Number Smoothies

or, How to Avoid Thinking in Math Class #2
(See Also Parts 1, 3, and 4)

This September, I gave my 7th-graders an elegant little problem about a 12-step staircase. You’re climbing from the bottom to the top, using combinations of single and double steps. The question is, how many ways can you do this?


I was stunned when some of my students offered answers almost immediately. “145!” one screamed, as if he had just gotten bingo. “Am I right?”

“Whoa, that was fast!” I said. “Why 145?”

“12 times 12, plus 1!” he announced. “Am I right?”

“But…” I hesitated. “But why 12 times 12? Why plus 1? Are we just doing random computations that sound like fun?”

He listened to my questioning with the same patience you’d give a friend’s mediocre guitar solo. Then he launched right back into his chorus: “So,” he said, “am I right?”

To him, at that moment, “doing math” meant “making a number smoothie.” You take the numbers in front of you, throw them all into the blender, and mash the “pulse” button until you get something.

The funny thing is, in our classes, this often works. You see a thick block of text; you pick out the numbers; you run them through the formula; and voila, you’ve got a solution, no thinking required!


It’s like I’m trying to teach you to make smoothies, and I always start by laying out exactly the ingredients you need. “Here’s a banana, a cup of strawberries, a cup of milk, and a cup of vanilla yogurt. Can you make a smoothie?”

Of course you can!


But what happens when I throw extra ingredients in front of you? “Here’s a banana, an onion, a cup of strawberries, paprika, dry pasta, cooked pasta, milk, vegetable oil, a cup of vanilla yogurt, and toothpaste. Can you make a smoothie?”

Maybe you can. Or maybe you’ll conjure up a foul paprika pasta paste.


The number smoothie is a classic way to avoid thinking. It is the blind mashing of a button. Whereas real mathematics is thoughtful, selective, and carefully considered, the number smoothie is precisely the opposite: indiscriminate, wanton, thoughtless.

It doesn’t mean you should never use formulas, any more than you should swear off smoothies. I’ve written before about the need to understand where formulas come from, but even after you’ve done that, a formula needn’t be a lifeless tool.

Learning a formula doesn’t need to be the terminus of thought.

Instead, try something like this:


For the blue, pink, and purple triangles, you’ve got too much information. It’s the mathematical equivalent of needing to sort out the onions and toothpaste from the real ingredients. For the green, you don’t have enough information—the mathematical equivalent of needing to root through the fridge to find the missing strawberries.

Or you could ask a question like this:


Now, there are no ingredients at all! You’ve got to go grocery shopping all by yourself. It’s a newfound freedom (and, for many students, a new level of cognitive challenge).

Or, you could ask questions like this:


Whereas before you just blended yourself a puree and called it quits, these questions demand a more sophisticated kind of cooking. Now, using the formula is just one small step along the path.

Or how about this:


You’ve been given an ingredient that looks like a strawberry, but isn’t. You’d better know your stuff if you want to recognize this poisonous imposter-berry before it’s too late.

Occasionally, we teachers grow frustrated with our formula-thirsty students. (Okay, more like “often” or “weekly.”) Sometimes, we even denounce formulas altogether, deriding them as “brainless plug-and-chug” or “not real math.”

Of course, that’s going too far. The intelligent use of formulas is an important part of mathematics. But we’re right about one thing: there’s a lot more to formulas than just throwing numbers into a blender.


30 thoughts on “Number Smoothies

  1. Well… I just got well and truly nerd-sniped. This is amazing.

    You never actually say if your student’s answer is right or wrong but I think it’s wrong.

    Here goes:

    I can vary a) the number of doubles I take and b) where I put them.

    Suppose I take one double. I take 11 step actions overall. Now i can either try and work out how many ways i can fit one double step between 10 singles (11) or I can say I have 11 actions total of which one is different, and find how how many ways I can arrange them (also 11. Of course). I chose the first way of looking at it, but had a terrible time trying to generalise my counting for other scenarios. The second way of looking at it makes you realise that its just a plain old Choose function which there is an easy evilformula for. (I did try to derive that formula from first principals, but failed, and the lure of Google was too strong).

    So, if i take one double, it’s 11C1 which is 11. If it take two doubles, that leaves me with 8 single steps for a total of 10 actions in total. 10C2 is 45. I carry on for if i take 3 doubles (9C3), 4, 5 and 6 doubles, and sum all of the combination calculations. I get 233.

    But wait… there’s more.

    What if I vary the total number of steps. Is there a general solution?

    Reworking the above thinking for smaller staircases and you get 2 steps (two possibilities) 3 steps (3), 4 steps (5), 5 (8), 6 (13), 7 (21) and OH MY GOD ITS THE FIBONACCI SEQUENCE!

    Generalising – for n steps, the answer is the (n-1)th Fibonacci number.

    This seems so right, and so natural, and yet I cant work out why, and i cant prove it.

