Difference between revisions of "2017 JBMO Problems/Problem 1"
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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. | 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. | ||
| + | |||
| + | In addition, exactly two of the six are multiples of <math>3</math> and need to be multiplied together. Exactly one of these two is even (and also the only one which multiple of <math>6</math>) and the other is odd. | ||
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>. | 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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D = { 1⋅6, 2⋅7, 3⋅8, 4⋅9 5⋅10, 6⋅11, 7⋅12} | D = { 1⋅6, 2⋅7, 3⋅8, 4⋅9 5⋅10, 6⋅11, 7⋅12} | ||
| + | |||
| + | Either 3⋅6 or 6⋅9 or 9⋅12 needs to be included in every solution set. | ||
And we can only have five cases: | And we can only have five cases: | ||
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<math> 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7</math> | <math> 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7</math> | ||
| − | Y_2 is the only solution set for this case. | + | Y_2 is the only solution set for this case. |
| − | Case 2: <math>x_1-x_2 = x_3 -x_4 = | + | Case 2: <math>x_1-x_2 =1, x_3 -x_4 = 3, x_5 -x_6 = 1</math> |
<math> 1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5</math> | <math> 1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5</math> | ||
| − | Y_1 | + | <math> 7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11</math> |
| + | |||
| + | Y_1 and Y_6 are the only solution sets for this case. | ||
Case 3: <math>x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 1</math> | Case 3: <math>x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 1</math> | ||
| − | No solution set for this case. | + | No solution set for this case, as the multiples of three need to be multiplied together. |
| − | Case 4: <math>x_1-x_2 = x_3 -x_4 = | + | Case 4: <math>x_1-x_2 = 1, x_3 -x_4 = 5, x_5 -x_6 = 1</math> |
| − | No solution set for this case. | + | No solution set for this case, as the multiples of three need to be multiplied together. |
| − | Case 5: <math>x_1-x_2 = | + | Case 5: <math>x_1-x_2 = 3, x_3 -x_4 = 5, x_5 -x_6 = 1</math> |
No solution set for this case. | No solution set for this case. | ||
Revision as of 09:32, 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
Every set which a solution is of the form 
Since they are consecutive, it follows that 
are even and 
are odd.
In addition, exactly two of the six are multiples of 
and need to be multiplied together. Exactly one of these two is even (and also the only one which multiple of 
) and the other is odd.
In addition, each pair of positive integers destined to be multiplied together can have a difference of either 
or or 
.
So, we only have to consider integers from up to 
. 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}
Either 3⋅6 or 6⋅9 or 9⋅12 needs to be included in every solution set.
And we can only have five cases:
Case 1: 
Y_2 is the only solution set for this case.
Case 2: 
Y_1 and Y_6 are the only solution sets for this case.
Case 3: 
No solution set for this case, as the multiples of three need to be multiplied together.
Case 4: 
No solution set for this case, as the multiples of three need to be multiplied together.
Case 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 | ||
