Problems & Puzzles: Puzzles

Puzzle 469. 5040 JM Bergot sent one more nice observation: One sees that 2*3*4*5*6*7 = 5040 = 7*8*9*10. Q. Find other natural numbers such that they are a product of consecutive numbers in two or more ways.

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Contributions came from Carlos Rivera, J.K.Andersen, T. D. Noe, Frederick Schneider, JC Rosa, Farideh Firoozbakht.

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Carlos Rivera wrote:

After I produced the four first solutions (120, 210, 720 & 5040) I came up to A064224 through google.

After this sequence (and in particular the infinitude character of it, mentioned by T. D. Noe) for me the most interesting remaining questions of this puzzle are:

a) To find the first example of a number having 3 or more representations as products of consecutive integers, supposing this example exist, or to prove that this case is impossible.

b) To discover more non-overlapping numbers, other than the already known ones: 210, 720, 175560, 17297280

210 = 5.6.7 = 14.15
720 = 2.3.4.5.6 = 8.9.10
175560 = 19.20.21.22 = 55.56.57
17297280 = 8.9.10.11.121.13.14 = 63.64.65.66

Overlapping numbers N can be constructed using this formula:

N = n.(n+1).......(Pk-1) = (n+k)...Pk
where Pk is the product of the first k numbers n, n+1, n+k-1

Example: if n=2 & k=3 then Pk=2.3.4=24

2.3.4....23 = 5....24 = 25852016738884976640000

One overlapping number M can be constructed from each non-overlapping number this way:

From: 210 = 5.6.7 = 14.15
You get: 259459200 = 5.6.7.8.9.10.11.12.13 = 8.9.10.11.12.13.14.15

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J. K. Andersen wrote:

1*...*n = 2*...*n = n! is a trivial solution for all n > 2.
Assuming that is disallowed, the solutions are in A064224.

Let a < b be any integers above 1 with a*...*b = m.
Multiply both sides by the numbers from b+1 to m-1:
a*...*b * (b+1)*...*(m-1) = (b+1)*...*(m-1) * m.
This gives the solution a*...*(m-1) = (b+1)*...*m.

Any solution with overlapping intervals can be changed to non-overlapping intervals by removing the overlapping numbers. This may leave the second "product" as a single number, for example in 2*3*4*5*6 = 4*5*6 where
removal of 4*5 leads to 2*3 = 6.

Assuming there must be at least two numbers in both products,
I only found 4 solutions with non-overlapping intervals:
5*6*7 = 210 = 14*15.
2*3*4*5*6 = 720 = 8*9*10. (By removing 7 from Bergot's observation)
19*20*21*22 = 175560 = 55*56*57.
8*9*10*11*12*13*14 = 17297280 = 63*64*65*66.

They are also the only such solutions listed in A064224.

(What about the existence of a number expressible in 3 ways?) I don't know whether there is a number expressible in 3 ways when 1*...*n = 2*...*n is disallowed. (T D Noe wrote the same: "I have no information about that")

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Frederick Schneider wrote:

Non-trivial solutions:

Solution found: 210=5*6*7=14*15
Derive another "overlapping pair" solution by multiplying both sides by: 13!/7! and get: 259459200=5*6*7*8*9*10*11*12*13=8*9*10*11*12*13*14*15

Solution found: 720=2*3*4*5*6=8*9*10
Derive another o. p. solution by multiplying both sides by 7!/6!=7 and get the author's sample solution: 5040=2*3*4*5*6*7=7*8*9*10

Solution found: 175560=19*20*21*22=55*56*57
Derive another o. p. solution by multiplying both sides by 54!/22! and get: 36055954861352887137197787308347629783163600896000000000 = 54!/18! = 57!/22!

Solution found: 17297280=8*9*10*11*12*13*14=63*64*65*66
Derive another o. p. solution by multiplying both sides by 62!/14! and get:
6244042313569035223343873483125151604764341428027427022254596874567680000000000000 = 62! / 7! = 66! / 14!


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There are an infinite number of trivial (or overlapping pair) solutions

For any number which can be expressed as a product of consecutive positive integers > 1, there is a number which can be expressed as the product of consecutive numbers two different ways.

Suppose a number y can be expressed as x*(x+1).

x*(x + 1) = y

Multiply both sides by (y-1)! / (x+1)!

x*(x+1)*(y-1)! / (x+1)! = y*(y-1)! / (x+1)!

