This question is linked to Numbers in series

If you write out the numbers the British way (using "and")

What is the smallest number that has all the vowels used exactly twice.

What is the largest 6 digit number that has the same property?

Bonus Question

How high you have to go to find for a number with all vowels used exactly 3 times? ( "and" can be used only once). Is there such a number?

  • $\begingroup$ Maybe a no-computer tag should be added? $\endgroup$
    – newbie
    Commented Apr 7, 2020 at 12:43
  • $\begingroup$ I feel I should clarify this before I actually answered, but as a non-native, is 1101 written as one thousand and one hundred and one (two ands)? Or just one and and one comma? $\endgroup$
    – newbie
    Commented Apr 7, 2020 at 14:05
  • 2
    $\begingroup$ Or, is there some official reference for the British way of writing numbers? $\endgroup$
    – newbie
    Commented Apr 7, 2020 at 14:44
  • $\begingroup$ Nothing official @newbie but here is one ref en.wikipedia.org/wiki/English_numerals $\endgroup$
    – DrD
    Commented Apr 7, 2020 at 17:45

1 Answer 1


I believe the answer is,

4085 (four thousand and eighty-five)
962000 (nine hundred and sixty-two thousand)

My approach

First I made a table of all words and vowel appearences. [word] aeiou a 10000 one 01010 ... full table can be found at https://paste.ubuntu.com/p/X8sfQnhVFK/

First question,

From the table it's pretty clear no number <100 could work. We show why numbers in 100~999 couldn't work.
Those numbers have format: [] hundred (and []-[]/[]) (here [] stands for a word) hundred 01001 and 10000 [need] 11221 We need 7 vowels, however one word (except the -teen case, where 2 words can be used then) contains no more than 2 vowels and we only have 3 slots.
Now 4 digit numbers. Plain [] thousand certainly won't work. Notice that thousand and and both contain a, so we can at most have one and.
Two cases left: [] thousand and [] hundred and [] thousand and []-[]/[].
First case: hundred 01001 thousand 10011 and 10000 [need] 01210 Looking through all digits, sadly there aren't two digits matching this need.
Second case looks more promising. thousand 10011 and 10000 [need] 02211 As before, we need two or three words here with 6 vowels in total. Our only bet is to have three words, all with 2 vowels.
The only -ty's with 2 vowels: seventy 02000 eighty 01100 ninety 01100 If we choose seventy, we're left with 00211 and sadly no match again. If we choose eighty instead, we're left with 01111. four and five is such a match.

Second question,

6 digits, so a few hundred thousands. [] hundred thousand won't work. As said, we can only have at most one and pairing with thousand. Three cases: [] hundred and [] thousand, [] hundred thousand and [] hundred, [] hundred and []-[]/[] thousand (There's also [] hundred thousand and []-[]/[] but is certainly smaller).
The second case has 3us so not valid and the first case is identical to the [] thousand and [] hundred discussed in the first question. The only bet is the third case. hundred 01001 thousand 10011 and 10000 [need] 01210 Two digits won't work. -teen's have two es. So we're going for the three word case.
To make the number as big as possible, we put the largest digit nine in the front. nine 01100 [need] 00110 Two vowels left. Again we choose the biggest possible number to put with one vowel, which is sixty. We're left with the one last o so the last digit is two.

On the bonus question,

The answer is yes. For example, 3000000000084000 (three quadrillion and eighty-four thousand). I'm not sure if it's the smallest one (hopefully not).

  • $\begingroup$ Are you sure about your second answer @newbie? $\endgroup$
    – DrD
    Commented Apr 7, 2020 at 14:30
  • $\begingroup$ Umm I've commented under the question, can you check that? @DEEM $\endgroup$
    – newbie
    Commented Apr 7, 2020 at 14:32
  • $\begingroup$ I think I'm pretty sure about the second answer. $\endgroup$
    – newbie
    Commented Apr 7, 2020 at 15:00
  • $\begingroup$ Very nice analysis. The computer search would still be useful but also incomplete without it. $\endgroup$
    – humn
    Commented Apr 7, 2020 at 16:32

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