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Using digits 1,2,3,4,5,6,7,8,9 only once how do you equal 1 million.

Adding, multiplication, subtraction and division

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  • $\begingroup$ Do you mean we have to get exactly one million ? $\endgroup$ – Joel Cohen Sep 30 '12 at 20:09
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    $\begingroup$ What about constructing numbers, such as 12345, from the digits? I'm pretty sure it won't be possible without that. $\endgroup$ – SiliconCelery Sep 30 '12 at 20:13
  • $\begingroup$ @gam3 multiplication by 10 you say? 1*10*10*10*10*10*10. And do we have to use each digit 1 time? $\endgroup$ – Dason Oct 1 '12 at 0:50
  • $\begingroup$ @gam3:Since $1+2+3+4+5+6+7+8+9=45$, we only have to reverse a sum of $22$, then multiply by $10$ six times. So (following Dason) (-1-2-3-4-5+6-7+8+9)*10*10*10*10*10*10 with many other similar solutions. $\endgroup$ – Ross Millikan Oct 1 '12 at 3:14
  • $\begingroup$ Anyone for a game of "Street Countdown"? Basically, it's like normal countdown, only it's played on the street. It can get very cold. $\endgroup$ – njr101 Oct 1 '12 at 11:36
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Assuming you can construct number from digits one way to do it the following $$625*4*8(19*3-7)=5^42^22^3(57-7)=5^42^5*50=5^4*2^5*5^2*2=10^6$$

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  • $\begingroup$ How do you find such a solution? $\endgroup$ – B Seven Oct 1 '12 at 1:01
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    $\begingroup$ Fiddle with it. Trial and error is a legitimate problem-solving tool. $\endgroup$ – ncmathsadist Oct 1 '12 at 1:49
  • $\begingroup$ Good answer! (25 votes) $\endgroup$ – robjohn♦ Oct 1 '12 at 19:32
  • $\begingroup$ @robjohn Yes, my first one:) $\endgroup$ – clark Oct 2 '12 at 10:47
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Without some more options of operations, I don't think you can get there, as $9!=362880$. Powers would make it easy: $(1+9)^{(2*3+4+5+6-7-8)}=(1+2*3+4+5-7-8+9)^6$

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    $\begingroup$ Actually you can get to $3\cdot9!/2$ by adding the $1$ to the $2$. $\endgroup$ – joriki Sep 30 '12 at 21:11
  • $\begingroup$ @joriki: true, but still too small. $\endgroup$ – Ross Millikan Sep 30 '12 at 21:14
  • $\begingroup$ Still too small, but showing that the 9! argument does not work. $\endgroup$ – Did Oct 3 '12 at 9:38
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As Ross Millikan notes, this can't be done using each digit as a complete number, so I assume that building numbers from the digits is allowed.

For example: $(7814\times2-3)\times(69-5)=1000000$

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Also assuming powers: $((-1\times3+6\times9+7-8)\times4\times5)^2$

Actually $1 + 2 + 3 + 4 + 5*6 + 7 + 8 + 9 = 64 = 1000000_2$

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    $\begingroup$ I must have missed the step where base-2 '1000000' equals one million. $\endgroup$ – Joren Oct 1 '12 at 11:47
  • $\begingroup$ @Joren If the question didn't say anything about 'one million' and just stayed that the answer should be 1000000 then it works :) $\endgroup$ – swish Oct 1 '12 at 14:30
  • $\begingroup$ @swish Oh, so if it said that, your solution would work? Does it say that? $\endgroup$ – Graphth Oct 1 '12 at 19:29
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    $\begingroup$ What are these strange symbols: $2,3,4,5,6,7,8,9$? Oh! they're that base-$1010$ encoding I've heard of. $\endgroup$ – robjohn♦ Oct 1 '12 at 19:40
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$$(1+2+3+4)^6 \times (7-5-9+8) = 10^6 \times 1 = 1000000.$$

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