NOTE: I wrote this essay in 2015, shortly after the events described. As with any sensitive document, I chose to wait several years to publish it, so as to protect the careers, reputations, and fragile egos of those involved (most of all my own). Please read it with the leering disrespect it deserves.
I discovered recently that I can’t multiply 2573 by 389.
Neither can you, I bet.
Really, go ahead. Try it. Grab a pencil and paper. Use the standard algorithm. Give it the level of focus you’d normally give such an operation (so if you wouldn’t usually quadruple-check your work, don’t do it now, you little cheater. We want a fair test). Then, when you’re done, check your answer (you can just type it into Google).
How did it go?
The background: I’m lucky enough to work in a sterling mathematics department. Our school routinely sends mathematicians to Oxford and Cambridge. We’ve got six kids taking the highest level of IB Mathematics; there are only 300 students doing that in the world. It is, in short, a really impressive department.
So the teachers ought to be able to multiply integers, right?
I tried the problem myself, and failed. An addition mistake was my kryptonite. The department was 0 for 1.
I threw the problem to one colleague, who lofted a skeptical eyebrow at my ineptitude, then tried and stumbled himself. 0 for 2.
Another colleague showed up. When he too made a mistake in his computation, he didn’t believe it, suggesting that perhaps I’d entered it into the calculator wrong. I laughed and compared him to a politician demanding a recount after a ten-point defeat.
0 for 3.
A third colleague, a brilliant scholar upon whom I lean for advice and wisdom, couldn’t do it either, and wondered if we were playing some trick, such as switching to a base other than 10.
The department was 0 for 4.
We gathered to lick our wounds. The problem, we agreed, was that there are just too many chances to slip up, too many ways to err. You’ve got to perform 12 multiplications and 15 additions, and you need to carry out most of the multiplication steps while holding a number from the last multiplication step in your memory.
That’s hard! It feels less like an efficient problem-solving method than like a task cooked up in a psychology lab to stretch the limits of short-term memory.
Or maybe this is all just sour grapes.
I told another colleague of our struggles. He shook his head. “Pitiful,” he teased.
“C’mon,” I said. “There’s obviously some number of digits that’s so large, everyone will make a mistake. No one will be able to correctly carry out, say, a 15-digit number times a 14-digit number.”
“Of course I can,” he said.
So I grabbed a piece of paper and set him a problem: 15-digits by 14-digits.
Thirty seconds in, he knew he’d bitten off more than he could chew. (To be fair, I was trash-talking loudly in his ear, to his justified annoyance.) Dutifully, though, he carried it out, eventually arriving at an answer.
A spectacularly wrong answer.
I suspect that if I’d given him the original problem at the start, he’d have managed it correctly. But now his swagger was shaken. He tripped all over himself, and erred on the 4-digit by 3-digit as well.
We were 0 for 5.
And you want to know the most damning fact, the one that had prompted me to launch this crazy multiplication witch-hunt in the first place? I’d drawn this question, this exact question, from a school test we’d given the previous November.
We expected our youngest students to do what we, apparently, could not.
With the dust settling, and our egos lying in rubble, I looked for takeaways. What does it mean that this question looks like a matter of brainless automaticity, and yet smart people with fancy math degrees can’t answer it?
At the extreme, you can conclude that teaching standard algorithms is pointless. After all, we’ve all got calculator apps in our pockets, and evidently, even the professionals can’t get the standard algorithms to produce a correct answer. Perhaps they’re no longer worth teaching.
Or you can run to the opposite extreme: the standard algorithms must be taught, and taught carefully. As our embarrassment shows, this computational stuff takes more effort and focus than you might expect. In our drive for conceptual depth, we mustn’t take computational fluency for granted.
To me, though, this isn’t a pro- or anti-algorithm parable. Instead, the whole episode is a master-level seminar in something simpler:
We need to walk in our students’ shoes.
It’s no big deal that we erred on the problem, and it’s forgivable that we underestimated its difficulty. (You probably did, too, before reading this post.) But these reasonable, well-meaning mistakes added up to something rather awful: us forcing our students to do something unnecessarily difficult and unenlightening, for no particular reason.
A class is like a mountain path. We set the trail that our students must hike. We choose the direction; we draw the map; and we leave the markers along the way so that they don’t get lost. Where they’ll face gaps, we build bridges. Where they’ll struggle to climb, we build stairs. Where they may stray, we build railings (or nets to catch them).
