Difference between revisions of "2017 JBMO Problems/Problem 1"
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<math>x_1 x_2 + x_3 x_4 = x_5 x_6</math> | <math>x_1 x_2 + x_3 x_4 = x_5 x_6</math> | ||
| − | Every set which a solution | + | Every set which is a solution must be of the form <math>Y_k = \{k, k+1, k+2, k+3, k+4, k+5\}</math> |
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, exactly two of the six integers are multiples of <math>3</math> and need to be multiplied together. Exactly one of these two integers is even (and also the only one which multiple of <math>6</math>) and the other one is odd. |
| − | + | Also, 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>. | |
So, we only have to consider integers from <math>1</math> up to <math>12</math>. Therefore we calculate the following products: | So, we only have to consider integers from <math>1</math> up to <math>12</math>. Therefore we calculate the following products: | ||
| Line 28: | Line 28: | ||
Case 1: <math>x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 3</math> | Case 1: <math>x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 3</math> | ||
| + | |||
| + | <math>k(k+3)+(k+1)(k+4) \leq (k+2)(k+5) \to k \leq 3 </math> | ||
We just have to look at set B in this case. | We just have to look at set B in this case. | ||
| Line 36: | Line 38: | ||
Case 2: <math>x_1-x_2 =1, x_3 -x_4 = 3, x_5 -x_6 = 1</math> | Case 2: <math>x_1-x_2 =1, x_3 -x_4 = 3, x_5 -x_6 = 1</math> | ||
| + | |||
| + | <math>k(k+3)+(k+1)(k+2) \leq (k+4)(k+5) \to k \leq 6 </math> | ||
<math> 1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5</math> | <math> 1 \cdot 2 + 3 \cdot 6 = 4 \cdot 5</math> | ||
Revision as of 14:02, 22 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 is a solution must be of the form 
Since they are consecutive, it follows that 
are even and 
are odd.
In addition, exactly two of the six integers are multiples of 
and need to be multiplied together. Exactly one of these two integers is even (and also the only one which multiple of 
) and the other one is odd.
Also, 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⋅1, 4⋅5, 8⋅7, 13⋅14, 10⋅11}
B = { 1⋅4, 2⋅5, 3⋅6, 4⋅7 5⋅8, 6⋅9, 7⋅10, 8⋅11, 9⋅12}
C = { 2⋅7, 5⋅10, 10⋅15}
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: 
We just have to look at set B in this case.
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 since no element of set C can be a solution.
See also
| 2017 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
