Eggstreme division
Image by Sister72
Awash with chocolate eggs and gift bags, confectionery delirium sets in and your mind starts to wander... Given you have 20 bags, what is the minimum number of eggs needed so that you have a different number of eggs in each bag? (And here's a hint: go against common sense and try putting all your eggs in one basket!)
This puzzle was taken from the regular Puzzle Corner column for the Gazette of the Australian Mathematical Society. Why not try your hand at the problems in the latest Puzzle Corner?
Solution
A straightforward answer: assuming that you can only have a whole number of eggs and you are allowing an empty bag, you can have zero eggs in the first bag, 1 egg in the second bag, 2 in the third, and so on until you have 19 eggs in the 20th bag. Therefore you need
0+1+2+3+...+19=190 eggs.
A trickier answer: you only need 19 eggs. Again assume that you can only have a whole number of eggs in each bag and that you can have an empty bag. Then the first bag has no eggs. Put bag 1, and one egg, in bag 2, so that bag 2 contains 1 egg. Put bag 2, and one egg, in bag 3, so that bag 3 contains 2 eggs. Carry on until you put bag 19, and one egg, in bag 20, so bag 20 contains 19 eggs — the total number of eggs.
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For some challenging mathematical puzzles, see the NRICH puzzles from this month or last month.
I dervied a general equation for it!!!
0.5 * (x^2+x)
where x= number of bags;
so for 20 bags and u need the total different number of eggs it will be
0.5 * ((20) ^ 2 + 20) = 210 eggs
"not considering the strange assumption of a zero egg bag!! , so bag 1 will have one egg and bag two will have 2 eggs and so on"
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i used calculus to derive it , it's cool :)