# using numbers 1 to 9 only once to equal 1 million

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

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Do you mean we have to get exactly one million ? – Joel Cohen Sep 30 '12 at 20:09
What about constructing numbers, such as 12345, from the digits? I'm pretty sure it won't be possible without that. – SiliconCelery Sep 30 '12 at 20:13
@gam3 multiplication by 10 you say? 1*10*10*10*10*10*10. And do we have to use each digit 1 time? – Dason Oct 1 '12 at 0:50
@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. – Ross Millikan Oct 1 '12 at 3:14
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. – njr101 Oct 1 '12 at 11:36

$$(1+2+3+4)^6 \times (7-5-9+8) = 10^6 \times 1 = 1000000.$$

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

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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Actually you can get to $3\cdot9!/2$ by adding the $1$ to the $2$. – joriki Sep 30 '12 at 21:11
@joriki: true, but still too small. – Ross Millikan Sep 30 '12 at 21:14
Still too small, but showing that the 9! argument does not work. – Did Oct 3 '12 at 9:38

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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How do you find such a solution? – B Seven Oct 1 '12 at 1:01
Fiddle with it. Trial and error is a legitimate problem-solving tool. – ncmathsadist Oct 1 '12 at 1:49