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Today my 9-year-old nephew told me that he can spell any integer in English using only 9 letters. This is how he's doing it:

    ...
-3: MINUS ONE MINUS ONE MINUS ONE
-2: MINUS ONE MINUS ONE
-1: MINUS ONE
 0: ONE MINUS ONE
 1: ONE
 2: ONE PLUS ONE
 3: ONE PLUS ONE PLUS ONE
    ...

The letters he's using are E, I, L, M, N, O, P, S and U.

Can we do better and spell any integer with less than nine letters? I think so!

Just please be sure to briefly explain any maths you are using in your method, so that my young nephew (and his uncle) can understand.

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    $\begingroup$ I,V,X,L,C,D,M - no negatives though. $\endgroup$ – Mazura Sep 11 '16 at 1:02
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    $\begingroup$ Of course, the question taken literally as asked can be done with eight letters - to spell the literal phrase (A N Y I T E G R). Granted, that's not what it meant to ask (and the answers are good)...but my interpretation isn't specifically forbidden in the question, either :) $\endgroup$ – Megha Sep 11 '16 at 2:37
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    $\begingroup$ @Mazura If you're going to stretch the definition of "spell" to include non-English spellings, then why not just use I? 10 is IIIIIIIIII for example. Then apply an encoding of zero and negative numbers so I is 0, II is 1, III is -1, IIII is 2, IIIII is -2, IIIIII is 3, ... $\endgroup$ – user253751 Sep 11 '16 at 7:42
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    $\begingroup$ This seems really ambiguous, and the variety of answers seem to support this. What exactly does it mean to "spell any integer in English"? $\endgroup$ – Jason C Sep 11 '16 at 23:30
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    $\begingroup$ I can do with zero letters, by simply using digits and minus sign instead, in the usual way. ;-) $\endgroup$ – celtschk Sep 11 '16 at 23:48

16 Answers 16

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  ...
-3: MINUS ONE MINUS ONE MINUS ONE
-2: MINUS ONE MINUS ONE
-1: MINUS ONE
 0: ONE MINUS ONE
 1: ONE 
 2: ONE MINUS MINUS ONE
 3: ONE MINUS MINUS ONE MINUS MINUS ONE
    ...

two letters fewer since we do not use PLUS any more.

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  • $\begingroup$ +1 Was about to post this when I read the question, but you beat me to it a few hours ago. $\endgroup$ – Kevin Cruijssen Sep 10 '16 at 13:15
  • $\begingroup$ Is this considered unambiguous? For example, the expression 1 - - 1 would generally not be considered valid (but 1 - (-1) would). $\endgroup$ – svick Sep 11 '16 at 10:40
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    $\begingroup$ You do have to be careful that ONE MINUS MINUS ONE MINUS MINUS ONE isn't misheard as $1 - (-1 - -1)$, but I think that's doable. $\endgroup$ – Ben Millwood Sep 11 '16 at 11:51
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    $\begingroup$ I think $a-b-c$ is well defined as $(a-b)-c$ (grouped left to right). $\endgroup$ – Kos Sep 12 '16 at 9:40
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    $\begingroup$ @yo' No, I think Kos is right. Writing $a-b-c$ is acceptable and is interpreted as $(a-b)-c$ even if this may not necessarily be true for operations other than subtraction. $\endgroup$ – Shuri2060 Sep 12 '16 at 12:17
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4 letters: E N O T

Explanation:

My idea is to use binary and marcoresk's idea of using NOT to represent the - sign, although I have to admit it might be slightly 'cheating' as I have to define things and work in a different base.

To represent a number you read out its binary representation using ONE and NONE to represent $1$s and $0$s.

Eg. $5$ would be: ONE NONE ONE

If the number is negative, add a NOT to the beginning.

Eg. $-6$ would be: NOT ONE ONE NONE

Extension:

You can represent any decimal which is terminating in binary with an extra letter D for a total of 5 letters which allows you to say DOT for the decimal point.

