I hope that this past week’s puzzle was as fun to solve as it was to write. Here is the puzzle as previously published:
Looking through my list of friends recently on my favorite social media platform, I noticed that people tended to fall into a small number of “sets.” That is, if you I knew you through work, I probably did not grow up with you, and if you fell into one of these two groups, I probably didn’t go to college with you.
However, looking at the characteristics of people’s first names, it was easy to construct multiple criteria that multiple had in common. For example, both “David” and “Penny” have not only 5 letters in their names, but also have a repeated letter, where “Debra” and “Adam” meet one or the other criteria, but not both, and “Mike” meets neither.
Below is a list of 4 criteria, and 47 different names. The four criteria individually each split the group exactly in half, and, when applied together, split the list into equal groups — almost.
This week’s Geekdad Puzzle of the Week: Which of the 16 groups defined by these 4 criteria only has two members (and not three like the rest)?
Criteria #1: An odd number of letters in their name.
Criteria #2: Their name sums to a prime (A=1, B=2,.. Z=26, so ADAM=19, etc.)
Criteria #3: Duplicates a letter (like DAVID or SCOTT.)
Criteria #4: Has an even number of vowels (A,E,I,O, and U only.)
List of names:
Given the list above, ADAM is one of the three names that fails Criteria #1, but meets #2, #3, and #4.
The challenge in this puzzle was scoring each name, and then keeping track into which group each name fell. The trick that I used was to track compliance with each criteria as a 0 or a 1, and represent each name’s “score” as a 4-bit binary number. In this way, the name ADAM, above, (failing Criteria #1, but meets #2, #3, and #4) would read as 1110 or 14. Note that the lowest order “bit” is the first criteria, and the highest is the fourth.
Accordingly, the groupings fall as shown below (each name followed by its Criteria #2 sum):
|0||0||0||0||0||SABINE (50), JULIET (77), TOBIAS (66)|
|0||0||0||1||1||MAX (38), HENRY (70), SHAWN (65)|
|0||0||1||0||2||DWIGHT (71), LAUREN (71), PORTIA (79)|
|0||0||1||1||3||BETSY (71), CARMELO (67), WENDY (71)|
|0||1||0||0||4||LILY (58), MATT (54), ANGELA (40)|
|0||1||0||1||5||ALLISON (82), SCOTT (77), PENNY (74)|
|0||1||1||0||6||TODD (43), JACQUELINE (97), MICHELLE (67)|
|0||1||1||1||7||CANDACE (31), CHRISTIAN (101), MILLICENT (97)|
|1||0||0||0||8||NORA (48), HUNTER (86), LOUISA (77)|
|1||0||0||1||9||RILEY (69), WILFRED (77), AMBER (39)|
|1||0||1||0||10||ASTRID (71), KATE (37)|
|1||0||1||1||11||JACKSON (73), DAMON (47), MANDI (41)|
|1||1||0||0||12||SCARLETT (98), ROBERT (78), CHIQUITA (88)|
|1||1||0||1||13||ELEANOR (70), CLAUDIA (51), DAVID (40)|
|1||1||1||0||14||BRITTANY (109), VICTORIA (97), ADAM (19)|
|1||1||1||1||15||CATHERINE (83), OCTAVIA (71), WANDA (43)|
In this format, it is clear to see that the segment with only two names was #10 (Even # of vowels, no duplicated letters, prime sum of letters, and even number of letters) that included KATE and ASTRID. Every other group included exactly three names.
Represented as a Venn Diagram, my friends fall as follows:
Many thanks to everyone that submitted a response, and congratulations to Brad Rochford for winning this week’s fabulous prize: a $50 Gift Certificate from the fine folks over at ThinkGeek. His solution was drawn at random from the set of correct answers submitted this week.