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find all positive integer pair of
and inWhere
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hi juantheron
I started with a prime decomposition to see what factors we have here.
2012 = 2 x 2 x 503.
So there's not going to be a long string of sequencial factors that lead to 2012.
But every positive whole number is the second number in that line of Pascal's triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
........................
In other words 2012 = 2012!/(1!.2011!) so n = 2012, r = 1 is one solution.
The other is at the other end of the row. Can you spot it?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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bob bundy I want more explanation for that.
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hi juantherod
I'll have a go.
Pascal's triangle and the combinations formula calculate the same numbers. But, if you haven't heard of Pascal's triangle, don't worry,
I can get to the same place with the combination formula.
Say, n = 20 and r = 7
So you get a long string of numbers one after the other to make the product.
But
503 is a prime so that's it, no more factors.
So you are not going to make a long string combination calculation from that.
But
so every positive whole number can be made if you choose r = 1
Any more answers? Well yes because you can also put n-r = 1, by making r = 2011. Then you get
I'm fairly confident there are no more answers.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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