We don’t have time to double-check every inch of terrain. But if we lead them up a steep, hard hill, then we’d better ensure that it’s worth the struggle. It’s easy to send kids down a fruitless path, especially when you’ve had years to forget the rocks and obstacles.
It’s easy to ask a question that looks simple, and that you can’t actually do yourself.
So that’s my takeaway, a simple reminder that a teacher is a trail guide. He needs to test the path periodically, even if—no, especially if—he’s such an expert now that he can no longer remember what it’s like to walk the road for the first time.
But I’m sure you’re still wondering: Did anyone carry out the multiplication successfully? Yes, indeed, one person: a trainee Latin teacher named Emily who last took math when she was 15.
82 thoughts on “I can’t multiply 2573 by 389.”
> You’ve got to perform 12 multiplications and 15 additions
Ah, but only using the “official” algorithm. In the case of 2573*389, my personal route was
2573*389 = 2573*400 – 25730 – 2573 = 2*2*2573*100 – (25730 + 2573), i.e. 3 additions, 1 subtraction and two bit-shift (plus a look at the number 389). Which is probably the way I would have computed it at school, too …
Maybe we should set maths problems of the kind “perform this computation using the minimum number of elementary operations”, which sounds like more fun for everyone involved.
I did exactly like you, but still got it wrong by one digit. I could bring excuses (I was surfing Facebook at 3am in the toilet), but the point of the article stands. It is easy to make a small mistake while performing “elementary” operations
I used a very similar algorithm, (2573 * 400) – (20 * 2573) + 2573)…. oh that’s what i did wrong, 400 – 389 is not 19. oh well.
I just did long multiplication and got the right answer
2573*389 = 2573*400 – 2573*11 = 1029200 – 28303 = 1,029,200 – 20, 000 – 8,000 – 300 – 3 = 1,000,900 – 3 = 1,000,897. There’s a neat trick to calculate any number multiplied by 11. For example 11*12 = 1 1+2 1 = 132. 11*29 = 2 2+9 9. Since 2+9 is 11, the 1 is carried over to the first digit 2, thus, we get 11*29 = 319. I used the same truck to calculate 2573*11 = 2 2+5 5+7 7+3 3 = 2 7+1 2+1 0 3 = 28303. It’s also faster to perform simple multiplication from left to right. For example, we I multiply 2573 by 400. I can break it down as first do 2573*4 then take the result and add 2 0s to it. 4*2573 is equivalent to 4*2000 + 4*500 + 4*70 + 4*3 = 8+(2 carried from 4*5 = 20) = 10 | 0+(2 carried from 28) = 2| 8+(1 Carried from 12) = 9 | 2 = 10292. Finally add 2 0s and we will get 1,029,200! after getting used to these tricks I can calculate 2573*389 in a matter of seconds. I learned all of this from a book called “Secrets of Mental Math”, it’s a very entertaining and useful book.
I wouldn’t have used the standard algorithm that I was taught in school (oh so very long ago), but I DID since that’s what you were asking about. And I got it right. Now I admit I was being EXTRA careful because I had been warned (I even wore my glasses), and if I hadn’t been, I probably would have been more cavalier about it and messed it up too.
I did use the standard algorithm and got it right. But I had been forewarned: I failed the cartoon example by one digit using the shortcut (300 x 72) + 30 x 72 -72… damn!
I recall learning how to do this in primary grades. Place 2573 over 389 aligning the digits as appropriate. Then multiply each number to find each product and finally add the three products together to find the solution. Is that not what we are talking about here… or am I missing something? Thank you.
I’m pretty sure that’s what it is – at least that’s what I did. I learned this method…60 years ago? No calculators to check your work!
Yes, that’s the process.
But the question isn’t can you explain how the process *should* work.
It’s can you actually follow the steps of the process and come up with the correct answer? (I failed.)
Good point about can you come up with the solution. And yes I arrived at the solution and checked it with Google. This was a great exercise. Thank you. As an aside… while working this I could hear the chalk on the board and the smell of wooden pencils… lol.
