Question 1100005: I'm really sorry, I don't study maths in English so I really have no idea where to put this one but please give it a try.
Prove that the product of any three consecutive integers is divisible by 6.
I have discovered that this could be solved with the Erdos�Ginzburg�Ziv theorem but unfortanetly I can't use that one.
Found 2 solutions by CubeyThePenguin, ikleyn:
Answer by CubeyThePenguin(3113)   (Show Source):
You can put this solution on YOUR website! integers: (n - 1)(n)(n + 1)
There will be 1 odd, 2 even numbers or 2 odd, 1 even number, so the product is divisible by 1.
Here are the possible remainders when the number is divided by 3:
0, 1, 2
1, 2, 0
2, 0, 1
There will always be at least one multiple of 3 in the product, so overall, the product is divisible by 6.
Answer by ikleyn(52787)  (Show Source):
You can put this solution on YOUR website! .
Hello, it is really SIMPLE statement, and you do not need to use heavy artillery as that referred theorem.
This statement is for those BEGINNERS who solve their first problems on divisibility numbers
and make their first steps in learning this subject.
Trace attentively my reasons below.
Of any three consecutive integer numbers, at least one is an EVEN number (divisible by 2)
and exactly one is divisible by 3.
These are self-evident facts.
Hence, the product of any three consecutive integer numbers is divisible by 2*3 = 6.
Solved, proved, explained and completed.
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For your education, look into the lesson
- Introductory problems on divisibility of integer numbers
in this site.
It is written specially for those BEGINNERS like you . . .
By the way, at the end of that lesson, you will find numerous links to other lessons on divisibility numbers.
So, you have this happy opportunity to read about the subject in English.
It is really a fascinating reading (!)
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