Mathematician Problem
Number Theory
Level
pending
A pucelana sequence is an increasing sequence of 16 consecutive odd numbers whose sum is a perfect cube. How many pucelana sequences are there with 3-digit numbers only?
The answer is 3.
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1 solution
The first thing is to know the formula for the sum of a sequence: n*(a(1)+ a(n))/2, where n is the number of items in the sequence, a(1) is the first term and a(n) is the last term. Then you want to find out the lower bound and upper bound of the 3-digit pucelana sequence sums. Then you can find out which of the sums in that range are perfect cubes, or better yet, which perfect cubes are pucelana sequences.
To find the sum of the lowest 3-digit pucelana sequence: The sum of a sequence is n(a(1)+ a(n))/2 and in this case, n=16.
The lowest odd number with 3 digits is 101.
Let a(1) = 101 then a(n) = a(16) = 131
Using n(a(1)+ a(n))/2 and substituting: 16 (a(1)+ a(16))/2 = 16 (101+131)/2
Simplifying: 8 (101+131) = 8 232 = 1856
To find the sum of the highest 3-digit pucelana sequence: Again, the sum of a sequence is n(a(1)+ a(n))/2.
The highest odd number with 3 digits is 999.
Let a(16) = 999 then a(1) = 969
Using n(a(1)+ a(n))/2 and substituting: 16 (a(1)+ a(16))/2 = 16(969+999)/2
Simplifying: 8 (969+999) = 8*1968 = 15744
The cubes we are looking for will be between our minimum sum 1856 and our maximum sum 15744. So, there are 435 different sums to check in this range to see which ones are perfect cubes. There must be an easier way.
Instead of looking for a few perfect cubes in the 435 sums of series, we can look for the right sums in a shorter list of the perfect cubes in this range.
Take the cube root of the two sums of our extremes: (1856)^1/3 = 12.289... (15744)^1/3 = 25.063...
It will be easier to look at only the 13 potential integer cube roots between 12.2 and 25.06.
Lets look at another way of simplifying the sum of this series: Notice that in a pucelana sequence, a(16)=a(1)+30.
16 (a(1)+ a(16))/2 = 16 (a(1)+ a(1)+30)/2 = 16 (2 a(1)+30)/2 = 16 2 (a(1)+15)/2 = 16*(a(1)+15) So the sum of a pucelana sequence has 16 as a factor, therefore any cube that does not divide evenly by 16 can be eliminated, and only cubes that divide evenly by 16 are pucelana sequences.
Some examples: Lets look at the first root and its corresponding cube:
13^3 = 2197 and 2197/16 = 137.3125 It does not divide evenly so it is not a pucelana sequence.
However, 16^3 = 4096 and 4096/16 = 256. It divides evenly so it is a pucelana sequence.
Here is a table of the 13 cubes to consider:
N (N^3) [N^3]/16
13 (2197) 137.3125
14 (2744) 171.5
15 (3375) 210.9375
16 (4096) 256. **
17 (4913) 307.0625
18 (5832) 364.5
19 (6859) 428.6875
20 (8000) 500. **
21 (9261) 578.8125
22 (10648) 665.5
23 (12167) 760.4375
24 (13824) 864. **
25 (15625) 976.5625
** Therefore, there are [3] pucelana sequences in the specified range.