I used that method, and I also got the wrong answer. The problem was that there were so many cases where you had to do some side math to check whether you had to carry a digit from the tens column, and then remember to add what you carried from the previous time you carried, and there were a bunch of opportunities to make little errors that lost you a couple hundred.
I used lattice and got it right
Ah, the lattice always works well!
Almost correct. You forgot one thing: each product but the first must be shifted one place to the left (basically, multiplied by 10) before the addition. Then it works fine. By the way, I’m a biologist who got it right first try (after agonizing for 30 s to remember how the algorithm went, haha), at the regular speed I suppose I’d use if I ever calculated anything by hand (i.e., as fast as I could). But I think I got lucky, was a bit surprised when my calculator agreed with me –I sure was not confident I hadn’t slipped up somewhere amid all those operations…
I used the standard algorithm and got it right. But I’m always pretty careful and I’m and engineer so I use math on a daily basis. I’m not a teacher, though I greatly admire those who are!
I did use the perfectly typical, ordinary, nothing-special algorithm I was taught in grade school. Sorry, but for me it worked fine.
This may be because I don’t feel especially invested in algebra, so I didn’t feel particularly put on the spot. Not being anxious, I just wrote the problem down and did it with a pencil and it was fine.
I did it in my head via (4×2573)×100 = 10029200, 11*2573 = 28303, 10029200 – 28303 = 1000897. It helped that there were two early zeros in 10029200, and that 29200 and 28303 are so close.
Then I did it with pencil and paper to see if that way was worse, and also got it correct.
This reminds me of Craig Barton’s book ‘How I Wish I’d Taught Maths’. He explains these ‘silly mistakes’, using ‘Cognitive Load Theory’.
The idea is we all have a limited amount of brain space, like computer RAM. It gets used up by everything we’re trying to think about at the same time. These mistakes happen when we’re holding so much in our head that we run out of room and something gets dropped.
So I can do the problem fine because I’m sitting at home in front of my laptop with nothing else to think about. But doing it in the teacher’s common room with people talking to you is impossible, because you’re listening to them, thinking about if you’re going to get it wrong, and so on.
The implication for students is that they can reduce how often they make this kind of mistake by simply practising more. Someone who isn’t very practised will have a lot on their mind, even if they understand how to solve the problem. They’ll be thinking of the overall logic of the problem, and what steps to carry out. So they won’t have as much brain space left to focus on the individual calculations. The more they practice, the less they’ll have to think about the overall problem, and the more brain space they’ll have for remembering to carry the 2.
You say part of what this multiplication tricky is having to remember the carry bits from the previous multiplications. However, the algorithm for multiplication I was taught in school, you write those carry bits down so you don’t have to remember them. You then scribble them out when you don’t need them anymore, so things become quite the mess, but I was able to successfully multiply the numbers. Perhaps I owe my trail guide a thank you.
This also caught my eye in the post, as I also write down the carry digits and cross them out in the next round. Exactly as you describe, I end up with a messy tower of tiny carry digits accumulating over the problem. But that made me think: why don’t we write the carry digits down below, and do the extra additions in the addition phase? You could even leave an extra line at the bottom of the addition block to write down any carry digits that arose there. It also separates the problem into a pure “multiplication phase” and “addition phase”, which I find nice.
An example of my suggestion (hopefully this lines up for everyone):
8537 — 2573 x 9
1462 — the carry digits of the above
60640 — 2573 x 8
1452 — the carry digits of the above
651900 — 2573x 3
12 — the carry digits of the above
222 — the carry digits from adding all the intermediate products
I got the wrong answer using the regular method. Looking it over I now see one mistake but there must be more.
My background is that I was a math major for two years of undergraduate college, but didn’t get a degree in math. (Eventually got a bachelors that I don’t set much store by, in a computer subject.)
I can’t vote in the poll because my company blocks Facebook.
Is age a factor? I immediately went to paper and pencil and worked the problem fairly quickly and was surprised to find, correctly. Of course, I was tipped off it was going to be a challenge which doesn’t make my experiment fair compared to yours. I can literally tell you the year when I noticed my students no longer could compute numbers in their heads – 2001. Before that year my students were trained in computation just like me and they could think with me through their high school mathematics. After that year there were fewer and fewer numbers and computations running around their brains. It was shocking to see and to deal with in the classroom. I am 71 years old having graduated from high school in 1969. No calculators in sight! My first calculator sighting was as a junior in college in 1971 and it was a large square box that only did fundamental computation. Before that time we carried our Chemical Rubber Company book of tables to all math classes and we were constantly using interpolation to gain accuracy. I somehow missed using a slide rule as my engineering peers had to do. How old is your faculty? Were any of you born BC (before calculators)? Interesting and thought provoking read. Thank you!
