Last night, at an auction for the kids’ school, we got an interesting auction ID number, as shown below.
Auction #68 — often bidding, rarely winning.
It was one of the few bidding numbers we saw with a hand-written “bottom bar” below the number. Our friend, Tom, with #12, also had a hand-written bar applied. Thinking about it, I can see why they applied the bar — in the heat of bidding, if I held my number less than upright, I could inadvertently apply a bid auction participant #89.
This week’s GeekDad Puzzle of the Week is simple: If there were 650 different participating parties in the auction, each with their own consecutively determined number, how many numbers required a bottom bar to avoid possible bidding confusion?
Variations in solutions to this week’s puzzle centered around which numbers could be rotated, and into which numbers do they turn. Any answer that was well-supported was entered into the random drawing, but the solution for the puzzle as originally envisioned had the following number mappings:
1 -> 1
2 -> 2
3 did not flip
4 did not flip
5 -> 5
6 -> 9
7 did not flip
8 -> 8
9 -> 6
0 -> 0
Other basic limitations are that numbers would never be printed with leading zeroes (i.e., 100 would not be confused with “001,” as 001 would never be printed) and that both numbers would be printed with a bottom bar. This second limitation is rather elegant — technically, only one of a rotating pair (say, 152 and 251) would need to have a bottom bar!
In any case, the solution in line with above gives some 98 numbers that would require a bottom bar. Those numbers are: 6, 9, 12, 15, 16, 18, 19, 21, 25, 26, 28, 29, 51, 52, 56, 58, 59, 61, 62, 65, 66, 68, 81, 82, 85, 86, 89, 91, 92, 95, 98, 99, 102, 105, 109, 112, 115, 119, 122, 125, 129, 152, 155, 161, 162, 165, 182, 185, 191, 192, 195, 201, 205, 209, 211, 215, 219, 221, 225, 229, 251, 255, 261, 262, 265, 281, 285, 291, 292, 295, 501, 502, 509, 511, 512, 519, 521, 522, 529, 551, 552, 561, 562, 565, 581, 582, 591, 592, 595, 601, 602, 605, 611, 612, 615, 621, 622, and 625.
Congratulations to Justin Weir, who answered correctly and was chosen at random from all of the correct (and any reasonably supported) answers submitted. This week’s fabulous prize, a $50 ThinkGeek Gift Certificate, is on his way to him. Many thanks to everyone that submitted a solution!
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