Once there was a song. It was maddeningly repetitive and endlessly parodied.

It endorsed “quantity over quality” in gift-giving. It advocated buying for our loved ones not a good book, or a nice pair of pajamas, or even a scented candle, but teeming masses of geese, each one actively laying eggs.

The song proposed, dubiously, that human people could be given as gifts—maids, pipers, drummers, even lords and ladies. The song’s human-trafficking aspect has received little attention to date.

The song pegged the length of Christmas at a mystifying 12 days. The religious holiday, so far as I know, lasts 24 hours, and the secular capitalist holiday lasts roughly two-thirds of the year. But I guess “The One Day of Christmas” wouldn’t make much of a song, and “The 247 Days of Christmas” might bring about the slow, irritating end of human civilization.

The song invited a lot of questions, but one in particular leapt out to the mathematical mind. (Or to the bored mind seeking stimulation—which is fundamentally the same thing.) The question: How many gifts, in all, did my true love give to me?

I first heard this question asked during bar trivia. There were 12 of us, spread across two tables—10 UC Berkeley mathematicians (including my wife), little-old-high-school-teacher me, and an engineer named Neel.

Advantage: mathematicians. Or so you’d think.

After a few moments, the 10 mathematicians (okay, PhD students, but close enough) threw down their pencils in triumph, and showed the following solution. It entails finding the total number of gifts for each day, and then adding up those figures:

It gets you the answer, sure. But it requires sigma notation, a formula for finding the n-th triangular number, and after all that, a fair bit of messy arithmetic.

Neel and I glanced at each other, then shared the solution we’d arrived at independently. It begins by finding the number given of each type of gift:

Then, noting that the second set of 6 gifts has exactly the same total as the first set of 6, we do a little arithmetic to arrive at our final answer:

No sigma notation, no formulas, and no nasty arithmetic. Even the mathematicians agreed it was a better method for finding the number.

“On the twelfth day of Christmas, my true love gave to me,” we later sang, “10 mathematicians, and 2 people who could actually do math efficiently.”

Ah, that’s right! Bringing back some Spanish-class memories of “Dia de Los Reyes,” which I think corresponds to the 12th day of Christmas (where Dec. 25th is the first).

I gave this problem to my 4th graders in my first year teaching. (Minimal prep! I was just barely keeping my head above water planning wise, and I was pretty certain this could fill at least 30-45 minutes on one of our final sugar rush days before December break. I doubt I had even bothered to solve for the exact answer before handing it to my students.)

Some of the kids took 30 minutes, and benefited from some prompting. “Hmm… how can we organize our thinking? What have we used before?” (COME ON, KIDS, WE USE TABLES EVERY DAY.)

My more precocious 9-year-olds solved the problem the same way you did, and did so quickly.

Every 4th grader solved it. Not a single one used sigma notation. Sometimes I really do think the young ones benefit from not being hindered by trained formalities. I have seen 3rd graders solve systems of equations faster than 8th graders. (That said, my 8th graders are MUCH better about using generalizable strategies…)

Yeah, well said. I think that’s exactly the tradeoff: Trained formalities can hinder our speed and intuition, but they give us certainty and generalizability.

I guess, broadly speaking, the best learning path is (1) build local intuition; (2) generalize to an algorithmic procedure; (3) learn when to apply the algorithm, and when to employ a shortcut method.

And yeah, Michael, I fear sigma notation in 4th grade is a dystopia not too far from our own…

The 12th day of Christmas is when epiphany falls, which is the last day of Christmas in the religious calender and the reason that it’s considered unlucky to leave your Christmas decorations up past January 6th

I’m oddly surprised. I didn’t see the whole post at first, and just whipped out a legal pad and solved this the same way you did. Never considered sigma notation or anything of the kind. Was hoping when I saw the tag “elegant solution” that there was something even less calculation intensive and definitely more deeply insightful to a pattern than what I’d done. I’ve done enough problems to see the symmetry right away – like with finding all the squares on a standard chess/checker board, another symmetrical problem, but where there IS a formula for the sum of the first n perfect squares that would make things go even more quickly. The mathematicians didn’t use it. They would have found that the left-hand sigma is the same as n*(n+1)(2n+1) divided by 6, which can be simplified in this case (n = 12) to 26*25 = 25^2 + 25 or 650. Their other sum is 78. Then add and multiply by 1/2 and you get 728/2 = 364. Not fabulously better, but I was surprised to see them actually add up those consecutive squares!

But I didn’t see anything that would be better than what you and I did. There is a progression: the first number is 12, and the numbers increase by 10, 8, 6, 4, 2 before the symmetry kicks in. I suspect that before I’d figure out a clever way to simplify this process, the clever 4th graders would have finished. 😉

Yeah, for the n = 12 case, I think our method is the quickest. (And clever 4th-graders seem to beat me in every aspect of life, so no surprises if they top us here, too!)