(y-1)! / (x-1)! = y! / (x+1)!

So, this number can be expressed two different ways:

The product of all numbers between x and y-1
The product of all numbers between x+2 and y

Example:

6 = 2 *3
Multiply both sides by (6-1)! / (2+1)! = 5!/3! = 4*5 and get:

120 = 2*3*4*5 = 4*5*6

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By similar logic, we can generalize this to any number of factors:

Let y = the product of consecutive positive integers from x to x + k

Multiply both sides by (y-1)! / (x+k)!

The resulting number can be expressed two different ways:

The product of all numbers between x and y-1
The product of all numbers between x+k+1 and y


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The next 4 examples:

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3*4 = 12:
19958400
=3*4*5*6*7*8*9*10*11
=5*6*7*8*9*10*11*12

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4*5 = 20:

20274183401472000
=4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19
=6*7*8*9*10*11*12*13*14*15*16*17*18*19*20

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2*3*4 = 24:
25852016738884976640000
=2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23
=5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23*24

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5*6 = 30:
368406749739154248105984000000
=5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23*24*25*26*27*28*29
=7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23*24*25*26*27*28*29*30

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JC Rosa wrote:

About this puzzle I have only found the following solutions:

120=4*5*6=2*3*4*5
210=14*15=5*6*7
720=8*9*10=2*3*4*5*6
5040=7*8*9*10=2*3*4*5*6*7
175560=55*56*57=19*20*21*22
17297280=63*64*65*66=8*9*10*11*12*13*14

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Farideh Firoozbakht wrote:

The set of solutions are infinite.
I found two following patterns for these numbers.
1. For n>2, (n!-1)! = (n+1)*(n+2)*...*(n!)
(Because both numbers are equal to (n!)!/n!).
Examples:
n=3 (3!-1)! = 4*5*6 (5!=4*5*6)
n=4 (4!-1)! = 5*7*...*24 (23!=5*7*...*24)
n=5 (5!-1)! = 6*7*...*120(119!=6*7*...*120)
...
2. For n>1,

n*(n+1)*...*(n*(n+1)-1)=(n+2)*(n+3)*...*(n*(n+1))

(Consider the two sets {n,n+1,n+2,...,n*(n+1)-1)} and
{n+2,n+3,...,n*(n+1)-1,n*(n+1)}, it's obvious that both
sets have the same product.)

Examples:
n=2 2*3*4*5=4*5*6
n=3 3*4*...*11=5*...*12
n=4 4*5*...*19=6*7*...*20

Two soloutions which aren't of the mentioned forms:

I: 5*6*7=14*15
II: (5*6*7)*(8*9*10*11*12*13)=(8*9*10*11*12*13)*(14*15)
Note that from I we obtain II

Four other solutions which they aren't of the mentioned two forms:


8*...*14 = 63*...*66 => (8*...*14)*(15*...*62) = (15*...*62)*(63*...*66)

( 14! / 7! = 66! / 62! => 62! / 7! = 66! / 14! )


19*...*22 = 55*56*57 => (19*...*22)*(23*...*54) = (23*...*54)*( 55*56*57)

( 22! / 18! = 57! / 54! => 54! / 18! = 57! / 22! )

It seems that your new question (a) is interesting and I also like
to know the first example of a number having 3 or more representations as products of consecutive integers.

The formula N = n.(n+1).......(Pk-1) = (n+k)...Pk ( Pk=n.(n+1). ... .(n+k-1) ) is very interesting.

The two formula that I wrote are special cases of this formula because:

1. If n=1 then Pk = 1.2. ... .k = k! so n.(n+1). ... .(Pk-1) = 1.2. ... .(k! -1) and (n+k). ... .Pk = (k+1).(k+2). ... .k!
hence (k! -1)! = (k+1).(k+2). ... .k! and this is the first formula that I found.

2. If k=2 then Pk=n.(n+1) so n.(n+1). ... .(Pk-1) = n.(n+1). ... .(n.(n+1) -1) and (n+k). ... .Pk = (n+2).(n+3). ... (n.(n+1))
hence n.(n+1). ... .(n.(n+1) -1) = (n+2).(n+3). ... .(n.(n+1)) and this is the second formula that I found.

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