Someone a (little) older than I am! I saw my first calculator in college as well, the size of a brick. An expensive brick. There was a big fight in a statistics class about whether it was fair that some students could afford calculators and others couldn’t. I would write out the answer in tests as the calculation you would need to do to get the right answer (because I couldn’t afford a calculator). I figured that was what they wanted to know, not whether I could do arithmetic. It ended up going to the head of the department and I WON!
I got my first calculator (an HP-45) as a graduation present from college in 1974. Before that I used tables and slide rules. I was a math major (got as far as an MS in math before switching to CS for my PhD).
I think it’s just the many times you have to “carry” that makes it hard. If you prohibit writing down the carried digit above the place it carries to (“and you need to carry out most of the multiplication steps while holding a number from the last multiplication step in your memory”) that makes it more difficult, for sure. If you are allowed to write down the carried digits – and later erase them, so old carried digits don’t get confused with new ones – the standard algorithm isn’t especially tricky anymore. But it’s inefficient for this pair of numbers, as Andreas points out. It’s a good example in favour of Andreas’ suggestion that we teach students how to find efficient ways to perform the calculation.
Maybe it’s less that you have Ph.Ds than that you’re all more comfortable with arithmetic. I have a Ph.D in math but am quite slow with arithmetic (though having grown up in the precalculator era, I look fast compared to my students), and I got it correct without more then the usual muttering the carries over and over again in my head so I don’t forget them.
Of course, as Sue notes, I was surely biased by the set up. 🙂
I once had a non-Euclidian geometry professor that had trouble counting lines on his figures. The further we go, sometimes the less attention we pay to the “easy” steps. – and, yes, I didn’t get it right the first time, either, but I expected that. I almost always have to correct my checkbook when I balance the statement.
I tried to get this mathematical content out of the curriculum. I did not succeed… My collegeas used arguments like:
– is a part of our culture
– kids like this part of math because they know how to act if they know the algoritm. It gives them a succes experience. (My argument that learning this algoritm did not touch the heart of math learning, did not convince them)
– its on the curricula of all the other countries
That made me think in another way and I asked my selve: ‘How can we make the study of this algoritm usefull for becoming a better mathematician in the future’.
We’re now looking for connections between this algoritm and properties of operations. Especially for operations with rational numbers, (very difficult operations!) you can use a lot of properties instead of carry out a technical algoritm.
Detecting witch properties you use, is a kind of exercise that prepares you to work with letters in a later stage.
It worked fine for me, but I don’t use the standard American algorithm where all carries are put on the top of the working (and can get messy when there are lots of them). I put the carries on the same line as the individual digit products.
To be honest, the big problem at the school level is the mixing of digit multiplications AND carries, having to keep track of everything. This is why I am a huge fan of lattice multiplication, where you just do digit multiplications first and then add. Much easier to get it right that way.
I prefer the grid method too.
What an interesting exercise. I didn’t realise until I tried it how rarely I do long multiplication “by hand” anymore. I’m slightly embarrassed to admit that I had to stop and think for a moment about the procedure. And then I managed to fluff it with an arithmetic error.
Once I learned the lattice method in primary school, I never looked back. It was a bit embarrassing while everyone was just writing numbers and here I was drawing shapes. I’m in my 30s now and I couldn’t do it any other way. I also managed to do this one in one shot! Here’s to dumbing things down and simplying those computational processes!
I got it right but only because I have been helping my 11 year old with her math lately, and have been getting recent practice with the standard algorithm.
In fact last night she had a “challenge” problem which involved computing 4 * 2,268 AND 9,072/7 AND sqrt(1,296). The help she needed was mostly just encouragment, because there were so many computational steps, and so many opportunities to make mistakes, she was intimidated. I explained the (new to her) concept of “partial credit” to her, and reassured her that in the future, if she showed her work neatly, she could probably get 9 out of a possible 10 points for a question like that even if she got the wrong answer. Because even people who are great at math and know what they are doing make mistakes on problems like that.
