There is a square cabbage patch.He told his sister that i have a larger patch than last year and hence 211 more cabbages this year. Then how many cabbages I have this year.?
RE: There is a square cabbage patch.He told his sister...
There is a square cabbage patch.He told his sister that i have a larger patch than last year and hence 211 more cabbages this year. Then how many cabbages I have this year.?
RE: There is a square cabbage patch.He told his sister...
I think the answer is wrong. Here is the correct answer. Its a square this year so the area is L * L. Last year it must have been L * oldL. The difference is 211.;L*L-L*oldL 211;L*(L-oldL) 211; OldL is less than current L by say x.L*(L-(L-x)) 211; L*x 211. 211 is a prime number. So either L or x must b 1. As we know that L is larger than x x must be 1 and L must be 211. So L*L 44521.Note: This assumes that L and x have to be natural numbers. We can have many solutions if its not natural numbers.
RE: There is a square cabbage patch.He told his sister...
ya the answer 106*106 is correct. the explanation is simple. the diff. btwn 106*106 and 105*105 is 211.the simple way to solve is (n+1)/2. which implies (211+1)/2 106.
