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View Full Version : The Added Factors - A FOFC numbers puzzle


QuikSand
12-27-2006, 07:38 AM
This isn't intended to be a competitive puzzle, but rather one of community contribution. We'll see how that goes.


Setup

Every whole number is either a prime number (evenly divisible only by itself and one) or can be expressed as a product of a unique series of prime numbers, which may include a given prime number more than once.

For example, the number 12 is not prime, as it breaks down as 2 x 2 x 3 = 12.


Example

The number 14 has an interesting property. When you look at the original number, and count the letters in its common English spelling, you get:

F-O-U-R-T-E-E-N = 8

...and when you look similarly at its prime factors, you get:

T-W-O = 3
S-E-V-E-N = 5

...and if you add up the total letters in its prime factors, you again get 8.


Puzzle

The goal of this puzzle is to find additional numbers with this same property -- where the number of letters in the original number is equal to the sum of the letters used, in common English, to spell out its prime factors. Any factor used more than once shoudl be counted as many times as it is used. Do not use the word "and" (or any other conjunctions or non-numeric terms) in the spelling of larger numbers -- so consider the number 123 to be "one hundred twenty three" for a total of 21 letters, not 24.

I'm honestly not sure if the challenge is to find the most such numbers, to find the largest one, or what. I suspect this could be automated, which would basically obviate the puzzle element of it, but eventually that's fine I guess.



Enjoy.

QuikSand
12-27-2006, 09:07 AM
ONEHUNDREDELEVEN = 16

- - -

THIRTYSEVEN = 11
THREE = 5


Hooray brute force!

QuikSand
12-27-2006, 01:02 PM
guess not then *shurg*

AnalBumCover
12-27-2006, 01:27 PM
TWOTHOUSANDTHIRTEEN = 19

cuervo72
12-27-2006, 01:29 PM
.

AnalBumCover
12-27-2006, 01:38 PM
seventhousandninehundredseven = 29

Toddzilla
12-27-2006, 01:40 PM
Found a few:

T W E N T Y T W O = 9
=
E L E V E N * T W O = 6 + 3 = 9


T W E N T Y E I G H T = 11
=
S E V E N * T W O * T W O = 5 + 3 + 3 = 11


T H I R T Y T H R E E = 11
=
E L E V E N * T H R E E = 6 + 5 = 11

AnalBumCover
12-27-2006, 01:42 PM
Ahh... I misread the rules.

Toddzilla
12-27-2006, 01:47 PM
Dola - I think that's it between 33 and 99, since you start to add multiple factors of 2, 3, and 5 which push the letter count up there.

100 = 10 letters, so there is hope for many more...

QuikSand
12-27-2006, 02:43 PM
Ahh... I misread the rules.

Hmmmm...

TWOTHOUSANDTHIRTEEN = 19

I think this still works, right?

THREE x ELEVEN x SIXTYONE = 5 + 6 + 8

Toddzilla
12-27-2006, 08:57 PM
Here are a few more, with 3 consecutive even numbers to boot!

TWOHUNDREDTWENTY = 16
=
ELEVEN x FIVE x TWO x TWO = 6 + 4 + 3 + 3 = 16


TWOHUNDREDTWENTYTWO = 19
=
TWO x THREE x THIRTYSEVEN = 3 + 5 + 11 = 19


TWOHUNDREDTWENTYFOUR = 20
=
SEVEN x TWO x TWO x TWO x TWO x TWO = 5 + 3 + 3 + 3 + 3 + 3 = 20

and finally

TWOHUNDREDTHIRTYTWO = 19
=
TWO x TWO x TWO x TWENTYNINE = 3 + 3 + 3 + 10 = 19