Problems & Puzzles: Puzzles

 Puzzle 433. Three consecutive integers Enoch Haga observed that are 3 consecutive integers that contain 5 consecutive prime factors 20 = 2^2*5 21 = 3*7 22 = 2*11 Carlos Rivera have found other trio of consecutive integers involving 7 consecutive prime factors: 440 = 2^3*5*11 441 = 3^2*7^2 442 = 2*13*17 Q1. Can you find another trio using more than 7 consecutive prime factor? What if we consider not all the prime factors but only part of them? Carlos Rivera found a trio of consecutive integers using 14 consecutive prime factors (from 2 to 43): ```498936009 = 3*11*13*17*37*43^2 498936010 = 2*5*19*23*29*31*127 498936011 = 7*41*53*32801``` Q2. Can you find another trio using more than 14 consecutive prime factor, considering only part of the prime factors of the trio? What if we only observe only the largest prime factor for each consecutive integer? 20 -> 5 21 -> 7 22 > 11 Here we have 3 consecutive integers having 3 consecutive largest prime factors. As a matter of fact there are several of these trios, for example this one calculated by Carlos Rivera: ```455941654 = 2*7*29*103*10903 455941655 = 5*13*643*10909 455941656 = 2^3*3^3*193*10937``` The only one example of four consecutive integers of this type is this one: 4 -> 2 5 -> 5 6 -> 3 7 -> 7 Q3. Can you find another set of k>=4 consecutive integers using k consecutive largest prime factors?

Contributions came from Jan van Delden & Frederick Schneider

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Jan van Delden wrote:

Q2:

Maximum p[i]=47, the smallest being:
5163068910 = 2*3*5*31*41*43*47*67
5163068911 = 7*11^2*13*19*23*29*37
5163068912 = 2^4*17*18981871

Maximum p[i]=53, the smallest (only one in this range):

794010643699 = 7^2*13*37*191*233*757
794010643700 = 2^2*5^2*23^2*53*173*1637
794010643701 = 3*11*17*19*29*31*41*43*47

I found the following article on the Internet, baring to this problem, available at JSTOR:

The Prime Factors of Consecutive Integers
D. H. Lehmer
The American Mathematical Monthly, Vol. 72, No. 2, Part 2: Computers and Computing (Feb., 1965), pp. 19-20

In it a product of the form n(n+1)...(n+k-1) is considered, such that there exists a k depending on p[t] such that no divisor of this product is smaller then p[t]. So a slightly different approach. It also includes a table in which the maximum Mp[t] is printed above which the divisors are >p[t], for h=2..6 and the primes until 41 (t until13).

Q1:

For k=3:
p=17 M=440, so a solution to Q1, as given already.
p=19,23 M=2430, factor 7 is missing, (and 23), way below my search limit, so no solution exists for Q1
The argument above also applies for the other p<=41, so there are no solutions for Q1 with p<=41. In fact the given maxima increase so 'slowly' that finding another solution to Q1 seems a futile attempt.

Q3:

Although the aforementioned table has no direct baring on the given problem checking the given bounds M shows that there are no other solutions for k=4,5,6 with small primes <=41. In this sense the given solution for k=4 is the only one.

Fred wrote:

Q2:Due to the Chinese Remainder theorem, there will always be a solution.
To find the minimum solution, for n primes (beginning with 2),
one just needs to iterate through 3^(n-2) cases to find the lowest
number. 2 and 3 are already present in the three numbers. There are 3
ways each to place the other n-2 primes.

Q3 for k = 4
Solution: 10=2*5, 9=3^2,8=2^3, 7=7
Solution!: 8=2^3, 7=7,6=2*3, 5=5)

Solution: 161054913158291=223*4259*10333*16411,
161054913158290=2*5*229*293*14621*16417,
161054913158289=3^2*353*601*5153*16369,
161054913158288=2^4*19*3469*9323*16381

Solution: 19125003452051570=2*5*31*59*457*28279*80911,
19125003452051569=17*47*5393*54851*80917,
19125003452051568=2^4*3*79*5653*11027*80909,
19125003452051567=7*1093*3209*9629*80897

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