Make a program/function which takes in an integer from 0 to 999999999999 (inclusive) as input and returns how many syllables it has when spoken in English. Make your code short.


  • A billion is a thousand million, a trillion is a million million, etc.
  • A million has three syllables -- it's mill-ee-on, as opposed to mill-yon. Pronounce all -illions like this.
  • Numbers like 1115 are 'one thousand, one hundred and fifteen'. The 'and' counts as a syllable.
  • 0 is zero. That's two syllables.

I know the specification seems mishmash in terms of which standards are used (British or American, mainly) -- it's what I and everyone I know use. I would change it so it's consistent, but someone's already answered...

Test Cases

> 0

> 100
(one hundred)

> 1000001
(one million and one)

> 1001000001
(one billon, one million and one)

> 7

> 28

> 78

> 11

> 1111
(one thousand, one hundred and eleven)

> 999999999999
(nine hundred and ninety-nine billion, nine hundred and ninety-nine million, nine hundred and ninety-nine thousand, nine hundred and ninety-nine)
  • 1
    \$\begingroup\$ @DJMcMayhem They're similar, but certainly not duplicates. \$\endgroup\$ Commented Aug 17, 2017 at 17:03
  • 1
    \$\begingroup\$ You were quite short on tags :/ \$\endgroup\$
    – Mr. Xcoder
    Commented Aug 17, 2017 at 17:09
  • 1
    \$\begingroup\$ IMHO, This specification is awful. "Billion" is used in American English, so if that's the standard you're using (it is, seeing your examples), then you shouldn't include and in the numbers, because that's grammatically incorrect, unless you're using British English, I think. So 1111 would be one thousand, one hundred, eleven). But British English uses "thousand million" instead of "billion". So you're mixing standards. \$\endgroup\$
    – mbomb007
    Commented Aug 17, 2017 at 18:10
  • 1
    \$\begingroup\$ @mbomb007 I was just using what I use in real life. I'm British. I say 1111 as 'one thousand, one hundred and eleven' (which makes logical sense). What does 'mixing standards' take away from the challenge, though? \$\endgroup\$
    – 0WJYxW9FMN
    Commented Aug 17, 2017 at 19:24
  • 1
    \$\begingroup\$ Possible duplicate of Count up by Syllables \$\endgroup\$
    – hyper-neutrino
    Commented Aug 17, 2017 at 23:46

1 Answer 1


Python 2, 232 bytes

def f(a):o=map(int,format(a,",").split(","));print(a%100>0)*(a>1000)+sum((k%100>0)*(k/100>0)for k in o[:-1])+sum((k>99)*2+`k`.count('7')+(k/10%10>0)+3-`k+1000`.count('0')for k in o)+sum((0<o[~k])*(5-k)*k/2for k in range(len(o)))or 2

Try it online!

This still has some potential for golfing, maybe in a lambda. The function splits the input into groups of three digits, each of which have a certain number of syllables. Luckily, every digit is 1 syllable except for 7. Every tens-place adds an extra 1 syllable to the number of syllables of the digit except for 1, which adds 0 (-teen or ten).


def f(k):
    if a==0:return 2
    while a>0:
    return \
int(o[-1]%100>0 and len(o)>1)        +len('+1 if first group of three have "and"'[0:0])+\
  [int(k%100>0 and k/100>0)        +len('+1 if rest have "and"'[0:0])\ 
for k in o[:-1]])+    
  [int(k>99)*2                      +len('have "hundred"'[0:0])\
  +str(k).count('7')               +len('+1 for "se-ven"'[0:0])\
  +(k/10%10>0)                    +len('+1 for "-ty" e.g. nine-ty'[0:0])\
  +3-str(k+1000).count('0')        +len('+1 for each number'[0:0])\
for k in o])+\
  [int(o[-(k+1)]>0)*([0,2,3,3][k])  +len('add 2 if "thousand", add 3 if "million", add 3 if "billion"'[0:0])\
for k in range(len(o))])

-12 bytes thanks to @Mr.Xcoder

  • \$\begingroup\$ 237 bytes, starter, much more left to golf I think \$\endgroup\$
    – Mr. Xcoder
    Commented Aug 17, 2017 at 19:33
  • \$\begingroup\$ Would a>1000 => a>1e3 work? \$\endgroup\$
    – Adalynn
    Commented Aug 17, 2017 at 20:02
  • \$\begingroup\$ @Zacharý Indeed, 231 \$\endgroup\$
    – Mr. Xcoder
    Commented Aug 18, 2017 at 22:06

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