    Boom – nerd sniped. I’m now wondering about taking three steps at once, and my employer is coming for you. 😉

    1. OK, write p(n) for the number of arrangements for n steps which end in 1
      and write q(n) for the number of arrangements for n steps which end in 2
      then p(n) = p(n-1) + q(n-1)
      and q(n) = p(n-1) (this is what you did for 1,2,3,4,5,6,and 7)
      then q(n-1) = p(n-2)
      so p(n) = p(n-1) + p(n-2)
      also p(n-1) = p(n-2) + q(n-2)
      and with what we have above already, q(n) = q(n-1) + q(n-2)
      add them together to get the total t(n) for each step
      to get t(n) = t(n-1) + t(n-2), the fibonacci equation

      this took me a while !

    2. Great problem, isn’t it? Got it from a colleague.

      I used exactly your approach at first, trying to solve it with combinatorics. (If you’re curious about the origins of that n-choose-k formula, I have a post called ‘the mathematics of my wedding playlist’ that covers it.)

      Howard’s got exactly the right approach when it comes to Fibonacci. A slightly simpler version of the same argument:

      There are two basic ways to climb N steps:

      (A) You can climb N – 2 steps, then finish with a double.
      (B) You can climb N – 1 steps, then finish with a single.

      In other words, the number of ways to climb N steps is the number of ways to climb N – 2, plus the number of ways to climb N – 1.

      That’s Fibonacci!

      1. It is indeed a great problem.

        Thanks so much for your responses Ben and Howard. Makes complete sense to me now and – as is so often the case with maths – I dont know why I didn’t see it before.

        I think that’s given me a few useful angles to think about on the version of the problem where you can jump more than 2 steps. We shall see.


        Btw – I had read the playlist post. It was magnificent. Please do keep writing.

      2. Beautiful problem and beautiful answer 🙂
        Regarding the version with three steps…
        If you try the same method again, you have
        (A) You can climb N – 1 steps, then finish with a single
        (B) You can climb N – 2 steps, then finish with a double
        (C) You can climb N – 3 steps, then finish with a triple
        … aaaand I just realized this is the “tiling an n x 1 checkerboard problem” you learn in discrete mathematics class, now with the added twist of tri-ominoes. Hmm…

      3. Combining the long way and the short way of doing this problem, we learn that if you sum a diagonal array of Pascal’s triangle (moving like a knight: up 1, over 2, or rather 1.5 since it’s a triangular grid) starting at the edge of row N, you get the Nth Fibonacci number (or N+-1th depending on how you’re counting). I wasn’t expecting to connect those two concepts today.

  2. It doesn’t take the students long to figure out that the typical word problem is a window dressed piece of arithmetic, so of course they don’t read the words. I think that the term “word problem” should be banned.

    1. Agreed! That’s exactly the topic I’m planning for two weeks from now, in fact. ‘Word problem’ is as silly a thing to say as ‘number problem.’

    2. The classic is that kids in younger grades are more likely to get some problems right, like

      Three ships leave at the same time from Boston to New York. The first ship takes 11 hours, the second ship takes 13 hours, and the third ship takes 14 hours. How long does it take for all three ships to arrive?

      The second graders say “duh, 14 hours” and the sixth graders say “all three means add, so 11+13+14 = 38 hours.”

        1. Ah, I love that result, Joshua! As a high school English teacher once told me: ‘Education is a perpetual process of poisoning and corruption.’

          *wanders off, clacking coconuts together*

  3. Kids aren’t always the ones to blame. My 7th grader gets Math – she is doing high-school Math at school. She enjoys it when we derive/prove results (formulas) from scratch – *at home*. Last time we did this, she went, “Why can’t we do Math like this at school? All we do is plug numbers into formulas.”

    I can recall several incidents, all of which indicate the teachers’ focus on “how” rather than on “why.” What can we expect when test scores are the measure of success, and the tests consist of formulas pretending to be questions?

    That may be the norm, but there are great teachers sharing their wisdom over the web via posts like this. I am really thankful for the difference they are making in so many lives.

  4. This advice is completely applicable to my university philosophy class. Just substitute ‘words’ for ‘formulas’ or ‘numbers’.

  5. One must ensure that students can analyse problem. Formula and tips can come later. Once we understand problem in math or other subjects and know the approach, we appreciate formula and sometimes when one forget the formula once can deduce it.

    Wonderful work .

  6. Vladimir Arnold once reported an area-of-a-right-triangle word problem, whose answer was long accepted as 30, before it was pointed out that there is no such triangle.
    “The hypotenuse of a right-angled triangle is 10 inches, and the altitude dropped onto it is 6 inches. Find the area of the triangle.”
    The difficulty is that two numbers whose sum is 10 can have a product at most 25, so a right triangle with hypotenuse 10 can have an altitude to the hypotenuse of at most 5; 6 is right out.
    A perfect smoothie, indeed. Just push the button.

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