Eg. $-5.5$ would be: NOT ONE NONE ONE DOT ONE

Other possibilities:

I understand that using NOT probably isn't the best idea to indicate -, so as I've mentioned in the comments, perhaps NEG can be used instead as it's an abbreviation for negative (I wasn't sure of this until Jason C pointed it out in his answer). This will still result in 4 letters being used (although 6 for the extension).

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  • $\begingroup$ You could define a negative bit and get rid of the T $\endgroup$ – Carl Sep 11 '16 at 4:22
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    $\begingroup$ I think NOT x should be interpreted as -x-1 rather than -x. $\endgroup$ – R.. GitHub STOP HELPING ICE Sep 11 '16 at 4:50
  • $\begingroup$ @R.. Positive: ~x is -(x+1), not -x. $\endgroup$ – EKons Sep 11 '16 at 9:20
  • $\begingroup$ I would think NOT is 0 or FALSE, not - $\endgroup$ – Raystafarian Sep 11 '16 at 12:28
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    $\begingroup$ Well — I was only putting a suggestion for defining NOT which sort of makes sense. I wasn't thinking about the actual Boolean definition. Would NEG work for negative instead? $\endgroup$ – Shuri2060 Sep 11 '16 at 13:17
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If we use LESS to indicate subtraction, then we have

-1 = ONE LESS ONE LESS ONE

0 = ONE LESS ONE

1 = ONE

2 = ONE PLUS ONE

etc., using only seven letters ELNOPSU.

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  • $\begingroup$ What about using punctuation? Say "ONE LESS: ONE LESS ONE LESS ONE" as 2 (1 less -1). That gets you to 5 letters. Or 6 if you count the colon, which you really shouldn't. $\endgroup$ – Hugh Meyers Sep 10 '16 at 18:52
  • $\begingroup$ @Hugh Punctuation strikes me as cheating. $\endgroup$ – Rosie F Sep 10 '16 at 20:48
  • $\begingroup$ I don't think you need punctuation; it's clear enough that "less" is being used as an operation. $\endgroup$ – Samthere Sep 12 '16 at 12:31
  • $\begingroup$ @Samthere "Less" is indeed being used as an operation -- I agree with you there. But Hugh's intended solution needs to distinguish between 1-1-1-1 = -2 and 1-(1-1-1) = 2. The 4 operands and 3 operations are the same in the two cases. But they are grouped in different ways in order to get different results. And it's the proposed means to do that which struck me as cheating. $\endgroup$ – Rosie F Sep 12 '16 at 12:36
  • $\begingroup$ @RosieF Oh, I see what you mean; Hugh's suggestion is for using your subtraction operator to achieve positive numbers instead of having an addition operator. I agree that it's cheating, or rather that to do it in the spirit of the question would require you to write "OPEN BRACKET" "CLOSE BRACKET" or something similar. $\endgroup$ – Samthere Sep 12 '16 at 13:04
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If any integer can be spelled with this structure

one some operation one some operation one

the trick could be simply this: reduce the numbers of letters needed for operations (or use O, N, E as much as you can!).

M,I,N,U,S introduces 4 new letters (N does not count) P,L,U,S introduces 2 other letters (P and L)

A simple (but symbolic) solution is to represent a operation with a single letter... but this is not a way to "say it in English".

So I propose this structure: PLUS becomes AND (only two more letters, A and D). Every integer number N bigger than 1 can be said as

N =  ONE AND ONE AND ONE AND ONE...

B) Define the number MINUS ONE as "NOT ONE" (let's say this will became our convention, this is the weakest part!) in order to introduce only one more letter (T)

Now zero becomes 

ZERO =  ONE AND (NOT ONE)

C) Since every negative numbers is equal to its absolute value multiplied by -1 e.g. -59 = 59 * (-1) I suggest to use a new word to introduce the operation of multiplication, as in class we often say A multiplied by B as "A dot B" (A dot is one possible symbol to show multiplication in advanced math, instead of x or * or others)

Note that now DOT does not introduce any new letter. So -3 could be 

(ONE AND ONE AND ONE) DOT (NOT ONE)

Notice we have used only 6 letters instead of 9 (and mantaining a speakable structure) O, N , E, D, T, A

Now the last problem: how to "say" parenthesis? With the rithm of speech! ONE AND ONE AND ONE AND ONE [after this sequence do a little pause] DOT [say DOT as an important word, then a little pause] NOT ONE [say NOT ONE fast, as if would be a single word]

Thank you for this pretty puzzle and sorry for my poor english!

