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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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Do you mean we have to get exactly one million ? –  Joel Cohen Sep 30 '12 at 20:09
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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

5 Answers 5

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
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Fiddle with it. Trial and error is a legitimate problem-solving tool. –  ncmathsadist Oct 1 '12 at 1:49
    
Good answer! (25 votes) –  robjohn Oct 1 '12 at 19:32
    
@robjohn Yes, my first one:) –  clark Oct 2 '12 at 10:47

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

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
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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

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

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