Determine all the sets of six consecutive positive integers

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sqing

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sqing

#1 h Jun 26, 2017, 12:05 PM • 2 Y

Y by adityaguharoy, Adventure10

Determine all the sets of six consecutive positive integers such that the product of some two of them . added to the product of some other two of them is equal to the product of the remaining two numbers.

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Borbas

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Borbas

#2 h Jun 26, 2017, 12:27 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

Hint

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RagvaloD

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RagvaloD

#3 h Jun 26, 2017, 12:36 PM • 2 Y

Y by Adventure10, Mango247

Let these numbers are $a,a+1,...,a+5$
It is true , that $a(a+3)+(a+1)(a+2) \leq (a+4)(a+5) \to a \leq 6$ .
Easy to check $\pmod 3$ ,so both numbers, that are divisible by $3$ must be in same product.
Case $a=1$ : $6*3+1*2+4*5$
Case $a=2$ : $3*6+2*5=4*7$
Case $a=3,4$ : check $\pmod 4$
Case $a=5$ : check $\pmod 5$
Case $a=6$ : $7*8+6*9=10*11$

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Orkhan-Ashraf_2002

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Orkhan-Ashraf_2002

#5 h Jun 27, 2017, 5:47 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Determine all the sets of six consecutive positive integers such that the product of some two of them . added to the product of some other two of them is equal to the product of the remaining two numbers.

It's an obvious problem.In my solution,i checked a lot of case.The answers are:{1,2,3,4,5,6},{2,3,4,5,6,7},{6,7,8,9,10,11}.

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Sapi123

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Sapi123

#6 h Jun 27, 2017, 10:03 PM • 2 Y

Y by Adventure10, Mango247

RagvaloD wrote:

Let these numbers are $a,a+1,...,a+5$
It is true , that $a(a+3)+(a+1)(a+2) \leq (a+4)(a+5) \to a \leq 6$ .
Easy to check $\pmod 3$ ,so both numbers, that are divisible by $3$ must be in same product.
Case $a=1$ : $6*3+1*2+4*5$
Case $a=2$ : $3*6+2*5=4*7$
Case $a=3,4$ : check $\pmod 4$
Case $a=5$ : check $\pmod 5$
Case $a=6$ : $7*8+6*9=10*11$

Why does this inequality hold ?

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sqing

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sqing

#7 h Jun 29, 2017, 10:39 AM • 3 Y

Y by Hopeooooo, Adventure10, Mango247

sqing wrote:

Determine all the sets of six consecutive positive integers such that the product of some two of them . added to the product of some other two of them is equal to the product of the remaining two numbers.

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john0512

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john0512

#9 h Feb 10, 2023, 6:06 PM

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Let the numbers be $a,a+1,a+2,a+3,a+4,a+5$ . If $a\geq 7$ , we have $(a+4)(a+5)< a(a+3)+(a+1)(a+2)$ , contradiction since the left side is the largest possible value of a product and the right side is the smallest possible value of the sum of 2 products.

Bashing all pairings (there are 15 for each $1\leq a\leq 6$ ) gives us that we have $a=1,2,6$ , so our answer is the sets of consecutive positive integers starting with either 1, 2, or 6.

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Adventure1000

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Adventure1000

#10 h Apr 28, 2025, 2:15 PM

Y by

In official solution mistake is there, n(n+3) must be in LHS not in RHS.
Proof: Suppose on the contrary that n(n+3) lies on the right hand side, then right hand side gives us 0 (mod 3). As n and n+3 are divisible by 3 in the solution, we can get that n-1,n+2 are congruent to 2 (mod 3) and n-2,n+2 are congruent to 1 (mod 3). Hence in LHS one term can be congruent to 1 or 2.
1)If one term in LHS is congruent to 1, then the product is in mod 3 either 2*2 or 1*1, then the second term can only be 1*1 or 2*2 respectively. So their sum of products in mod 3 is 2, which is a contradiction
2) If one term in LHS is congruent to 2, then the product is in mode 3 either 1*2 or 2*1 (doesn't matter), then the second term can only be 1*2. So their sum of products in (mod 3) will be 2+2 which is congruent to 1 (mod 3), which is a contradiction

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