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  • $\begingroup$ How about defining NOT as an operator which multiplies by -1? You wouldn't need DOT in C), although it doesn't make a difference to the letter count. $\endgroup$ – Shuri2060 Sep 10 '16 at 13:04
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    $\begingroup$ Welcome to Puzzling! $\endgroup$ – IAmInPLS Sep 10 '16 at 13:08
  • $\begingroup$ @QuestionAsker your doubt is right, thank you for asking. I made this choice basically because i wanted to force my answer to be "the more speakable" as possible (it was the request after all). And introducing the DOT I didn't add any letter, so I preferred a longer but more understandable (I hope!) solution $\endgroup$ – marcoresk Sep 11 '16 at 9:30
  • $\begingroup$ Why not have ZERO = NONE? $\endgroup$ – SQB Sep 12 '16 at 10:55
  • $\begingroup$ @SQB I mainly use the zero to exemplify the NOT operator, needed for negative integers, and simply because the using of NONE does not reduce the 6 letters ;) $\endgroup$ – marcoresk Sep 12 '16 at 12:07
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ONE ON ONE = 2

ONE OFF ONE = 0

ONEF: total 4 letters

by replacing plus and minus with on and off which mean the same.

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    $\begingroup$ How and why do on and off mean the same as plus and minus?!? $\endgroup$ – curiousdannii Sep 10 '16 at 16:38
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    $\begingroup$ @curiousdannii If you take one dollar off your bank account balance you have one less dollar in your bank account; if you put one on it you have one more dollar there; if your balance is, at some point, zero and you take one off you are in debt by a dollar. $\endgroup$ – Jonathan Allan Sep 10 '16 at 17:01
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    $\begingroup$ @Jonathan I know how to count, but in English "one off one" is not the same as "one minus one". "One off one" does not have any idiomatic meaning. $\endgroup$ – curiousdannii Sep 10 '16 at 17:04
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    $\begingroup$ in idiomatic terms one wouldn't call one plus one instead of two at the first place $\endgroup$ – Ali Humayun Sep 10 '16 at 17:15
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    $\begingroup$ For the minus, this would be different: x OFF y is y - x. $\endgroup$ – EKons Sep 11 '16 at 9:22
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Since

"O" is in the Collins English dictionary as "used to mean zero"
(pronounced the same as "owe", as in "I owe MackTuesday a coconut")

...and, as pointed out by @MackTuesday, "a" means one
(as in "I owe MackTuesday a coconut")

We can

Use just two letters: O, and a
(I previously used three with One rather than a, which may be less confusing)

If we use base 2, or binary with a signed magnitude representation by using our first bit as an indication of "is this number negative" and the time we stop speaking to represent our most significant bit for example:
$-32 =$ a O O O O O a $=- (0\times 2^0 + 0\times 2^1 + 0\times 2^2 + 0\times 2^3 + 0\times 2^4 + 1\times 2^5)$

$-7 =$ a a a a
$=- (1\times 2^0 + 1\times 2^1 + 1\times 2^2)$

$-4 =$ a O O a
$=- (0\times 2^0 + 0\times 2^1 + 1\times 2^2)$

$0 =$ O O (or a O)
$=+ (0\times 2^0)$ (or $=- (0\times 2^0)$)

$17 =$ O a O O O a
$=+ (0\times 2^0 + 0\times 2^1 + 0\times 2^2 + 0\times 2^3 + 1\times 2^4)$

Note that

$2^n$ just means $2$ multiplied by itself $n$ times, so $2^3$ is $2\times 2 \times 2=8$
...and that the identity element of the multiplication group is $1$ so $2^0$ is $2$ multiplied by itself $0$ times, which must be $1$.

