Uketivity

The first learning session for the “Ukulele and Ice Cream Therapy” summer program took place last night. Like last month’s ukulele sing-in it was organized by Jean MacLeod of the Greater Syracuse School of Music. I didn’t count heads but I think there were about 15 or so of us there, with an age range covering, I’d guess, about 70 years or so (The girl sitting to my right I would say was no more than 6. She’s a violinist too.) Jean said she knew of five or so more people who were interested but couldn’t make it last night. Most were absolute beginners, a few had more experience. One girl apparently has been teaching herself and was playing “left handed” even though she’s right handed. She tried switching to the standard position for a while and then, invoking the spirit of Libba Cotten, switched back. (On a reentrant tuned uke I suppose playing upside down is less problematic than on guitar.) The one person (Kevin?) who clearly had the most uke experience led the evening, teaching the C, G7, and F chords and getting the group through four songs. The plan in future meetings is to have the more advanced less beginning players do some separate work on slightly less simple songs.

The instruments too were varied in age; one dated from, I think the owner said, the 1940s. Sopranos predominated but other sizes were present including a couple baritones providing their own challenges to their players. I brought my Kala soprano and my Makala tenor, and ended up loaning the latter to a girl who’d come without an instrument.

The learning sessions continue on Wednesdays until June 20 after which we take our show on the road to various ice cream stands through August 15. Wednesdays are also Binghamton Morris Men practice nights so I’ll have to skip one in favor of the other each week.

Working the goals list

Does putting together a cheap prefab Adirondack chair count as “creating cool things”? Probably not. I’m going to claim it for “home improvement” though.

Does taking the fence down in the back yard count as “home improvement”? I’m going to claim it does, but am prepared to face objections on the grounds that keeping it from falling down would have been a harder job.

Replacing the very weird outlet on the patio with a modern outdoor outlet would have counted, and I would have done it if the cable had been grounded. Instead I just opened up the box, said, “Huh. 110V, no ground,” and closed it up again.

Building a catapult with Kenny counts as “creating cool things”.

Under the heading of “exploring new places with Kenny” I assert that Venus, Mars, Saturn, Castor, M3, M31, M97, and M81 count as eight of them. New for both of us. I don’t think I’ve ever looked through a scope that large (12 inches, I think); I know I’ve never seen Saturn’s rings clearly, nor gotten a glimpse of Mars’s polar ice cap before, and those last four represent a quintupling, I think, of my Messier sightings.

Gearing up for June’s music project is borderline, but I’m calling it “making music” even though most of what I’m doing is really just laying the groundwork and getting tools in order.

My cousin Joss

Hm, according to this page Joss Whedon is a descendant of John BABCOCK, who appears to be a son of James and Sarah BABCOCK (or BADCOCK), my 8th great grandparents via another son, James.

Also, this intersection is almost exactly one mile from my house.

 

Syracuke

The Greater Syracuse School of Music organized a “ukulele sing-in” this afternoon. There were about 35 people there by my count, maybe about a third of them absolute beginners, a few accomplished enough to stand up and perform a song or three. That included four members of a reported 20 or so in the Fayetteville-Manlius high school’s ukulele club, who did Leonard Cohen’s “Alleluia”. Other than that there was someone teaching enough basic chords to get through four songs, and then a fifth song for those who already had some uke playing experience. Bernunzio Uptown Music brought a bunch of ukuleles and related stuff from their store in Rochester. (I was good; I didn’t buy anything.) I guess they have the biggest uke selection in upstate New York?

So they’ll be running a series of “Ukulele and Ice Cream Therapy” sessions on Wednesday nights this summer, meeting at various ice cream stands and playing together. Which strikes me as a great concept. It’ll be interesting to see where this goes.

Protected: Selected G’ville photos

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Another MB

Besides the MouseBot, after Free Day I got a MintyBoost kit. Put it together today.

Herbie

We built a bot.

I got a Solarbotics Herbie the Mousebot kit from Sparkfun after last year’s free day and Kenny and I finally got around to putting it together yesterday and today.

It was a fun project and it ended up working great!

There was just one big problem. There was just no way the battery would fit as intended. Even though I think we followed the instructions carefully, the IC in its socket and the sleeve on the tail both stand too high off the PC board to allow clearance for the battery at the angle needed to clip it in.

