a farmer has 15 pigs and wants to put them into 4 pens and have a odd number in each pen with no pigs left ove?
8 Responses
bharvey47
09 Feb 2010
ignant_slob
09 Feb 2010
Too bad. Can’t be done.
Two odd numbers added together always equals an even number. Four odd numbers added together is simply doing this twice. 15 is an odd number. You can’t get it by adding four odd numbers together, then.
Dav M
09 Feb 2010
not possible
TheMathemagician
09 Feb 2010
The farmer can create three very small pens and put one pig in each.
Then the farmer can create a large pen, which contains the three small pens inside of it, and put twelve pigs inside the large pen but outside each of the smaller pens.
Then the small pens each contain 1 pig, and the largest pen contains 15 pigs (the 12 pigs outside the three smaller pens, and the 3 pigs inside the smaller pens, which are also inside the large one).
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There’s no way to solve this problem without having some pens contained in another or having pens overlap in some way like this, because the sum of four odd numbers can’t be odd.
Leaving a pen empty won’t work, because then that pen contains 0 pigs, and 0 isn’t an odd number.
chapelgirl26
09 Feb 2010
3 in one, 5 in the second, 7 in the third, and the fourth is empty. You can’t split an odd number by an even number without having one even number (ex. 5 = 2+3 or 4+1. There is an even number in both solutions… whereas 5= 3+1+1. Since there are an odd number of numbers adding up, they are all odd numbers adding up to the odd number.) I think I just came up with a new (yet stupid) rule in math. Lemme announce it to the world and make millions! Just kidding.
Wakey
09 Feb 2010
he’s going have to buy another pig then
Farhan R
09 Feb 2010
YOUR MOTHER IN BED.
Kylie
09 Feb 2010
This is impossoible, it may not be done.
When 2 numbers are added they equal an even number. Fourn odd numbers being added is doing it twice. The number 15 is unequal. You cannot get it by adding 4 odd numbers together.

+——————–+———+
*…. pen 1…….* pen 3 *
*…+———+….*……….*
*…* pen 2 *….+———+
*…+———+….* pen 4 *
*………………..*……….*
+——————-+——–+
Pen 2 = 1 pig
Pen 3 = 5 pigs
Pen 4 = 5 pigs
Pen 1 = 4 pigs which also includes pen 2 for 5 pigs