I don't understand what you mean by "Greg wants to get 2010". Do you mean he wants to get the triple 2010,2010,2010? But that can't be right.
3 identical integers greater than 1 are written. one of the numbers is crossed off and replaced by a number which is 1 less than the sum of the other two. This is repeated a number of times. The number crossed off is always chosen so that it is different from its replacement.
Greg wants to get the number 2010 from as many starting triplets a, a, a as possible. One way of course is to start with 2010, 2010, 2010.
What are all the other ways?
Thanks.
hv
Say we start with 7,7,7 and target number is 25:
7,7,7
13,7,7
13,19,7 ; this one has 2 possibilities:
13,19,31 OR 25,19,7
Ok...BUT suppose target number was 50 instead of 25;
then you'd have to continue from the 2 choices above:
BUT which one? The one you hit on first?
I see no formula possibility for this...
unless the erasing is left to right, left to right, ..... ; like
7,7,7
13,7,7
13,19,7
13,19,31
49,19,31
49,79,31
49,79,127
and so on...
Hi Wilmer
I've worked out that if a target number is an even number then the starting triplets should be even. There must be a way so that we can get a large number such as 2010 the easier way instead of trial and error. Imagine how long they will take! Isn't maths supposed to be fun and simple? That's what I've been told at school.......
Cheers, hv