Finally I gave up and soldered a wired battery clip to the board, clipped the battery to that, and held it in place with tape and rubber bands. It works, but it ain’t pretty.

Fun though.

 Herbie the Mousebot video

Hey a blog

I started another blog.

Enigma 1692

Spoiler for New Scientist Enigma 1692:

Clever logic should enable you to find the nine-figure number that I have in mind. It consists of the digits 1 to 9 in some order, and in the number each digit is next to another that differs from it by one.

In just one case a digit has both neighbours differing from it by one. Furthermore, the solution is exactly divisible by more than three-quarters of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

What is the nine-figure number?

The number can’t be divisible by 10, because it doesn’t end in 0. However, it has to be even if 10 of the first 12 integers are to be factors. So it can’t be divisible by 5 either. Having ruled out 5 and 10, the remaining numbers 1, 2, 3, 4, 6, 7, 8, 9, 11, and 12 must be factors.

If 11 is a factor then the sum of the digits in the odd places (call this so) minus the sum of the digits in the even places (call this se) must be a multiple of 11. In fact, since so+se=45, so and se must have opposite parity so must differ by 11. Therefore so = 28 and se = 17.

1 and 2 can’t both be in even places (or odd); neither can 8 and 9. So, looking at sets of four digits that sum to 17 and contain exactly one of 1 and 2 and exactly one of 8 and 9, there are only three possible even place / odd place divisions: 1349/25678, 1358/24679, or 2348/15679. But the first of these can be ruled out because 7 has to be next to 6 or 8, and likewise the last because 3 must be next to 2 or 4.

So the split must be 1358/24679. 9 must be an odd-place digit; 8 must be next to 9; and since 7 cannot be next to 6 it must be next to 8. Therefore 8 is the one digit for which both neighbors differ from it by one.

For divisibility by 4, and noting that the last two digits must differ by 1, that 1 and 2 must be adjacent, and that 7 cannot be next to 6, the last two digits must be 12 or 56. Then for divisibility by 8 the last 3 digits must be 312, 512, 712, 912, 256, 456, or 856. Of these 312, 512, and 856 are not possible with the odd-even place split required. 456 is ruled out because 5 differs from both 4 and 6 by one. So our number ends with 712, 912, or 256.

If 712: We have to end with 98712. The two preceding digits, and the two before them, can be 43 or 65, but 6543 isn’t allowed, so the only candidate is 436598712.

If 912: Similarly there is only one candidate, 436578912.

If 256: The preceding digit is 1. Then the first five digits must be the group 789 or 987, and 43 or 34. The four candidates are 437891256, 439871256, 789341256, and 987341256.

Of these six candidates only 436578912 is divisible by 7. Divisibility by 1, 2, 3, 4, 6, 8, 9, 11, and 12 has been guaranteed by our procedure, so this is the solution.

 

Enigma 1691

Spoiler for New Scientist Enigma 1691:

The number 3120 is divisible without remainder by each number in the following set: 3, 312, 12, 120, 2 and 20. Each of these numbers will be called a visible proper divisor (VPD) of 3120 because each is visible either as a single digit or as a group of adjacent digits in 3120. (1 and 3120 have been excluded as being improper, in order that any prime number has no VPDs).

ENIGMA represents an odd six-digit number in which all the digits are different. The set of all VPDs of ENIGMA is E, NI, IG, G, GMA, M, MA, and A. What is ENIGMA?

If ENIGMA is odd, then so are all its factors. This means E, I, G, M, and A must be odd digits. None but I can be 1 (the rest must be proper factors). ENIGMA is divisible by 5, so A must be 5. The other three must be 3, 7, and 9.

ENIGMA is divisible by 9, so its digits must sum to a multiple of 9. The sum is 1+3+5+7+9+N = 25+N, so N must be 2.

There are then six possibilities of the form E21GM5. Of these only one, 921375, is divisible by 7. Of course it is also divisible by NI = 21 since it’s divisible by 3. It’s obviously divisible by 25 and divisibility by 3 means it’s divisible by MA = 75. We have to check separately for IG = 13 and GMA = 375. It passes.