Difference between revisions of "2017 JBMO Problems/Problem 1"
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'''Method 1:''' Using a graph | '''Method 1:''' Using a graph | ||
| − | 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. So, there should be | + | 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. So, there should be eight such cycles. Only three of them will be the valid solution sets: |
<math> 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7</math> | <math> 2 \cdot 5 + 3 \cdot 6 = 4 \cdot 7</math> | ||
| Line 35: | Line 35: | ||
<math> 7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11</math> | <math> 7 \cdot 8 + 6 \cdot 9 = 10 \cdot 11</math> | ||
| − | Rejected: { 1⋅4, 2⋅5, 3⋅6 }, {3⋅6, 4⋅7 5⋅8}, { 3⋅6, 4⋅5, 8⋅7 }, {6⋅9, 4⋅5, 8⋅7}, {3⋅6, 2⋅7, 4⋅5 | + | Rejected: { 1⋅4, 2⋅5, 3⋅6 }, {3⋅6, 4⋅7 5⋅8}, { 3⋅6, 4⋅5, 8⋅7 }, {6⋅9, 4⋅5, 8⋅7}, {3⋅6, 2⋅7, 4⋅5} |
'''Method 2:''' Taking cases | '''Method 2:''' Taking cases | ||
| Line 69: | Line 69: | ||
'''Case 3:''' <math> |x_1-x_2| = 3 |x_3 -x_4| = 5, |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> | ||
| − | <math>k(k+5)+(k+1)(k+4) \leq (k+3)(k+5) \to k \leq | + | <math>k(k+5)+(k+1)(k+4) \leq (k+3)(k+5) \to k \leq 2 </math> |
| − | Rejected: {3⋅6, 2⋅7, 4⋅5 | + | Rejected: {3⋅6, 2⋅7, 4⋅5} |
No solution set for this case since they were all rejected. | No solution set for this case since they were all rejected. | ||
Revision as of 14:13, 27 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 
since 
(see inequalities below). 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}
C = { 2⋅7, 5⋅10}
In any case, either 3⋅6 or 6⋅9 needs to be included in every solution set. {7⋅10, 8⋅11} cannot be part of any potential solution set, but they were included here just for completeness.
Method 1: Using a graph
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. So, there should be eight such cycles. Only three of them will be the valid solution sets:
Rejected: { 1⋅4, 2⋅5, 3⋅6 }, {3⋅6, 4⋅7 5⋅8}, { 3⋅6, 4⋅5, 8⋅7 }, {6⋅9, 4⋅5, 8⋅7}, {3⋅6, 2⋅7, 4⋅5}
Method 2: Taking cases
Alternatively, we can have five cases at most (actually only three) :
Case 1: 
We just have to look at set B in this case.
Rejected: { 1⋅4, 2⋅5, 3⋅6 }, {3⋅6, 4⋅7 5⋅8}
is the only solution set for this case.
Case 2: 
Rejected: { 3⋅6, 4⋅5, 8⋅7 }, {6⋅9, 4⋅5, 8⋅7}
and 
are the only solution sets for this case.
Case 3: 
Rejected: {3⋅6, 2⋅7, 4⋅5}
No solution set for this case since they were all rejected.
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 | ||