Were they really teaching square root algorithm or just guess-and-check?
Guess and check, but with systematic guesses.
I got it right, but then I do Word Arithmetic (long division) puzzles for fun. I considered multiplying 2573 by 390 and then subtracting, but that’s not the “standard algorithm.”
I tried to vote; it said “Thank you for voting!” but didn’t appear to record it.
I’m going to riff on this in a different direction. Pre-calculators – society required calculations like this to be done and done accurately.
So its definitely possible. I like thinking about what systems were used to make sure errors didn’t creep in. Everything from “casting the nines” to slide rules.
Casting out nines! Haven’t heard about THAT for a long time! And I still have my slide rule as well as my teensy book of trigonometric functions. When I was volunteering in an elementary school (pre-pandemic), I would bring in my slide rule and the kids were just flabbergasted at it.
I used to curse my grade four teacher who taught us all to “hold” the carry over number on our fingers, or write them above the top number to remember them.Such a babyish way to deal with numbers as an adult. But that lesson happened over 50 years ago. Today I said a quiet “thank you” to Mrs Mellors. BTW, I got it right, even though I’m the sad person who has been wailing here that I just can’t understand the concept of multiplying two negative numbers results in a positive number. You win some you loose some. I’d rather have understood concepts more than mechanical computation. But both are useful. Especially when building a cabin without electronic devices.
What’s -3 times 3? -9.
What’s -3 times 2? -6.
What’s -3 times 1? -3.
What’s -3 times 0? 0.
What’s -3 times -1? Well, what’s the next number in the sequence -9, -6, -3, 0? It’s 3.
Or another way: What’s -3 times 0? 0. But that’s -3 × (1 + -1) = -3 × 1 + -3 × -1 = -3 + ? = 0. So -3 × -1 must be 3.
Since spending a few years as a primary teacher, 20 years ago, I now much prefer the grid method that I learned and passed-on then.
In this case, you draw a grid 4 across and 3 down. You write 2000, 500, 70, 3 across the top squares and 300, 80, 9 down the side. You then work out the mutiplication in each box, and add all the boxes up at the end. Perfect.
Yes, I got this right, but then I am 73. When I was in junior school we were taught the standard algorithm because being able to calculate quickly and accurately was an important life skill – many jobs required it, and some required nothing else. So we practised it a lot, and then used it a lot, in other subjects as well as in maths.
What I couldn’t do was get the poll to work, to find out how other people got on. Perhaps that is also because I am 73. It would be interesting to correlate success with age.
I did it correctly, but I also acknowledge the point of the thought experiment. if I threw in a few more digits, then it would probably too difficult. I’d imagine that number is different for different people. I’ve always done a lot of work by hand because I lost a lot of calculators when I was younger, and when I teach, sometimes I leave my calculator at home or in my office.
I am pleasantly surprised I got it right. (My computational abilities have degraded with age, alas.) There is some merit to knowing the algorithms well and to having computational skills. I struggle with figuring out how much emphasis that deserves. See, for instance, the anecdote about Serge Lang told here:
(You’ll have to search for “Serge” in the text, I don’t see how to link directly to the relevant paragraph.)
I’m a (pure) mathematician at an unselective American university. It’s very difficult to teach differential equations to students who can’t reliably complete the square (with simple numbers), or solve a linear equation with small, integer coefficients in their head.
I got it right, but I used the matrix method. I learned the matrix method after one of my students used it a few years ago, and I use it whenever I have big multiplication questions, including expanding crazy polynomials, sometimes.
I answered in the poll, but when I look at poll results it says zero responses.
Anyway, I did it the way I learned in elementary school, and I got the right answer.
Looking over other responses and seeing they’re overwhelmingly “I got it right”, I think perhaps you might look at them and say, “Bah, the majority who got it wrong don’t want to admit it and the few who got it right want to brag about it”, especially if the poll’s not working. I don’t think that’s what’s going on, or at least I don’t think the bias is that big. I am fairly shocked you found only one person able to do the arithmetic, and I think a broader survey conducted carefully would show most people who do mathematical or numerical things for a living — I include scientists, engineers, accountants, etc. — can multiply 4 digit numbers pretty reliably. I don’t know if you just got unlucky with your informal survey group, or if there was some confounding factor going on, or, I suppose one other possibility, it’s just the scientists and engineers and accountants who are answering “yes”, and the mathematicians, who may think a lot about numbers (or not) but don’t necessarily calculate with them much, are more out of practice.
