Jack and Jill each have marble collections. The number in Jack's collection in a square number.
Jack says to Jill, "If you give me all your marbles I'll still have a square number." Jill replies, "Or, if you gave me the number in my collection you would still be left left with an even square."
What is the fewest number of marbles Jack could have?
(In reply to
re(7): Lost their marbles (full solution) by SilverKnight)
hmm. lets first assume that even infers perfect squares. so.. (assume for everything that x is jack's original amnt of marbles and y is jill's)
{x|x=perfect square} (this is because of what jill said)
{y|y not < 0} (the convo doesn't really set limits to y)
x+y > 0 (it doesn't say that the sum has to be an even square)
x-y= perfect square (according to what jill said, this has to be even)
in addition because (x - y) must be not less than 0 (everyone can understand why this must be), x must be greater than y
because x > y, x cannot be 0 also, because if x is 0, then there is no possible y, with the y cannot be negative restraint., thus x cannot be 0 either.
Well the least number y could be is 0 here because x is already a perfect square (now is abbreviated as PS)and a PS - 0 = PS and PS + 0 = PS, PS > 0. This then means the smallest perfect square that is higher than y (0) would be 1. so x=1 and y=0.
HOWEVER, ravi also said they both had collections, which can make you believe that x > 0 and y > 0
so now, you can assume x-y=0, because if both cant be zero, they can be the same exact number, and if they are the same exact number, jack can give jill everything he got and still end up with an even square, 0. This would then infer x=y.
so the smallest even number higher than 0 is 1, so x=1 and y=1.
to be continued. gotta do sum things. comment and criticize this as much as u want.
solutions for some possibilities part 1
