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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

  • 0
    Do you mean we have to get exactly one million ?2012-09-30
  • 3
    What about constructing numbers, such as 12345, from the digits? I'm pretty sure it won't be possible without that.2012-09-30
  • 0
    @gam3 multiplication by 10 you say? 1*10*10*10*10*10*10. And do we *have* to use each digit 1 time?2012-10-01
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    @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.2012-10-01
  • 0
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  • 0
    Using the numbers 1,2,3,4,5,6,7,8,9 only once how can you construct the number 1 million using only addition, multiplication, subtraction and division. Intermediate numbers can be created using addition and multiplication by 10, but these numbers can not contain factors of 10.2012-10-01

5 Answers 5

33

Assuming you can construct number from digits one way to do it the following $$625*4*8(19*3-7)=5^42^22^3(57-7)=5^42^5*50=5^4*2^5*5^2*2=10^6$$

  • 0
    How do you find such a solution?2012-10-01
  • 8
    Fiddle with it. Trial and error is a legitimate problem-solving tool.2012-10-01
  • 0
    Good answer! (25 votes)2012-10-01
  • 0
    @robjohn Yes, my first one:)2012-10-02
13

Without some more options of operations, I don't think you can get there, as $9!=362880$. Powers would make it easy: $(1+9)^{(2*3+4+5+6-7-8)}=(1+2*3+4+5-7-8+9)^6$

  • 4
    Actually you can get to $3\cdot9!/2$ by adding the $1$ to the $2$.2012-09-30
  • 0
    @joriki: true, but still too small.2012-09-30
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    Still too small, but showing that the 9! argument does not work.2012-10-03
4

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

3

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

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

  • 4
    I must have missed the step where base-2 '1000000' equals one million.2012-10-01
  • 0
    @Joren If the question didn't say anything about 'one million' and just stayed that the answer should be 1000000 then it works :)2012-10-01
  • 0
    @swish Oh, so if it said that, your solution would work? Does it say that?2012-10-01
  • 2
    What are these strange symbols: $2,3,4,5,6,7,8,9$? Oh! they're that base-$1010$ encoding I've heard of.2012-10-01
2

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