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Fair and Square (Posted on 2003-11-03) Difficulty: 3 of 5
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?

See The Solution Submitted by Ravi Raja    
Rating: 2.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
part 2 of the mayhem of my previous post | Comment 20 of 25 |
(note: read part 1 first) now lets consider that all of these numbers are perfect squares and when ravi says even squares he meant perfect squares that is a multiple of 2.


(for my third scenario assume that collections can equal 0)this would then make the following restrictions:


{x|x=perfect square & even & integral} (infered by Jill)


{y|y=non-negative and integral} (must be integral for both x and y because of the reason stated in part 1)


x+y = perfect square, integral.


x-y = perfect square, integral, and not less than 0. this then leads you to conclude that x > y


x > 0 (reasons stated in part 1)


with the following restraints, x = 4 (the smallest perfect square > 0) and y = 0


now like scenario 2 in part 1, we must now assume that collections does mean x and y are both more than 0. so like in part 1 we can also assume than x-y = 0, so x=y.


so, with these restraints, x= 4 and y= 4


  Posted by Victor Zapana on 2003-11-03 19:20:53

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