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Difference between revisions of "2017 JBMO Problems/Problem 1"

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== Solution ==
 
== Solution ==
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<math>x_1 x_2 + x_3 x_4 =  x_5 x_6</math>
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Every set which a solution is of the form  <math>Y_k = \{k, k+1, k+2, k+3, k+4, k+5\}</math>
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Since they are consecutive, it follows that <math>x_2, x_4 x_4,  x_6</math> are even and <math>x_1, x_3, x_5</math> are odd.
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In addition, each pair of positive integers destined to be multiplied together can have a difference of either <math>1</math> or <math>3</math> or <math>5</math>.
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So, we only have to consider integers from <math>1</math> up to <math>12</math>. Therefore we calculate the following products:
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 +
A = { 2⋅3, 4⋅5, 6⋅7, 8⋅9 10⋅11, 11⋅12}
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B = { 2⋅1, 4⋅3, 6⋅5, 8⋅7 10⋅9, 11⋅13}
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C = { 1⋅4, 2⋅5, 3⋅6, 4⋅7 5⋅8, 6⋅9, 7⋅10, 8⋅11, 9⋅12}
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D = { 1⋅6, 2⋅7, 3⋅8, 4⋅9 5⋅10, 6⋅11, 7⋅12}
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And we can only have five cases:
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Case 1: <math>x_1-x_2 = x_3 -x_4 =  x_5 -x_6 = 3</math>
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<math> 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7</math>
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 +
Y_2 is the only solution set for this case.
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Case 2: <math>x_1-x_2 = x_3 -x_4 = 1,  x_5 -x_6 = 3</math>
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<math> 1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5</math>
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Y_1 is the only solution set for this case.
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 +
Case 3: <math>x_1-x_2 = x_3 -x_4 =  x_5 -x_6 = 1</math>
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No solution set for this case.
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Case 4: <math>x_1-x_2 = x_3 -x_4 = 1, x_5 -x_6 = 5</math>
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No solution set for this case.
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Case 5: <math>x_1-x_2 = 1, x_3 -x_4 = 3, x_5 -x_6 = 5</math>
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No solution set for this case.
  
 
== See also ==
 
== See also ==

Revision as of 09:14, 21 April 2018

Problem

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.

Solution

$x_1 x_2 + x_3 x_4 = x_5 x_6$

Every set which a solution is of the form $Y_k = \{k, k+1, k+2, k+3, k+4, k+5\}$

Since they are consecutive, it follows that $x_2, x_4 x_4, x_6$ are even and $x_1, x_3, x_5$ are odd.

In addition, each pair of positive integers destined to be multiplied together can have a difference of either $1$ or $3$ or $5$.

So, we only have to consider integers from $1$ up to $12$. Therefore we calculate the following products:

A = { 2⋅3, 4⋅5, 6⋅7, 8⋅9 10⋅11, 11⋅12}

B = { 2⋅1, 4⋅3, 6⋅5, 8⋅7 10⋅9, 11⋅13}

C = { 1⋅4, 2⋅5, 3⋅6, 4⋅7 5⋅8, 6⋅9, 7⋅10, 8⋅11, 9⋅12}

D = { 1⋅6, 2⋅7, 3⋅8, 4⋅9 5⋅10, 6⋅11, 7⋅12}

And we can only have five cases:

Case 1: $x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 3$

$2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7$

Y_2 is the only solution set for this case.

Case 2: $x_1-x_2 = x_3 -x_4 = 1, x_5 -x_6 = 3$

$1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5$

Y_1 is the only solution set for this case.

Case 3: $x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 1$

No solution set for this case.

Case 4: $x_1-x_2 = x_3 -x_4 = 1, x_5 -x_6 = 5$

No solution set for this case.

Case 5: $x_1-x_2 = 1, x_3 -x_4 = 3, x_5 -x_6 = 5$

No solution set for this case.

See also

2017 JBMO (Problems • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4
All JBMO Problems and Solutions
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