For the record, I think the general feeling among the mathematicians was: “I know there’s a formula for the sum of the first n squares, and I know it’s cubic, and maybe involves a six, but I don’t remember what it is and am too lazy to derive it right now.”

I wrote in part, “There is a progression: the first number is 12, and the numbers increase by 10, 8, 6, 4, 2 before the symmetry kicks in” What I meant is that what you add to the previous number increases by that count. So it’s 12 + (12 + 10) + (12 + 10 + 8) . . . + (12 + 10 + 8 + . . . + 2), which is still 12 + 22 + 30 + 36 + 40 + 42 no matter how you slice it.

It is indeed a nice and simple way to have young students understand! I wouldnt say its more efficient than using summations though, cuz if you are calculating for 247 days of christmas then your method will be tiring =( summations will still give the ans with just the board you use =)

Definitely! My and Neel’s method is simpler for a small number of days, but for a large number of days, you’d eventually want to resort to sigma notation. (Our method can be written using sigma notation as well, but I’m pretty sure it will wind up being computationally equivalent.)

Has anyone else ever seen this the way I do, that the SONG is recursive but not gift giving? I guess that would eliminate the need to do all this math and wouldn’t allow the media to rehash, every Christmas season as if it was the first time anyone ever thought of the idea, the “actual” number of presents given by the True Love. 78 gifts in total has been my take on this since pretty much forever, ONE set of each item.
This is from a guy who opted out of his 6th grade Secret Santa, so take me for the Scrooge I so obviously am.

Your reading makes a lot more sense. I think that was the way I first interpreted it, but with time the commercialistic, gift-maximizing spirit of the holiday must have overtaken me.

Wait doesn’t the hockey-stick identity kill this problem? The total number of gifts given on day $n$ is (n+1) choose (2) and you add all of those terms up to get 14 choose 3, or 364

Not sure what any of this has to do with hockey-sticks, but indeed summing (i chose k) for i from 1 through n, for fixed k, gets you (n+1 chose k+1); so sum(i*(13-i)) = sum(i*12 -i*(i-1)) = sum((i chose 1)*12 -2*(i chose 2)) = (n+1 chose 2)*12 -2*(n+1 chose 3) = 13*12*12/2 -13*12*11/3 = 13*(72 -44) = 13*28 = 364. Which is just enough presents for one per day, except February 29th and Ash Wednesday (or any other day of your chosing, but the start of Lent seems an apt day to “fast” from these presents).

The religious holiday does indeed last (or at least used to last, or something) 12 days! http://en.wikipedia.org/wiki/12_Days_of_Christmas

Ah, that’s right! Bringing back some Spanish-class memories of “Dia de Los Reyes,” which I think corresponds to the 12th day of Christmas (where Dec. 25th is the first).

Also, ‘Twelfth Night’ as in Shakespeare’s play. But that’s January 6th, so Christmas is actually the zeroth day of Christmas!

I gave this problem to my 4th graders in my first year teaching. (Minimal prep! I was just barely keeping my head above water planning wise, and I was pretty certain this could fill at least 30-45 minutes on one of our final sugar rush days before December break. I doubt I had even bothered to solve for the exact answer before handing it to my students.)

Some of the kids took 30 minutes, and benefited from some prompting. “Hmm… how can we organize our thinking? What have we used before?” (COME ON, KIDS, WE USE TABLES EVERY DAY.)

My more precocious 9-year-olds solved the problem the same way you did, and did so quickly.

Every 4th grader solved it. Not a single one used sigma notation. Sometimes I really do think the young ones benefit from not being hindered by trained formalities. I have seen 3rd graders solve systems of equations faster than 8th graders. (That said, my 8th graders are MUCH better about using generalizable strategies…)

I’m shocked no 4th grader used sigma notation. Wait until the 3rd revision of the Common Core. I’m sure that will be mandatory.

Yeah, well said. I think that’s exactly the tradeoff: Trained formalities can hinder our speed and intuition, but they give us certainty and generalizability.

I guess, broadly speaking, the best learning path is (1) build local intuition; (2) generalize to an algorithmic procedure; (3) learn when to apply the algorithm, and when to employ a shortcut method.

And yeah, Michael, I fear sigma notation in 4th grade is a dystopia not too far from our own…

Hell yes we kicked ass that trivia! Perhaps this is a sign that we should resuscitate that tradition. Happy holidays my friend!

I’d totally be down for some Red Tomato trivia next semester. (Kipp’s is a little late for me, as it is for all sane citizens of the world.)