I’ve been reading this exchange on my phone along with the replies. I’ve looked and don’t see any survey, so maybe desktops are seeing the survey.
What did you consider to be the highest level of mathematics?
I got the right answer, as I expected, ‘cos I’m a Secondary Maths teacher at Spain (grades 7 to 12), and I usually teach this algorithm, more frequently in the context of decimal numbers. My question is: did you and your colleagues fail by not remembering the algorithm steps or because you made trivial wrong products? Because those two are very differente mistakes, and I’bet teachers consider these mistakes differently when evaluating pupils.
Oh, and I also answered the poll and saw the zero responses.
there is no question that this can be a tedious process if you do it using the late elementary school algorithm. when i think about some of the ways that i do computataions, for instance that 389 = 400 – 11 = 400 – 10 – 1, then i know that i can use the distributive property if i want. as a high school math teacher for over 20 years, i have found that students generally seek instant results and are not good thinking arithmetically or performing arithmetic computations without relying on technology and that has led to them having a lesser understanding of mathematical concepts in general. this is not soething i would expect them to do in their heads, but it does not have to be as complex or overanalyzed as it seems to me.
Oof, I tried it and made *two* mistakes.
I did it in my head and got it right, using similar methods to those above (389=400-11). To be fair though, mental arithmetic was always my thing, and now that I’m a teacher, I occasionally use it to inspire curiosity (how did he do it so fast? What does he know that we don’t?). Depending on how that conversation goes I either turn it towards coming up with better methods (work smarter not harder), or the fact that practice, though often underestimated, really is what makes the difference.
I did not use the standard algorithm, but did what I normally would do (treating 8 as 10-2 and 9 as 10-1). Despite the simplification of the problem, I did make one arithmetic mistake in my mental arithmetic and got an incorrect answer: in doing 7+2+carried 1, I forgot about the 2 and just did the carried 1. Had I followed my normal procedure of checking my work, I would have found the mistake, but I decided to cut out the checks to more closely mimic the experiment you wanted done.
I agree that giving students large arithmetic problems to work on is not a very good test of their understanding.
Editing to add: I did simplify the arithmetic by cancelling a + 25730 and a -25730, so in effect I did 400-10-1, like many others here.
as a math teacher i wouldn’t necessarily give them a calculation like that just to do,, but if it is part of what comes up in a problem to solve they should be able to work it without much problem.
First try. Seems trivial.
I’m curious what the ‘standard algorithm’ is. I did it on paper, using a method I haven’t used in at lest 30 years (I did a ballpark check in my head: 2500 x 400). I’m slightly smug to have gotten it right, which I didn’t expect!
The method I used is the one I learnt in Germany: you start by multiplying the first number with the last digit of the second, (first ones then tens then hundreds). On the next line, you shift across to multiply the next digit etc. Any number I carry I write at the top of its slot.
When I’m done, I add up the numbers; any number I carry over is written at the bottom of the bottom line.
This is a good time to introduce a German word: Flüchtigkeitsfehler. This is the type of mistake you make when your attention wanders momentarily – a fleeting mistake, if you want – where you know the thing in principle, but suddenly end up with 7×9 = 61 or forgetting to carry over a number. To a perfectionist, every error is a catastrophe; I wish I’d learnt earlier than this type of error is annoying-but-insignificant. (There’s a reason why ‘show your working’ is standard practice.)
This is the standard method. I like that word for a fleeting mistake.
if only i could pronounce german…i’d teach the word to my students tomorrow. that’s such an awesome word. and proves our humanity in computation as well.
2573x(400-10-1) is the way for me
I was off by *20*. I’ve never been so annoyed.
I got the right answer, doing it with a pen and paper. I do a lot of paper calculations in my job (I do building maintenance, and I frequently have to figure out how much paint or carpet to buy for a particular job.) There is a calculator on my phone, but I can figure out the answer longhand faster than I can find the right app. I also play tabletop RPGs, which involve a lot of arithmetic. So I am probably more in practice than the average person.
thank you. my students used to complain about me being able to compute faster than they could even find their calculators. i may show this to them…though they’ve gotten better about it as well.