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  • $\begingroup$ Could you get away with using I for one? :p $\endgroup$ – Shuri2060 Sep 10 '16 at 16:06
  • $\begingroup$ @QuestionAsker I thought about it, and it is in Collins but with the parenthesised prefix "Roman numeral", so I'm not so sure. $\endgroup$ – Jonathan Allan Sep 10 '16 at 16:10
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    $\begingroup$ The article "a" is also used to mean 1, as in "a banana". $\endgroup$ – MackTuesday Sep 10 '16 at 16:35
  • $\begingroup$ @MackTuesday your solid grasp of English must be unsurpassed, who knows such obscure words?! I updated the answer; thank you! $\endgroup$ – Jonathan Allan Sep 10 '16 at 16:46
  • $\begingroup$ I think this stretches OP's wording of what it means to spell something too far. $\endgroup$ – user1717828 Sep 11 '16 at 19:41
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6 Letters: T I M E S U

-1: i times i
0: sum i i times i times i
1: i times i times i times i
2: sum i times i times i times i i times i times i times i
3: sum i times i times i times i i times i times i times i i times i times i times i
-2: i times i times sum i times i times i times i i times i times i times i

i being the mathematical constant for the square root of -1. i squared ("i times i") is -1 and i to the fourth power ("i times i times i times i") is 1.

If the pause between the items in the set it not clear enough, you could also use the notation:
3: sum set item i times i times i times i item i times i times i times i item i times i times i times i

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    $\begingroup$ Wow, nice idea! $\endgroup$ – Rand al'Thor Sep 12 '16 at 18:27
  • $\begingroup$ You could also use T I M E S L to spell LESS instead of SUM. And you could write 0 less verbosely as $i + i^3$ (sum i i times i times i). Other than that, +1. $\endgroup$ – GOTO 0 Sep 12 '16 at 20:39
  • $\begingroup$ @GOTO0 good point, edited for the 0 calculation $\endgroup$ – Kys Sep 13 '16 at 17:19
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Another way of using only seven letters (A, D, E, L, N, O, S):

    ...
-3: ONE LESS ONE LESS ONE LESS ONE LESS ONE
-2: ONE LESS ONE LESS ONE LESS ONE
-1: ONE LESS ONE LESS ONE
 0: ONE LESS ONE
 1: ONE
 2: ONE AND ONE
 3: ONE AND ONE AND ONE
 4: ONE AND ONE AND ONE AND ONE
    ...
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    $\begingroup$ For a moment there I thought you were spelling my name $\endgroup$ – Adelin Sep 11 '16 at 8:18
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If we are allowed any encoding we like, we only need one letter. We choose some way of enumerating integers (say, in the order 0, 1, -1, 2, -2, 3, -3 ...) and then we encode the $n^\mathrm{th}$ integer as $n$ repetitions of our letter.

I don't think this is in the spirit of the question, which means that many of the other answers also aren't in the spirit of the question ("spell any integer in English"; I interpret this to mean, if you need to explain what number you are spelling, you have already lost). But if you're going to give an unspirited answer, you might as well give the optimal one :)

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    $\begingroup$ Hurrah for unary notation! ;-) $\endgroup$ – azurefrog Sep 12 '16 at 21:27
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The rules to this seem a little ambiguous but does this count? 4 letters:

{O,N,E,G}

Where all positive integers are just a that many ones:
1: one
2: one one
3: one one one


And negative integers are indicated by "neg" (an English abbreviation):
-1: neg one
-2: neg one one
-3: neg one one one


And zero is just an o:
0: o

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    $\begingroup$ what are you doing down here? lol $\endgroup$ – lois6b Sep 13 '16 at 6:08
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5 (O N E L S)

For example: 110=ONE ONE ONE LESS ONE=111-1 Just do the lowest integer above your target value consisting of only ones, then subtract down. Works for any positive or negative number.

