Mental arithmetic: Add numbers by combining numbers that add up to 10

Question

I'm reading a book on mental arithmetic, and came across this paragraph:

You were taught - or should have been taught - at school that speed in addition is acquired by combining pairs of successive numbers that add up to 10.

I've never been taught this, what does this mean? If it means what I think it means, then I have included some exercises from the book (see image attached). How is searching through, finding pairs of numbers which adds up to 10, remembering which pair, trying to remember which ones you've missed out, etc... somehow quicker than just adding them up normally?

There has got to be something I'm missing.

Answer 1

An example will be fruitful, and moreover let's take one of the more well-known examples. A common story is that Gauss was punished in primary school by having to add up all of the numbers from $1$ to $100$ in his head, but defeated it by calculating it in only a few seconds. How might he have done this, assuming no knowledge of the famous formula (fittingly called "little Gauss" by some)

$$1+2+\cdots+n = \frac{n(n+1)}{2}$$

where $n=100$?

Let's consider, then, instead of taking the usual straightforward approach of adding $1+2+3+4+\cdots$ and so on in sequence, using the equivalent sum

$$(1+99) + (2+98) + (3+97) + (4+96) + \cdots + (49+51) + 50+100$$

It should be no difficulty to see we have $49$ parenthetical expressions here, each summing to $100$, plus $150$ on the side. This implies

$$1 + \cdots + 100 = 49(100) + 150 = 5050$$

This has the same principle as your given exercise, but more generally: it is somewhat faster to do addition in ways that help complement multiples and powers of $10$.

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Let's also work an example from your sheet for simplicity. We have the list of numbers below:

Each blue pair we see on the list is one that sums up to $10$, and there's also a quick triplet you can find that sums up to $20$. The remainder I haven't matched up clearly up to $12$. You can thus use this means to quickly deduce the sum is $62$.

Usually when I would be in this sort of situation working by hand, I would just cross out pairs on the list, keeping track of the running total in my head. So I would cross out $3,7$ and think $10$. And then the pair of $5$'s and think $20$, and continue going with the un-crossed-out pairs.

Answer 2

Consider:

5 + 4 + 3 + 6 + 7 + 2 + 9 + 1 = ??

Group by:

$5 + \color{red}4\color{black} + \color{blue}3\color{black} + \color{red}6\color{black} + \color{blue}7\color{black} + 2 + \color{green}9\color{black} + \color{green}1\color{black} = ??$

Easy to see: total = $37$

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As an aside: While mental arithmetic may be an amusing diversion, you will almost never use it in real life... and in the future even fewer people will use it even more rarely. It is like using an abacus.

As a response to @fleablood: Well, you can always hand-pick some particular arithmetic problem where some special trick you sort-of learned decades ago might or might not save you a few seconds, but this is (in my view) a complete waste of time. I spoke @fleablood's multiplication problem into Siri and got the correct answer in less than a second. I spoke fleablood's addition problem into Siri and got the answer in less than a second. Both less time than it took @fleablood to type a single three-digit operand. Computing devices are ubiquitous... why oh why fill eight lines of error-prone calculation when the answer is immediately and instantly available!?! (Everyone will always have a cellphone handy.) Watch this to see a clearer refutations as to the claim these one-off tricks give "deeper" understanding of "the basics." They don't.

From my perspective, there are so many better things to focus on than mental arithmetic... such as memorizing basic integrals... that you'd do better to devote your time to those.

Answer 3

What I presume they mean is the following: Suppose you wish to calculate $$7 + 5 + 6 + 5 + 4 + 3.$$

Instead of trying to calculate sequentially, you can observe that there are three pairs of numbers that add up to $10$ and thus, the total is $30$. Finding a pair that adds up to a multiple of $10$ is fairly easy (or rather, it has become easy over time).

Then, one can just count the multiples of tens (and thus, add them) and then add anything remaining.

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For example, in exercise 1, you can quickly start pairing numbers and then striking them off and keeping count of how many pairs you've encountered so far. This surely is quicker than adding all of them up one by one.

Answer 4

I was never taught this but I read a book in the fourth grade and it's a huge help.

Imagine adding $124 + 589 + 276 + 985 + 231$.

If I see the $4$ and the $6$ that's $10$ and can forget it. I can see the $9$ and the $1$ and that's $10$ and I have $20$ so for. All that's left is the $5$ for $25$. I note the $5$ and carry the $2$.

Notice that is a lot easier then doing $4 + 9$ is $13$ and $13+6$ is $19$ and $19+5$ is $24$ and $24+1$ is $25$ so I not the $5$ and carry the $2$.

Then when we do the next column, we have the $2$ we carries and $2,8,7,8,3$ and the $2+8$ is $10$ and the $7+3$ is $10$. That leaves $2$ and $8$ and that's $10$. So that's $30$. We write $0$ and carry $3$.

Note: that is easier than $2+2$ is $4$. $4+8 =12$. $12+7=19$. $19+8=27$. $27+3=30$.

ANd the we have $3,1,5,2,9,2$ so $9+1=10$ and we have $3,5,2,2$. Okay, a bit of creativity. $2+2 = 4$ and $4+5 = 9 =10-1$. We can "borrow" from the $3$ and $4+5 +1 = 10$ and that leaves us with $2$. So that's $22$

The answer is $2205$.

Thing is... it doesn't make things worse. And what ever gives us more flexibility will sometimes make things better and never make things worse.

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Addendum:

On the other hand we have what David Stork wrote:

As an aside: While mental arithmetic may be an amusing diversion, you will almost never use it in real life... and in the future even fewer people will use it even more rarely. It is like using an abacus.

From my perspective, there are so many better things to focus on than mental arithmetic... such as memorizing basic integrals... that you'd do better to devote your time to those.

I can't argue with that. If tricks help, use them. But if the just confuse and give you things to frustrate you.

To my mind this is similar to the the tricks of multiplication:

$379\times 513 = $

$(400 - 21)(500 + 13) =$

$400*500 - 21*500 + 13*400 - 21*13 =$

$(4*5)10,000 - (20*500 + 1*500) + (10*400 + 3*400) - (20+1)(15-2)=$

$200,000 - (2*5*1000 + 500) + (4000 + 1200) - [(20*15 +15-2*20 -2)]=$

$200,000 - 10,000 -500 +5000 + 200 - [30*10 + 15 - 40 - 2]=$

$200,000 -5,000 -300 - [300- 25 -2]=$

$200,000 -5,000 - 600+27=$

$195,000 - 600 + 57= 194,427$

So should you do it this way. Well, only if intuitively it makes sense and is easy for you do it that way. It certainly isn't anything I'd feel comfortable teaching someone to do.

You could also do $379\times 513 = (400 - 25 + 4)(500 + 25-12)= ((\frac 12*10-1)*100 -\frac 14*100 + (\frac 12*10-1))(\frac 12*1000 + \frac 14*100 - (\frac 2*10-2)(\frac 12*10-1))$

and ... in theory that should be really easy. It's just place holding. But should you do it this way... Well.... probably not....I wouldn't.

But I take great comfort in knowing that I can device these ways of doing things. And that I understand why they work.

Answer 5

I just look for tens or twenties.

Starting from the top of the first column, $7+8=15$ and then there's a $5$ a few numbers down, so that's $20$. And ... oh look, $4+6=10$ in between. Scanning the rest I see $5+5$ and $8+2$ for $20$ more, and what's left is $3+1+5=9$, so the sum is $59$.

It's a lot faster than $7+8=15$, $15+4=19$, $19+6=25$, ...