clock puzzler

mathematrucker

Last week's puzzler (divide the clock with two lines so the sums of the numbers in each segment are all equal), is a special case of what might be an interesting math problem: for which positive integers m and n is it possible to divide an m-hour clock with n lines (none of which have 1 through m as an endpoint) so the sums of the numbers in each segment are equal? The puzzler shows it can be done when m=12 and n=2.

Mechaniker

∀ m>2, ∃ n = FLOOR( (m-1)/2 ) that satisfies your math problem.

Hokiedad

As long as there is an even number of integers from 1 to n, there will be pairs of integers whose sum is equal to n+1 (e.g., 12+1, 11+2, 10+3...). So diving them into equal sums becomes trivial. A clock face can be divided into three equal sums or six equal sums.

mcolvin

All the foregoing is much too complicated for me. Maybe it amounts to the same way that I figured it out, but the obsfucation obliterates the similarity. Here's my take:

Given that the sums of the sections must all equal, I thought it prudent to find the outside parameters of those sums. The total of the digits on the clockface is 78.

If you draw 2 straight lines on the clockface, they will either intersect and create 4 sections; or not, and create 3 sections.

So, the next step would be to find out what the sums of the numbers in the sections

mcolvin

Obvously, I am not adept at email communications. I don't even know what I did to get kicked out of the comment box.... Moving on:

So, the next step would be to find out what the sums of the numbers in each of the sections would be. If there are 4 sections, then each section would have 78/4 as its sum. If there are 3 sections, then each section would have 78/3 as its sum.

Lucky for us, 78/4 is not an integer, which narrows the sum down to one value:78/3, or 26.

Then it is just a matter of finding how many ways there are to do this. The obvious place to start is with the 12, and see which numbers have to travel with it to equal 26
The only combination that works is 11+12+1+2=26.

That makes the rest easy. I just went to the 6, since it would have to travel in a different section than the 12 if the lines don't intersect. Turns out that section has to have 8+7+6+5 to =26.

That leaves the middle section of 9+10+4+3=26.

VOILA! I thought it was elegant and cool, and all that other stuff. I thought sure that Ray would be explicating this or a similar approach to the solution. As he didn't, I offer it for your perusal.

I listen to CARTALK on Sundays @ 4 pm MST,on KNAU, FLAGSTAFF, AZ, although I live just down the hill 75 miles away in Cottonwood, Az (Flagstaff is at 7000+, Cottonwood 3600 feet). Thanks. I always enjoy your show when I get a chance to hear it. Marvin

Mechaniker

mcolvin —

VOILA! I thought it was elegant and cool ...

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Cool in mathematics is giving the solution in one short line.

littlemouse

I've been following this one because the puzzler is often my favorite part of the show. It seems to me that no one is addressing the OP's question: is there a generic formulation of and solution to the problem of which the puzzler is a particular instance.

∀ m>2, ∃ n = FLOOR( (m-1)/2 ) just says that for any number m greater than 2 there is a number n that is either equal to or one integer to the left on the number line from half of m-1. It's really just saying that for any number m>2 there is a number that is smaller. I don't see how this addresses the OP's post.

I think that Hokiedad is right that it works with an even number of hours. It doesn't seem to work with an odd number of hours (5,7, etc.).

Mccolvin's method of solution is the one I used, but would be time-consuming with a 100-hour clock.

Seems to me that cool in mathematics is proving the Riemann Hypothesis and elegant is giving the solution in one short line.

bobs1947

Why answer the puzzler here? Everyone will see the answer. Or is only the first answer picked?

littlemouse

Because the fun thing is trying to solve the puzzle, not being the one person who wins the $25 prize.

JPC

If there is no requirement for the straight lines to run all the way across, you can split into two sections of 39 i.e.

Straight line 1: Between 9/10 to the middle
Straight line 2: Between 3/4 to the middle

Area 1 : 10+11+12+1+2+3 = 39
Area 2: 4+5+6+7+8+9 = 39

PaulMarshall

@bobs1947: Or is only the first answer picked.

According to the folks at cartalk, a winner is picked at random from among the correct answers received. I would expect instead that entries are picked at random until a correct one is found. Usually, as in this case, the pool of entries was closed and the winner picked ten or more years ago when the puzzler was first used.

littlemouse

"According to the folks at cartalk, a winner is picked at random from among the correct answers received. I would expect instead that entries are picked at random until a correct one is found."

That's the same thing.

geeta719

Mechaniker: One line solutions do not necessarily always give the complete answer.
In this case, for eg;,
(1) your 1-liner assumes even values for m, but doesn't say it.
(2) n has non-unique values. Your method gives just the max value for n, that sums 2 numbers (max and min of those left) at a time.

Solutions for even m in general: n is the result minus 1 of division by x>m of sum(1 thru m) such that the sum is integrally divisible by x.
So when m=12: n= (78/x - 1); x being 13, 26 and 39 (78 being the trivial case)

Mechaniker

geeta719 —

Mechaniker: One line solutions do not necessarily always give the complete answer.
In this case, for eg.,
(1) your 1-liner assumes even values for m, but doesn't say it.

My equation is valid for all values of m.
Consider m = 15 ☟

According to my equation the clock face can be divided by
FLOOR( (m-1)/2 ) = 7 lines (horizontal lines in this case) giving 8 groups.
⠀⠀⠀⠀⠀⠀⠀15⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀☜ See diagram to left.
⠀⠀⠀⠀14⠀⠀⠀⠀⠀1
⠀⠀13⠀⠀⠀⠀⠀⠀⠀⠀⠀2
⠀12⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀3
11⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀4
⠀10⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀5
⠀⠀ 9⠀⠀⠀⠀⠀⠀⠀⠀⠀6
⠀⠀⠀⠀ 8⠀⠀⠀⠀⠀7
Horizontal lines group the numbers in the fashion
{15}, {14,1}, {13,2}, {12,3}, {11,4}, {10,5}, {9,6}, {8,7}. The sum of the numbers in each group is 15.

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(2) n has non-unique values. Your method gives just the max value for n, that sums 2 numbers (max and min of those left) at a time.

It is a simple matter of getting other groupings. If every other separating line is removed, then the groups become
{15,14,1}, {13,2,12,3}, {11,4,10,5}, {9,6,8,7}. Each group sums to 30.

Finally, if two more lines are removed, one has
{15,14,1,13,2,12,3} and {11,4,10,5,9,6,8,7}, both of which sum to 60.

The original problem did not ask for all sets of lines that divide the clock face into equal sums, but there it is. Frankly, I think the problem is trivial.

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One line solutions do not necessarily always give the complete answer.

They give the most aesthetically pleasing solution. That is why everyone in the universe knows of Einstein's equation E = mc².

littlemouse

Sorry, Mechaniker, "∀ m>2, ∃ n = FLOOR( (m-1)/2 )" still just says "for any number m greater than 2, there is a number n that is less than m".

judge

As I was in the shower this morning listening to your lousy radio show I was forced to hear the theorys of time saving which led to a discussion on the revolution of the clock industry. An idea immediatly leaped into my head of which I would buy the first ten for myself and all my friends and family. A clock radio which immediatly changes the station for a period of one hour when it recognizes the first notes of the Car Talk theme song. I will never have to step out of the shower dripping wet again and lay my hand on an electric appliance at the same time............