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Given 577 x 423 with grade school algorithm you calculate

577 x 3 = 1731

577 x 2 = 1154

577 x 4 = 2308

These are 3 multiplications of a number by a single digit.

Then, you go on and add

1731 + 11540 = 13271

and then

13271 + 230800 = 244071 to get the result

These are 2 additions.

At the bottom line, this makes ~ 5 operations (of course depending on the number of digits you take, e.g., if you conceive the whole operation, it is just one multiplication, but this set aside).

In my copy of CC by Arora and Barak (you can have a look at page xx) it says for this "3 multiplications of a number by a single digit and 3 additions". Is this an error? Thank you.

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up vote 3 down vote accepted

Apparently this is an error. See Figure 1 on page 4 here: only 2 additions are required.

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Thank you! You are right. The same error is printed there on page 3 last sentence of the second paragraph. Wondering why nobody spotted it till 2006... – panny Jul 13 '11 at 10:48
Note also that the $243071$ in Figure 1 should be $244071 \,(=423 \cdot 577)$. – Shai Covo Jul 13 '11 at 10:54
This is true. In the final version of the book (as you can see from the amazon link), the result is corrected. So, someone must have looked over it, but did not correct the number of additions. – panny Jul 13 '11 at 10:58

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