Difference between revisions of "2017 JBMO Problems/Problem 1"
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C = { 2⋅7, 5⋅10, 10⋅15} | C = { 2⋅7, 5⋅10, 10⋅15} | ||
| − | + | One could also construct a graph G=(V,E) with the set V of vertices (also called nodes or points) and the set E of edges (also called arcs or line). The elements of all sets A,B,C will be the vertices. The edges will be the possibles combinations, so the candidate solutions will form a cycle of exactly three vertices. | |
| − | + | In any case, either 3⋅6 or 6⋅9 or 9⋅12 needs to be included in every solution set. So we can only have five (actually only three) cases: | |
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> | ||
| Line 47: | Line 47: | ||
Y_1 and Y_6 are the only solution sets for this case. | 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 = 3, x_3 -x_4 = 5, x_5 -x_6 = 1</math> |
| + | |||
| + | No solution set for this case since no element of set C can be a solution. | ||
| + | |||
| + | Case 4: <math>x_1-x_2 = x_3 -x_4 = x_5 -x_6 = 1</math> | ||
No solution set for this case, as the multiples of three need to be multiplied together. | No solution set for this case, as the multiples of three need to be multiplied together. | ||
| − | Case | + | (This case is actually not realistic and it was included here just for completeness.) |
| + | |||
| + | Case 5: <math>x_1-x_2 = 1, x_3 -x_4 = 5, x_5 -x_6 = 1</math> | ||
No solution set for this case, as the multiples of three need to be multiplied together. | No solution set for this case, as the multiples of three need to be multiplied together. | ||
| − | + | (This case is actually not realistic and it was included here just for completeness.) | |
| + | |||
| − | |||
== See also == | == See also == | ||
Revision as of 00:16, 23 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 is 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}
One could also construct a graph G=(V,E) with the set V of vertices (also called nodes or points) and the set E of edges (also called arcs or line). The elements of all sets A,B,C will be the vertices. The edges will be the possibles combinations, so the candidate solutions will form a cycle of exactly three vertices.
In any case, either 3⋅6 or 6⋅9 or 9⋅12 needs to be included in every solution set. So we can only have five (actually only three) 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 since no element of set C can be a solution.
Case 4: 
No solution set for this case, as the multiples of three need to be multiplied together.
(This case is actually not realistic and it was included here just for completeness.)
Case 5: 
No solution set for this case, as the multiples of three need to be multiplied together.
(This case is actually not realistic and it was included here just for completeness.)
See also
| 2017 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 | ||
| All JBMO Problems and Solutions | ||
