In how many ways can 5^13 be expressed as a product of three natural numbers?
Also repeat the problem for 5^12.
If the order of the multiplicands counts for different ways, such as 5 * 625 * 390625 counting as a different way to 390625 * 625 * 5, then there's a simple way of calculating the ways: If the factors are represented as their respective power of 5 (such as 1,4,8 or 8,4,1 in the above examples) then the total powers of 5 has to be 13. These numbers can be represented by groups of one 1, four 1's and 8 1's. All together, these 13 1's, together with two separators (to separate which go into which power) make 15 items of which two are separators. There are C(15,2) ways of doing this, or 105. Similarly, for 5^12, there are C(14,2) = 91 ways.
However if order of factors is not considered to create a new way, then I don't see an alternative to enumerating them. For 13 and 12, the below tables enumerate, first the powers (exponents) of 5 themselves, for simplicity, and then the numbers themselves. The latter tables show also the number of ways of rearranging each, with the total of those matching the combination method outlined above.
0 0 13
0 1 12
0 2 11
0 3 10
0 4 9
0 5 8
0 6 7
1 1 11
1 2 10
1 3 9
1 4 8
1 5 7
1 6 6
2 2 9
2 3 8
2 4 7
2 5 6
3 3 7
3 4 6
3 5 5
4 4 5
21 combinations
1 * 1 * 1220703125 3
1 * 5 * 244140625 6
1 * 25 * 48828125 6
1 * 125 * 9765625 6
1 * 625 * 1953125 6
1 * 3125 * 390625 6
1 * 15625 * 78125 6
5 * 5 * 48828125 3
5 * 25 * 9765625 6
5 * 125 * 1953125 6
5 * 625 * 390625 6
5 * 3125 * 78125 6
5 * 15625 * 15625 3
25 * 25 * 1953125 3
25 * 125 * 390625 6
25 * 625 * 78125 6
25 * 3125 * 15625 6
125 * 125 * 78125 3
125 * 625 * 15625 6
125 * 3125 * 3125 3
625 * 625 * 3125 3
105 total permutations
0 0 12
0 1 11
0 2 10
0 3 9
0 4 8
0 5 7
0 6 6
1 1 10
1 2 9
1 3 8
1 4 7
1 5 6
2 2 8
2 3 7
2 4 6
2 5 5
3 3 6
3 4 5
4 4 4
19 combinations
1 * 1 * 244140625 3
1 * 5 * 48828125 6
1 * 25 * 9765625 6
1 * 125 * 1953125 6
1 * 625 * 390625 6
1 * 3125 * 78125 6
1 * 15625 * 15625 3
5 * 5 * 9765625 3
5 * 25 * 1953125 6
5 * 125 * 390625 6
5 * 625 * 78125 6
5 * 3125 * 15625 6
25 * 25 * 390625 3
25 * 125 * 78125 6
25 * 625 * 15625 6
25 * 3125 * 3125 3
125 * 125 * 15625 3
125 * 625 * 3125 6
625 * 625 * 625 1
91 total permutations
So for 5^13, if order counts, there are 105 ways; if order doesn't count, then there are 21 ways.
For 5^12, if order counts, there are 91 ways; if order doesn't count, then there are 19 ways.
For the general formula for the order-not-counting result, see Sloane's A001399:
a(n) = nearest integer to (n+3)^2 / 12 = [(n+3)^2 / 12 + 1/2], where the square brackets represent the floor function. For various powers of 5 (or any prime) the following table shows the results:
1 1
2 2
3 3
4 4
5 5
6 7
7 8
8 10
9 12
10 14
11 16
12 19
13 21
14 24
15 27
16 30
17 33
18 37
19 40
20 44
21 48
22 52
23 56
24 61
25 65
26 70
27 75
28 80
29 85
30 91
31 96
32 102
33 108
34 114
35 120
36 127
37 133
38 140
39 147
40 154
Programs used:
DEFDBL A-Z
g = 13
FOR a = 0 TO 4
FOR b = a TO a + (g - a) / 2
c = g - a - b
IF c >= b THEN
PRINT a; b; c
ct = ct + 1
END IF
NEXT
NEXT
PRINT ct
PRINT
FOR a = 0 TO 4
FOR b = a TO a + (g - a) / 2
c = g - a - b
IF c >= b THEN
PRINT INT(5# ^ a + .5); "*"; INT(5# ^ b + .5); "*"; INT(5# ^ c + .5),
ct = ct + 1
IF a = b AND b = c THEN
inc2 = 1
ELSEIF a = b OR b = c THEN
inc2 = 3
ELSE
inc2 = 6
END IF
PRINT inc2
tot2 = tot2 + inc2
END IF
NEXT
NEXT
PRINT tot2
For last table:
FOR i = 1 TO 40
PRINT USING "## ######"; i; INT((i + 3) ^ 2 / 12 + .5)
NEXT
Edited on November 22, 2009, 3:39 pm
Edited on November 22, 2009, 3:41 pm
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Posted by Charlie
on 2009-11-22 15:38:14 |
solutions (computer used) (spoilers)