For real fun, see who can multiply 7′ 4 1/4″ by 6′ 5 3/8″.
I guess I don’t understand this post. I dutifully got a sheet of paper and multiplied these two numbers the way I was taught as a youngster. I checked via lattice multiplication and had the same answer again. I then checked on a calculator and got the same thing. Am I missing something? How could an entire math department of math instructors get this wrong? I am a math professor in my 4th decade but not particularly amazing at arithmetic. The algorithm works fine. It isn’t the most fun thing to do that I can think of, but I can do it, and I would expect students to do it correctly too with sufficient practice too.
This reminds me of Isaac Asimov’s short story “The Feeling of Power”. Google it 🙂
On the very first day at university, more than 44 years ago, I was told “All physicists are arrogant bastards.” Coincidentally, that was the approximate era of my last attempt at pen-and-paper multiplication. But since I’m an arrogant bastard (autocorrect has problems with both of those) I couldn’t resist.
After one false start (“how do I do this again?”) I got the right answer. But it wasn’t pretty.
Let’s not forget the impact aging has on the average human brain. Using the grade-school method I learned way back in the early ’50s, I flubbed this simple problem on the first try owing to carelessness, but got it the second time around. However, I doubt very much that I’d do as well as once I did with more difficult problems. For example, once upon a time I’d have been able to prove that e is irrational without straining. Today, I pay someone else to do my income taxes.
That’s why I teach lattice multiplcation to my year 5, 11-year-old students.
With pen and paper, it took 45 seconds and I had the right answer. I thought it was trivial, actually. But, reading through the comments, I noticed that several others who are ‘older’ also got the correct result while many younger folks stumbled. In the 1960s, when I learned multiplication, it was drilled into kids with oral repetition. We hated it, but learned it. I can still picture the Grade 2 teacher leading the chorus of little voices, “Seven times six is…”
I realized that this unpleasant gift was no longer distributed among young people when my children, born in the 1980s, appeared stunned recently as I quizzed them “Quick! What’s six time eight?” They stared at me, blankly, then counted out the answer. Of course, they could solve today’s problem (2573 times 389) in 15 seconds using their ever-present phone/calculators.
It also took me a minute. With the default algorithm.
I was born in 1976. I was taught to by hearth all the “tabelline”, multiplication of single digits (7*6=42) as you.
Those knowledge is basic. It cannot be avoided.
It’s like skying. Before teaching you carving you must master snowplough turn which is easier.
Before managing black piste you must be at ease on blue and red ones (see https://en.wikipedia.org/wiki/Piste#Ratings ), otherwise you will hurt yourself.
I did it correctly in my head by doing 2500*400 + 73*400 – 2573*11 (which are all fairly easy steps) but I would certainly have gotten this wrong for a “worse-looking” pair of numbers.
Got it right by hand, flubbed checking it on the calculator!
I feel really perplexed by this post. I tried the multiplication at the beginning, and as I expected I got it right on the first try without being especially careful, and more to the point I didn’t really see how or why many people would get it wrong. In the post you say “you need to carry out most of the multiplication steps while holding a number from the last multiplication step in your memory”, which I guess could explain it — do you not write down the tens digit you get from a multiplication, usually written small and hovering above the digit where you’ll need to add it later? If not… why not? If none of your school’s math teachers can multiply, maybe try it?!?!
Right answer but not with the standard algorithm. I wrote (2500+73)*(400-11), which makes things a lot easier, I do see others n the comments who did similar things. I’ll also admit to playing this https://arithmetic.zetamac.com/ with pairs of 3 digit numbers and no paper semi-regularly. I’m working on the theory that it’s helpful for staying able to keep track of a few things in my head at once, though I get those spectacularly wrong pretty freaking often.
I did it two ways. First I multiplied (2573)(400-11) and got the right answer. Then I did it the grade school way and was off by 90. (I thought that 9*7 = 72)
Maybe we should be teaching the non-standard algorithms. Parents will hate it though.
Tried to do the 2573*(400-11) trick in my head, screwed it up the first time (but that’s what mod 9 and mod 11 checksums are for) then patched it up.
maybe kids would enjoy learning a lot of these tricks- — shortcuts are good!