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I got this question from a last year's olympiad paper.

Compute $999,999\cdot 222,222 + 333,333\cdot 333,334$.

Is there an approach to this by using pen-and-paper?

EDIT Working through on paper made me figure out the answer. Posted below.

I'd now like to see other methods. Thank you.

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Place your calculator on a paper, and use your pen to hit the keys of the calculator... Have fun! –  Awal Garg May 21 at 8:35

7 Answers 7

My observation suggests that we may simplify it somewhat like this assuming $x = 111,111$: $$\begin{align*} 9x\cdot2x+3x(3x + 1) &= 9x^2 +18x^2 + 3x \\ &= 27x^2 + 3x \\ &= 3x(9x + 1) \end{align*}$$

So this means: $$\begin{align*} 3x(9x + 1) &= 333,333\cdot (999,999+1) \\ &= 333,333\cdot1,000,000 \\ &= \boxed{333,333,000,000} \end{align*}$$

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11  
Don't forget to accept this nice answer! :) –  Ilya Jan 11 '13 at 15:02
1  
@Ilya: I think it'd be rude to not entertain other solutions if posted. :) –  Parth Kohli Jan 11 '13 at 15:03
    
Another potentially useful thing to remember (less so for this question) is that $111111 = \frac {10^6-1}{9}$, which could make the multiplication easier. See for example this. –  Calvin Lin Jan 11 '13 at 15:22

It's about the same:

$$\begin{aligned}&999,999 \cdot 222,222 + 333,333 \cdot 333,334 \\ = &333,333 \cdot 666,666 + 333,333 \cdot 333,334 \\ = &333,333 \cdot 1,000,000\end{aligned}$$

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@DumbCow more generally (2/3)A * (3/2)B = A * B –  Dan Neely Jan 11 '13 at 18:37

An algebra free way: the expression is $$(10^6 -1)\bigg({2 \over 9}(10^6 - 1)\bigg) + {10^6 - 1 \over 3}{10^6 + 2 \over 3}$$ $$= {10^6 - 1 \over 3}\bigg({2 \over 3}(10^6 - 1) + {10^6 + 2 \over 3}\bigg)$$ $$= \bigg({10^6 - 1 \over 3}\bigg)10^6$$ $$= 333,333*1,000,000$$ $$= 333,333,000,000$$

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I think that this is a nice answer. –  dwarandae Jan 12 '13 at 4:11

My first (useful) though was factoring out 333,333:

$$\begin{align*} &999,999*222,222+333,333*333,334 =\\ &= 333,333*(3*222,222+333,334) =\\ &= 333,333*(666,666+333,334) =\\ &= 333,333*1,000,000 =\\ &= 333,333,000,000 \end{align*}$$

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My immediate thought was to compute this as $$1,000,000\times222,222 - 222,222+111,111\times1,000,002$$

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For what it's worth, my thought, different from the others so far:

$1000000\times(0.999999\times 222222 + 0.333333\times 333334)$

This directly leads to an approximation of the answer, if we pretend that 0.9999 is 1, and 0.3333 is 1/3.

That inspiration leads to (where M = $10^6$):

$(M-1)222222 + (M/3 - 1/3)33334$

Which rearranges to:

$222222M - 222222 + (333334M - 333334)/3$

(In the second term, we take the 1/3 out, and bring the 33334 in). The subtractions we can evaluate quite easily with mental calculations, because the minuends end with six zeros, and the digits are repeating. For instance by analogy with subtraction from 100 we know that 100 - 22 is 78, and so 100...00 - 22...22 is 77...78.

$222221777778 + 333333666666/3$

Of course the division by 3 is easy:

$222221777778 + 111111222222$

And the addition is also trivial. On the lower half, we are adding back the 2222222 digits that we subtracted in the first place to make 777778, which makes a million again, and the 1 which carries out of that bumps up the 222221 upper half to 222222, which adds with 111111 to make 333333, hence:

$333333000000$

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$$999,999\cdot 222,222 + 333,333\cdot 333,334=333,333\cdot 666,666 + 333,333\cdot 333,334$$ $$=333,333 \cdot 1,000,000$$

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