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 5^13 as product of 3 numbers (Posted on 2009-11-22)
In how many ways can 5^13 be expressed as a product of three natural numbers?

Also repeat the problem for 5^12.

 See The Solution Submitted by Vishal Gupta No Rating

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 solutions (computer used) (spoilers) | Comment 1 of 2

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     1210     1411     1612     1913     2114     2415     2716     3017     3318     3719     4020     4421     4822     5223     5624     6125     6526     7027     7528     8029     8530     9131     9632    10233    10834    11435    12036    12737    13338    14039    14740    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
 Posted by Charlie on 2009-11-22 15:38:14

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