefore, 1 is the correct answer. As listed in a hint, the US coins are used: penny (1¢), nickel (5¢), dime (10¢), quarter (25¢), half dollar (50¢), and dollar (100¢).

Another example:

cab = 3 + 1 + 2 = 6 cents = 5 cent coin + 1 cent coin = 2 coins

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    $\begingroup$ If you guys want to try this method out, use this code (JavaScript): function SOLVE(v){x="abcdefghijklmnopqrstuvwxyz".split("");v=v.split("");for(i=0,a=0;i<v.length;i++)a+=x.indexOf(v[i])+1;return c(a);}function c(a,f=Math.floor){S=f(a/100);a-=S*100;T=f(a/50);a-=T*50;U=f(a/25);a-=U*25;V=f(a/10);a-=V*10;W=f(a/5);a-=W*5;S+T+U+V+W+a;} and enter SOLVE("<your word>"). This code disregards any non-lowercase, alphabetical, English letters. $\endgroup$ – Conor O'Brien Jun 7 '15 at 23:10

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