Problems & Puzzles: Puzzles

Puzzle 284.  A+B=C | A.B.C = pk#

This week I tried to find solutions to the simple equation A+B=C such that A.B.C = pk#

Example:

247 + 110 = 357  13.19 + 2.5.11 = 3.7.17 (using the first 8 consecutive primes)

As a matter of fact I have obtained all the solutions using k consecutive primes for 3<=k<=13 (except for k=12)

k Qty of sol's Solutions Prime expression
3 1 3 + 2 = 5 3 + 2= 5
4 1 7 + 3 = 10 7 + 3= 2.5
5 3 11 + 10 = 21
15 + 7 = 22
33 + 2 = 35
11 + 2.5 = 3.7
3.5 + 7 = 2.11
3.11 + 2 = 5.7
6 1 42 + 13 = 55 2.3.7 + 13 = 5.11
7 3 165 + 17 = 182
210 + 11 = 221
221 + 10 = 231
3.5.11 + 17 = 2.7.13
2.3.5.7 + 11 = 13.17
13.17 + 2.5 = 3.7.11
8 2 247 + 110 = 357
385 + 57 = 442
13.19 + 2.5.11 = 3.7.17
5.7.11 + 3.19 = 2.13.17
9 2 874 + 231 = 1105
1311 + 119 = 1430
2.19.23 + 2.7.11 = 5.13.17
3.19.23 + 7.17 = 2.5.11.13
10 1 1495 + 1463 = 2958 5.13.23 + 7.11.19 = 2.3.17.29
11 1 22971 +374 = 23345 3.13.19.31 + 2.11.17 = 5.7.23.29
12 0  Not found Not found
13 4 57646 + 49335 = 106981
127687 + 16523 = 144210
213486 + 6479 = 219965
219965 + 6118 = 226083
2.19.37.41 + 3.5.11.13.23 = 7.17.29.31
7.17.29.37
+ 13.31.41 = 2.3.5.11.19.23
2.3.7.13.17.23
+ 11.19.31 = 5.29.37.41
5.29.37.41
+ 2.7.19.23 = 3.11.13.17.31
 
       

Questions:

1. Can you extend the Table?

2. Can you develop a smart approach in order to calculate the solutions for a given k value?

2. Would you expect to exist solutions for 'large' values of k? If so, please find a solution for a k larger than 50.


Contributions came from Luke Pebody and Adam Stinchcombe

Pebody wrote:

You seem to have ignored the possibility of A=1:

k=1 -> 1+1=2
k=2 -> 1+2=3
k=3 -> 1+5=6
k=4 -> 1+14=15
k=7 -> 1+714=715

For k=13, you missed 1235+495726=496961

There are none for k=14,15,16,17,18,19,20,21. There are almost certainly no more solutions for puzzle 284:

The probability that a random integer N is of the form AB(A+B) is no more than 1/N^(1/6):

For a given A, the probability that N is of the form AB(A+B) is no more than 1/N^(1/2), and there are N^(1/3) possibilities for A given that A<=B.

Therefore the expected number of solutions with k>=13 is less than the sum of the reciprocals of the 6th roots of the primorials for k>=13.

...

Choose a random integer N. The probability that N is of the form AB(A+B) is less than 1/N^(1/3).

Let s_n be the nth primordial. So s_1=2, s_2=2*3=6, s_3=2*3*5=30 and so on. Let p_n be the nth prime.

Hand-waving claim: "The probability that a random large integer N is of the form AB(A+B) is no more than 1/N^(1/6)"

Hand-waving proof: "For a fixed A, the probability that N is of the form

AB(A+B) is no more than 1/N^(1/2)." Further, with A<=B, there are fewer than N^(1/3) possible values of A. QED.

Now, s_19=7858321551080267055879090>=10^24. Further, for n>=19, p_n>=64. Therefore for n>=19, s_n>=(10^24)*64^(n-19). Therefore, s_n^(1/6)>=(10^4)*(2^(n-19)).

Therefore "the expected number of solutions to AB(A+B)=s_n with n>=19 is smaller than 1/(10^4)+1/(10^4)*2+...=2/10^4=0.0002.

Also, there are no new solutions with n<100 and A<=1000000.

***

Stinchcombe wrote:

I get the values {495726, 496961, 1235}... with the unlisted one being

5*13*19+17*23*31*41=2*3*7*11*29*37

no affirmative results for k=14, 15 nor 16 with exhaustive testing

 

The approach I used was to divide the set of primes into subsets, calculate the product within the subsets.  If a+b=c with positive numbers a,b,c, then c has to be the maximum of the three candidate products.  I used Maple to enumerate all subsets and then just test the equality a+b=c.  This is unfeasible for larger sets, so other approaches would be necessary, for instance for the goal of 50 primes.

***

So, it seems that no more solutions will ever come...

***

 



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