2 = 3 - 1
3 = 3
4 = 3 + 1
5 = 9 - 3 - 1
6 = 9 - 3
7 = (9 + 1) - 3
8 = 9 - 1
9 = 9
10 = 9 + 1
11 = (9 + 3) - 1
12 = 9 + 3
13 = 9 + 3 + 1
14 = 27 - 9 - 3 - 1
15 = 27 - 9 - 3
16 = (27 - 9 - 3) + 1
17 = 27 - 9 - 1
18 = 27 - 9
19 = 27 - 9 + 1
20 = (27 - 9) + 3) - 1
21 = (27 + 3) - 9
22 = ((27 + 3) - 9) + 1
23 = 27 - 3 - 1
24 = 27 - 3
25 = 27 - 3 + 1
26 = 27 - 1
27 = 27
28 = 27 + 1
29 = (27 + 3) - 1
30 = 27 + 3
31 = 27 + 3 + 1
32 = (27 + 9) - 3 - 1
33 = (27 + 9) - 3
34 = ((27 + 9) - 3) + 1
35 = (27 + 9) - 1
36 = 27 + 9
37 = 27 + 9 + 1
38 = (27 + 9 + 3) - 1
39 = 27 + 9 + 3
40 = 27 + 9 + 3 + 1
- Jim
Jeffry Houser wrote:
> First, we must make the assumption that everyone orders things in "whole
>numbers".
>
> Second, we know that this formula must hold true:
>
> a + b + c + d = 40
>
> Then I made the assumption that one of the weights must be equal to 1, so ..
>
> a = 1
>
> Then I saw the answer and stopped working on the problem.
>
> Using 1, 3, 9, and 27 can you get all the numbers between 1 and 40?
>
> 1 = 1
> 2 = 3 - 1
> 3 = 3
> 4 = 3 + 1
> 5 = 9- 3 - 1
> 6 = 9 - 3
> 7 = ??
> 8 = 9 -1
> 9 = 9
>10 = 9+1
>11 = 9 + 3 -1
>12 = 9 + 3
>13 = 9 + 3 + 1
>14 = ??
>15 = ??
>
> I'm going back to work now..
>
>At 01:05 PM 11/3/2003 -0500, you wrote:
>
>
>>Subject: Math Puzzle
>>From: "Randell B Adkins" <[EMAIL PROTECTED]>
>>Date: Mon, 03 Nov 2003 12:51:52 -0500
>>Thread:
>>http://www.houseoffusion.com/cf_lists/index.cfm/method=messages&threadid=10303&forumid=5#94330
>>
>>that is what I am asking. I understand that 1,3,9, and 27 have a common
>>factor
>>of 3 however not sure how one would have come up with that and if in
>>fact
>>there was an easy way of finding it out.
>>
>>This is truly a puzzle of the minds...
>>
>>
>
>
>
>
>--
>Jeffry Houser, Web Developer <mailto:[EMAIL PROTECTED]>
>Aaron Skye, Guitarist / Songwriter <mailto:[EMAIL PROTECTED]>
>--
>AIM: Reboog711 | Phone: 1-203-379-0773
>--
>My Books: <http://www.instantcoldfusion.com>
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>
>
>
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