Leftovers
Not food. Puzzles. Before I start new puzzles for the fall, I want to clear out the old ones from the Spring. There are three on this site that didn’t have solutions posted. Here they are. I will post solutions next week.
To see them click –>
A prime puzzle: Can you create a 3-digit number such that the number and all other numbers formed by premutations of its digits, are prime? (eg, 127 is prime, and 271 is prime, but 172, 217, 721, and 712 are not…) How many such numbers are there? (May 27)
Photographer Puzzle: photographer takes about 15 seconds to line up a group of people and take their photo. Faced with a group of 4 people, and a little extra time, he decides to take their photo in every arrangement possible (this guy always lines them up in 1 row, and never leaves gaps). How long will it take to take all the photos? Guess before you do the math, always more fun that way. 
Four more people come along, so he decides to take pictures of the whole group, again in every possible one-row, no-gap arrangement. How long will this take? (guess first!)
Finally, four more people come along. Now with twelve people, how long will it take for him to take all the photos? (June 3)
Barcelona Barber: A Barcelona barber usually gives good hair cuts, but today he was spectacular. Ten in a row, absolutely perfect. He hires a photographer to take pictures of the pleased clients in different arrangements.The photographer is no dummy. He … wants to get home in a reasonable amount of time, so he asks for some constraints.
It’s good with the barber. “Take them in two rows of five each. Let each man be taller than the man to his left, and let each man be taller than the man immediately in front of him. And take all the arrangements like that!”
The photographer agrees. How many photos will he take? (June 3)
JD2718 
Huh, I just realized an interesting connection between question 3 and the research I’ve been doing — cool :-)
For number 1, trying to count as cleverly as possible:
The only allowable digits are 1, 3, 7, 9. We can’t have all 3s and 9s or all 1s and 7s or the arrangements will be divisible by 3. So we need at least one of {1, 7} and at least one of {3, 9}. All such possibilities are:
{1, 3, 1}, {1, 3, 3}, {1, 3, 7}, {1, 3, 9}
{1, 9, 1}, {1, 9, 7}, {1, 9, 9}
{7, 3, 3}, {7, 3, 7}, {7, 3, 9}
{7, 9, 7}, {7, 9, 9}
So, only 12 possibilities — not bad. Divisibility by 11 lets me wipe out some of them: 319 (and 913), 737 and 979 are all divisible by 11 and so not prime. This leaves me with 9 sets to check:
{1, 3, 1}, {1, 3, 3}, {1, 3, 7}, {1, 9, 1}, {1, 9, 7}, {1, 9, 9}, {7, 3, 3}, {7, 3, 9}, {7, 9, 7}
Now, 301 = 210 + 91 is divisible by 7, so 371 is as well. Similarly 910 is divisble by 7 so 917 is, as is 903 and thus 973. Also 133 = 140 – 7, and 119 = 140 – 21, so those are divisible by 7 as well. I think that’s as far as I get with this gimick, but it’s useful. Now I have only left 4 options, and each has only 3 permutations.
{1, 3, 1}, {1, 9, 9}, {7, 3, 3}, {7, 9, 7}
The first three of these lead to three prime numbers. 977 and 797 are both prime, but 779 = 19*41, so that one is out. (It also tells me I didn’t lose anything by not trying divisibility by 13.) And we’re done: there are 3 such numbers (plus permutations).
I wonder what the largest number of digits is such that there exists such a prime. Also, what this might look like in bases other than base 10.
For some reason, I can’t seem to get my posts to go through on the first try any more. Do you think you could check your spam filters for a solution to the prime puzzle?
Best,
JBL
Sorry about the filters. I need to check and delete more often.
Clever counting deserves much more attention than it gets. I intend to teach much more systematic counting than I have in the past (can’t be a clever counter without being a tolerable regular counter first, or so I claim). Anyway,
OK, now my turn to go think.
As a start, look at base 2 primes.
101, but 110 is not. Why not? Because anything ending in 0 is even. But that means our prime must be made of only 1’s, iow, 2^n – 1. Look familiar?
Base 3?
12. 21. OK, if unimpressive.
102, nope. Once we have a 0, it can go in the 1’s place and give us a multiple of 3.
111. Trivial.
212. 122. 221. Fails.
(112 is no good, because if the sum of the digits is even, in base 3, then the number is even)
Try some 4 digit permutations: {1,1,1,2} and {1,2,2,2} (there are not a lot of possibilities).
1112, 1121, 1211, 2111 (41, 43, 49, 67) fails.
2221 (79, 77, 71, 53) fails.
Try some 5-digit permutations: {1,1,1,2,2} and {1,2,2,2,2}
11122, 11212, 11221, 12112, 12121, 12211, 21112, 21121, 21211, 22111. The first (11122 = 25) fails.
12222, 21222, 22122, 22212, 22221. The first (12222 = 161) fails.
Perhaps for base 3 (12, 21) is the only non-trivial solution. But how would you check? I am calling 111 a trivial solution, but are there larger numbers of the form of the sum from 0 to n of 3^n that are prime?
I like this question, and will keep playing.
Thank you.
“are there larger numbers of the form of the sum from 0 to n of 3^n that are prime?”
No general approach, but the answer is “yes”: 1111111 = 1093 is prime. I suspect, in fact, that there is some behavior similar to that of the base-2 case. The number of 1s would have to be prime, for instance.
And, resorting to my usual tricks (Mathematica and the Encyclopaedia of Integer Sequences) I’ve come up with http://www.research.att.com/~njas/sequences/A028491 , the list of numbers for which 11….11 = (3^n – 1)/2 is prime.
So, bombarding this thread with comments: there might be ways to show that certain non-trivial combinations don’t work. For example, {1, 2, 2, 2, …, 2} with an odd number of 2s will never work because if the number of 2s is 1 mod 4, 12…2 is divisible by 5, while if n is 3 mod 4, 2…21 is divisible by 5.
I don’t have any idea how far arguments like this can be extended.
It might be interesting to hammer a few bases, and see what commonalities fall out. I should have some quiet time for 4 tomorrow.
btw, the spam filters are still catching your comments. Odd strings of numbers, maybe?
Integer sequences is one of my links. I should try it some time.