8
$\begingroup$

As far as I know, all numbers have a root of 4. What I mean by this is as follows:

Starting with any number, for example 384, I take the number of letters in that number. Then I repeat this process until it infinitely repeats the number 4.

Three Hundred Eighty Four = 22

Twenty Two = 9

Nine = 4

Four = 4

So I challenge you to find any number which does not follow this rule.

| improve this question | |
$\endgroup$
  • 1
    $\begingroup$ I believe this is impossible. However, it is so strange that I know it is possible to not be :). $\endgroup$ – Rigidity Feb 15 '19 at 21:36
18
$\begingroup$

There are some trick solutions

For example, in Spanish, 4 and 6 are cuatro and seis respectively, so starting at either one of those will lead to an infinite cycle of 4-6-4-6-4... Thus, I'll begin with the assumption we are just using English.

With that in mind

Negative number and decimals both have a positive, whole number of letters, so after one iteration, every number is a positive, whole integer. Every number below ten has, at most, 5 letters within it. Thus, numbers from 20-29 will contain no more that 6+5=11 letters, below the lower bound for this range. A similar argument can be applied for every other number above 29. Every number from 10-19 has, at most, 9 letters in it. This is also below the lower bound of 10. With this, it is easy to see any number above ten will quickly drop below ten.

So, for the final step

One, two, six and ten lead to three.
Three, seven, eight lead to five.
Zero, four, five and nine lead to four.
This accounts for all of the numbers. We have shown every number reaches below ten, and every number up to ten reaches four within its third step. Thus, every number will eventually reach four.

To answer your question

There is no number that does not follow this rule.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Ah, I figured such. This is such an interesting fact in English. I suppose the next thing to look for would be a language in which this doesn't end up in an infinite loop between 1 or 2 numbers haha. $\endgroup$ – Rigidity Feb 16 '19 at 21:08
  • 1
    $\begingroup$ @Rigidity: Go with French, there's a four-long loop of 3 (trois) - 5 (cinq) - 4 (quatre) - 6 (six) - 3 - 5 - 4 - 6... $\endgroup$ – ZanyG Feb 16 '19 at 21:32
  • $\begingroup$ That's true. Nice. $\endgroup$ – Rigidity Feb 16 '19 at 22:48
6
$\begingroup$

There's no such number. Sooner or later you'll end up with a positive integer equal to its own number of letters, which can only be 4; or a loop where it regularly increases and decreases. Only the non-negative integers 0, 1, 2 and 3 are less than the number of their letters due to the way all integers are written, and even they end up giving 4, so a loop is impossible.

| improve this answer | |
$\endgroup$
4
$\begingroup$

Since the OP asks for any number, we can extend the definition to constant numbers like pi or e (Euler's Constant)

Both these cannot be counted (till infinity??)

So technically they do not end with four??

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You could argue that infinity has eight letters. $\endgroup$ – ZanyG Feb 15 '19 at 23:53
  • $\begingroup$ Infinity is not the pronounced nummber in pi and e. the pronouncing is so long it will never end $\endgroup$ – DrD Feb 15 '19 at 23:59
  • $\begingroup$ But you can describe these numbers using English. No matter how you choose to do so, that description has an integer number of letters. $\endgroup$ – aschepler Feb 16 '19 at 15:48
  • $\begingroup$ If you label decimals like this: Three point one four one five.... even if you go to 500 decimal places, you will end up with 4. But it is not certain, since infinity can't be measured. Rip. $\endgroup$ – Rigidity Feb 16 '19 at 21:10
3
$\begingroup$

Technically you could say

'Seven billion billionths' (always add those last two words)

The last two words make certain that you won't get to 4.

There's also:

'times one' 'not the number four' 'is a silly solution'

and many more!

   Wikipedia says this is an unsolved problem in mathematics, so yeah.
   I realized 'eight billion billionths' might be the second number alphabetically...
| improve this answer | |
$\endgroup$
  • $\begingroup$ If confusion were a color, and I were a chameleon, well lets just say I don't quite understand what you mean bu the billion billionths? $\endgroup$ – Rigidity Feb 16 '19 at 21:12
  • $\begingroup$ Take a billionth of something. Then have a billion of those billionths, which is basically one. So, seven billion billionths is just 7 ones, or just 7. $\endgroup$ – Alto Feb 17 '19 at 3:31
1
$\begingroup$

There are 10 digits in decimal number. Each have the number of letters between 3 to 5.

  • 0=zero
  • 1=one
  • 2=two
  • 3=three
  • 4=four
  • 5=five
  • 6=six
  • 7=seven
  • 8=eight
  • 9=nine

Any of this number would initially reach to 3, 4 or 5. 4 is not needed.

For 3 3~5 5~4.

For 5 5~4

Hence there are no such numbers in English.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.