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2 answer depends on the rule

write numbers in binary representation (9 letters o,n,e,z,r,m,i,u,s) 1 = one
-1 = minus one
0 = zero
2 = one zero

or

or maybe just 6 letters (m,i,n,u,s,o) if we can write 1 as i and 0 as o (maybe invalid) 1 = i
-1 = minus i
0 = o
2 = io

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We can refer to the number 1 using the less common word UNIT. Using MINUS adds two letters. Then using the word SUM in place of PLUS adds no new letters. Unlike PLUS, which is an operation between two operands, SUM naturally operates on all following operands.

-3: MINUS SUM UNIT UNIT UNIT
-2: MINUS SUM UNIT UNIT
-1: MINUS UNIT
 0: UNIT MINUS UNIT
 1: UNIT
 2: SUM UNIT UNIT
 3: SUM UNIT UNIT UNIT

If you're unhappy with the SUM operator taking an arbitrary number of operands, we can have it apply to only two operands. I've summarised below, using Polish notation for both SUM x y=x+y and MINUS x y=x-y.

-3: MINUS UNIT SUM SUM SUM UNIT UNIT UNIT UNIT
-2: MINUS UNIT SUM SUM UNIT UNIT UNIT
-1: MINUS UNIT SUM UNIT UNIT
 0: MINUS UNIT UNIT
 1: UNIT
 2: SUM UNIT UNIT
 3: SUM SUM UNIT UNIT UNIT

Thus we have 6 unique letters: IMNSTU

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Five letters: {Z, E, R, O, N}

No maths required, but you have to understand binary. We spell out the bits involved in a binary representation of the desired integer. The first bit is the sign bit (ZERO for positive, ONE for negative).

Examples:

-3: ONE ONE ONE
-2: ONE ONE ZERO
-1: ONE ONE
 0: ZERO ZERO
 1: ZERO ONE 
 2: ZERO ONE ZERO
 3: ZERO ONE ONE
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  • $\begingroup$ The idea here seems to be similar to that of Question Asker's answer, but using 5 letters instead of 4. $\endgroup$ – Rand al'Thor Sep 10 '16 at 15:11
  • $\begingroup$ Or, on the topic of using 0 and 1 as signs, you could just represent 0 as NONE, hence three characters! NONE ONE ONE = -2 for example. edit: should I make this a community wiki answer? :P $\endgroup$ – Florrie Sep 10 '16 at 17:17
  • $\begingroup$ Or even use O instead of ZERO so you can remove Z & R $\endgroup$ – Thomas Ayoub Sep 11 '16 at 10:17
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    $\begingroup$ The phrase "no maths required, but you have to understand binary" has me scratching my head :) $\endgroup$ – Ben Millwood Sep 11 '16 at 11:53
  • $\begingroup$ Question Asker's answer uses a special prefix to indicate negative numbers; this one uses two's complement, forcing the extra bit. That makes it different enough IMO. $\endgroup$ – Sebastian Redl Sep 12 '16 at 14:23
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I used only SUB instead of PLUS and MINUS.

Using only six letters (B, E, N, O, U, S):

-3 : SUB (ONE ONE ONE)

-2 : SUB (ONE ONE)

-1 : SUB (ONE ONE)

0 : (ONE SUB ONE)

1 : SUB (SUB ONE)

2 : SUB (SUB ONE SUB ONE)

3 : SUB (SUB ONE SUB ONE)

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  • $\begingroup$ Your answer relies on additional punctuation to avoid ambiguity, which feels like it's outside of the scope of good answers. Not a bad idea but it relies on interpreting or encoding language as well; we can obviously recognise the intended meaning of "sub" but it isn't standard. $\endgroup$ – Samthere Sep 12 '16 at 12:29
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Looking at all the number in binary give use the word zero and One which results in 5 characters.

8 = 1000 = one zero zero zero.

15 = 1111 = one one one one.

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    $\begingroup$ How is your answer any different from the ones already posted here before you? In addition you don't deal with the negative cases. $\endgroup$ – Shuri2060 Sep 12 '16 at 13:18

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