The 12th day of Christmas is when epiphany falls, which is the last day of Christmas in the religious calender and the reason that it’s considered unlucky to leave your Christmas decorations up past January 6th

Whoa, it’s unlucky past January 6th? That explains some things about my childhood…

I’m oddly surprised. I didn’t see the whole post at first, and just whipped out a legal pad and solved this the same way you did. Never considered sigma notation or anything of the kind. Was hoping when I saw the tag “elegant solution” that there was something even less calculation intensive and definitely more deeply insightful to a pattern than what I’d done. I’ve done enough problems to see the symmetry right away – like with finding all the squares on a standard chess/checker board, another symmetrical problem, but where there IS a formula for the sum of the first n perfect squares that would make things go even more quickly. The mathematicians didn’t use it. They would have found that the left-hand sigma is the same as n*(n+1)(2n+1) divided by 6, which can be simplified in this case (n = 12) to 26*25 = 25^2 + 25 or 650. Their other sum is 78. Then add and multiply by 1/2 and you get 728/2 = 364. Not fabulously better, but I was surprised to see them actually add up those consecutive squares!

But I didn’t see anything that would be better than what you and I did. There is a progression: the first number is 12, and the numbers increase by 10, 8, 6, 4, 2 before the symmetry kicks in. I suspect that before I’d figure out a clever way to simplify this process, the clever 4th graders would have finished. 😉

Yeah, for the n = 12 case, I think our method is the quickest. (And clever 4th-graders seem to beat me in every aspect of life, so no surprises if they top us here, too!)

For the record, I think the general feeling among the mathematicians was: “I know there’s a formula for the sum of the first n squares, and I know it’s cubic, and maybe involves a six, but I don’t remember what it is and am too lazy to derive it right now.”

I wrote in part, “There is a progression: the first number is 12, and the numbers increase by 10, 8, 6, 4, 2 before the symmetry kicks in” What I meant is that what you add to the previous number increases by that count. So it’s 12 + (12 + 10) + (12 + 10 + 8) . . . + (12 + 10 + 8 + . . . + 2), which is still 12 + 22 + 30 + 36 + 40 + 42 no matter how you slice it.

It is indeed a nice and simple way to have young students understand! I wouldnt say its more efficient than using summations though, cuz if you are calculating for 247 days of christmas then your method will be tiring =( summations will still give the ans with just the board you use =)

Definitely! My and Neel’s method is simpler for a small number of days, but for a large number of days, you’d eventually want to resort to sigma notation. (Our method can be written using sigma notation as well, but I’m pretty sure it will wind up being computationally equivalent.)

Has anyone else ever seen this the way I do, that the SONG is recursive but not gift giving? I guess that would eliminate the need to do all this math and wouldn’t allow the media to rehash, every Christmas season as if it was the first time anyone ever thought of the idea, the “actual” number of presents given by the True Love. 78 gifts in total has been my take on this since pretty much forever, ONE set of each item.

This is from a guy who opted out of his 6th grade Secret Santa, so take me for the Scrooge I so obviously am.

Your reading makes a lot more sense. I think that was the way I first interpreted it, but with time the commercialistic, gift-maximizing spirit of the holiday must have overtaken me.

Reblogged this on tglennb and commented:

Try this at 11:00pm on New Year’s Eve!

I’m surprised the mathematicians were so slow.

It’s just Sigma(n(13-n)), so that’s 13*Sigma(n) – Sigma(n^2), which any mathematician ought to be able to read off as (13×78) – 630 = 384.

Perhaps you would also be surprised at how many mathematicians are terrible with numbers.

I know I was!

I think you ended up with 20 too many… The answer in the post is 364….

This counting problem connects well with Pascal’s Triangle. See the cool video https://www.youtube.com/watch?v=fC8W4s6N9HQ

Wait doesn’t the hockey-stick identity kill this problem? The total number of gifts given on day $n$ is (n+1) choose (2) and you add all of those terms up to get 14 choose 3, or 364

http://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-t-0n-binom-tk-binomn1k1

Not sure what any of this has to do with hockey-sticks, but indeed summing (i chose k) for i from 1 through n, for fixed k, gets you (n+1 chose k+1); so sum(i*(13-i)) = sum(i*12 -i*(i-1)) = sum((i chose 1)*12 -2*(i chose 2)) = (n+1 chose 2)*12 -2*(n+1 chose 3) = 13*12*12/2 -13*12*11/3 = 13*(72 -44) = 13*28 = 364. Which is just enough presents for one per day, except February 29th and Ash Wednesday (or any other day of your chosing, but the start of Lent seems an apt day to “fast” from these presents).

http://www.chaos.org.uk/~eddy/math/sumplex.html (albeit in an eccentric notation)

12 * 13 * 14 / 6 = 364.