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Question from Martha, a parent:

Cody has a collection of baseball cards. Is it possible for him to place the cards into 5 boxes so that exactly 3 of the boxes contain an odd number of cards and the total number of cards in two of the boxes is the same as the total number in the other 3 boxes? Explain why or why not.

Suppose that boxes A,B and C have an odd number of cards then they contain, in total, an odd number of cards. The given statement that exactly three boxes had an ODD number of cards means that boxes D and E each have an even number of cards. Thus the total number of cards must be odd (an odd + an odd + an odd + an even + an even = an odd) However if two boxes contain the same number, in total, as the other three boxes in total, then the total number of cards is an even number a contradiction.

the total number of cards in two of the boxes is the same as the total number in the other 3 boxes?

two boxes contain a total of x; the other three contain a total of x; grand total 2x, an even number.

Penny

 

Martha,

Look at the sentence

Is it possible for him to place the cards into 5 boxes so that exactly 3 of the boxes contain an odd number of cards.

Does it mean that the total number of cards is odd or even?

Now ask yourself the same question about the other sentence:

The total number of cards in two of the boxes is the same as the total number in the other 3 boxes.

Claude

 

I'm assuming that we are not looking for "trick" solutions involving (eg) putting some cards in a box inside another box. (Your child may consider this possibility if his/her teacher has a good sense of humor, but don't say I said to do it.)

Now, boxes A,B,C each contain an odd number of cards; D and E contain an even number. We will also use these letters to refer to these numbers.

Is the total number of cards even or odd? Can they be divided into two equal batches?

Good Hunting!
RD

PS: I assume the phrase "use cases" referred to a solution in which (eg) the following three cases were analyzed separately

total in 3 odd boxes = total in 2 even boxes
2 odd boxes + 1 even = 1 odd + 2 even
1 odd box + 2 even = 2 odd

but as you have (I hope) seen this is unnecessary.

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