using numbers 1 to 9 only once to equal 1 million

Question

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

Adding, multiplication, subtraction and division

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? 1101010101010. And do we have to use each digit 1 time? — Dason, Oct 01 '12 at 00:50
@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)101010101010 with many other similar solutions. — Ross Millikan, Oct 01 '12 at 03:14
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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. — gam3, Oct 01 '12 at 21:27

score 34 · Answer 1 · answered Sep 30 '12 at 20:40

34

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

answered Sep 30 '12 at 20:40

clark

15,327

How do you find such a solution? – B Seven Oct 01 '12 at 01:01
9

Fiddle with it. Trial and error is a legitimate problem-solving tool. – ncmathsadist Oct 01 '12 at 01:49
1

Good answer! (25 votes) – robjohn Oct 01 '12 at 19:32
@robjohn Yes, my first one:) – clark Oct 02 '12 at 10:47

Ross Millikan · Answer 2 · 2012-09-30T20:55:18.543

13

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

edited Sep 30 '12 at 20:55

answered Sep 30 '12 at 20:40

Ross Millikan

374,822

4

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 03 '12 at 09:38

robjohn · Answer 3 · 2012-10-01T19:25:41.363

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

edited Oct 01 '12 at 19:25

answered Oct 01 '12 at 19:19

robjohn

345,667

swish · Answer 4 · 2012-10-01T10:43:47.657

3

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

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

edited Oct 01 '12 at 10:43

answered Oct 01 '12 at 10:36

swish

850

4

I must have missed the step where base-2 '1000000' equals one million. – Joren Oct 01 '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 01 '12 at 14:30
@swish Oh, so if it said that, your solution would work? Does it say that? – GeoffDS Oct 01 '12 at 19:29
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. – robjohn Oct 01 '12 at 19:40

score 2 · Answer 5 · edited Oct 03 '12 at 15:43

2

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

edited Oct 03 '12 at 15:43

Henry T. Horton

18,318
5
60
77

answered Oct 03 '12 at 09:30

piyush

21

using numbers 1 to 9 only once to equal 1 million

5 Answers5