In April, in the first of my age-guessing series of posts, I discussed the magic trick known as the “Age Cards.”

In that post, I focused solely on their use in finding someone's age, but today I'm going to look at the trick from several other perspectives.

As a quick note before we proceed, you should be familiar with the basics of the Age Cards from that original post to follow most of this post.

Binary is important in this trick, because all the numbers are being uniquely identified by whether or not each of the powers of 2 are involved in adding up to that number.

The numbers are distributed on the Age Cards in accordance with their binary equivalent. For example, 53 in binary becomes 110101. The 1s in 110101 appear only in the 1s, 4s, 16s, and 32s places, so 53 will only appear on the cards with 1, 4, 16, and 32 in the upper left corner. This is why you can add the numbers in the upper left corner, 1+4+16+32 in this case, to get the person's number.

What if the range of numbers were smaller? If I limit your choice to the numbers 1 through 7, I would only need 3 cards, since 7 is 111 in binary. However, the trick would seem far less impressive, as it wouldn't take that long to just look through the cards and find a number common to the cards containing the number, and making sure those numbers aren't on the remaining cards.

This seems to suggest that the trick becomes less impressive with smaller sets. However, this is actually only true when you're using cards filled solely with numbers. What if the numbers were represented with something else?

Let's try using months. January will replace 1, February will replace 2, and so on up to December replacing 12. How many cards will we need for the month version? The largest number we'll deal with is 12, which is 1100 in binary, a 4-digit number, so we'll only need 4 cards.

How do we distribute the months? We can test each bit using the AND operation with the first 4 powers of 2 to work out where they go. With Wolfram|Alpha, this is achieved using the BitAnd command.

Testing each of the numbers from 1 to 12 by performing the AND operation against each of the first 4 powers of two, we get the rather odd-looking result of {{1, 0, 0, 0}, {0, 2, 0, 0}, {1, 2, 0, 0}, {0, 0, 4, 0}, {1, 0, 4, 0}, {0, 2, 4, 0}, {1, 2, 4, 0}, {0, 0, 0, 8}, {1, 0, 0, 8}, {0, 2, 0, 8}, {1, 2, 0, 8}, {0, 0, 4, 8}}. What does this mean?

This is the distribution of months we need! Formatted more clearly, the answer would look like this:

January (1) = {1, 0, 0, 0}
February (2) = {0, 2, 0, 0}
March (3) = {1, 2, 0, 0}
April (4) = {0, 0, 4, 0}
May (5) = {1, 0, 4, 0}
June (6) = {0, 2, 4, 0}
July (7) = {1, 2, 4, 0}
August (8) = {0, 0, 0, 8}
September (9) = {1, 0, 0, 8}
October (10) = {0, 2, 0, 8}
November (11) = {1, 2, 0, 8}
December (12) = {0, 0, 4, 8}

In other words, if we consider the various cards as 1, 2, 4, and 8, as we would were we using just numbers, November would appear on cards 1, 2, and 8, while May would only appear on cards 1 and 4. The final lists, then, would look like this (each column represents a singe card):

January        February       April          August
March          March          May            September
May            June           June           October
July           July           July           November
September      October        December       December
November       November