Arithmetical Problem

[Magic1]

Using all the digits 0,1,2,3,4,5,6,7,8,9, once only create a sum that will add up to1000 ? eg: 601+243+98+ 57 = 999 You can 0nly use one addition

Magic1

[don1]

Magic1 can you clarify please, you say you can only use one addition?

[Magic1]

Don1  Thank you for your prompt reply.  My mistake, the last sentence should read "you can only use addition " delete the word "once"                                                                          

I suppose this means you cannot use subtraction  division etc I hope this will clarify ?? Thanks

[sellotape59]

Where did this question originate? Do you know that it can be done? Have you copied the question word for word? 

[Magic1]

Sellotape 59. Thanks for your interest. Question exact word for word: Target 1000-you have ten digits 0,1,2,3,4,5,6,7,8,9, (one of each).  Use all of them to make an addition. How can you get to the answer 1000? I have got to 999 twice, one answer (example ) is shown in my original question.  There must be a methodology or proof but I do not know one. The probability of guessing it on a random basis is not high. Source my grandsons maths teacher who says that it can be done !???

[sellotape59]

PROOF IMPOSSIBLE: lets agree to begin with thay 9 does not divide into 1000. we are going to show that if we were able to construct an answer then this would imply that 9 does divide 1000 and thus we reach a contradiction. suppose then, for arguments sake, that such a sum exists i.e. there are numbers:

123 + 456 + ... = 1000 (e.g.)

pick number from {0,1,2,3,4,5,6,7,8,9}, lets take 7 by means of example:

this 7 is either in the 'hundreds column', the 'tens column' or the 'units column'. We say that 7 is the 'coefficient' of 1,10 or 100. This means that the corresponding contribution that 7 makes to the sum is either 7, 70 or 700.

- ok,: 1 = 1 , 10 = 9 + 1 , 100 = 99 + 1.

that is: 7 = 7 ,  70 = 7 x ( 9 + 1 ) , 700 = 7 X ( 99 + 1 )

thus 7 = 7,  70 = (7 x 9) + 7 , 700 = (7 x 99) + 7 = (7 x 11) x 9 + 7

i.e: thing = 9n + 7, where 'hing' refers to 7,70 or 700. n also just some number (here would be 0,1 or 77)

7 is arbitrary, could equally well chosen 6, 5, 2,8 or anything, always able to write things like:

60 = 6x9 + 6 , 500 = 55x9 + 5

2 = 0x9 + 2 , 80 = 8x9 + 8

and so on for ANY number...

THUS, we break our original sum up into constituent parts:

123 + 456 +...=1000, = 100 + 20 + 3 + 400 + ... now:

100 = n[1]x9 + 1  ,  20 = n[2]x9 + 2

etc, here: n[1] = 11,n[2] = 2.

so no matter what our original sum was (was never literally 123+456, that was just for example) we can write it as:

1000 = (n[1]x9 + 1) + (n[2]x9 + 2) +  .... + (n[8]x9 + 8) + (n[9]x9 + 9).

1000 = 9x(n[1] +  ... + n[9]) + (1+2+...+9)

= 9(n[1] + ...+ n[9]) + 45

= 9(n[1] + ...+ n[9]) + 9x5 = 9(n[1] + ...+ n[9] + 5)

implies 1000 divisible by 9, gives us contradiction -ends proof.

[sellotape59]

to further clarify, even if you don't fully understand the proof:

12 + 34 + 56 + 78 + 90 = 270 - divisible by 9

123 + 456 + 79 + 80 = 738 - divisible by 9

1234 + 567890 = 569124 - divisible by 9

45 + 321 + 907 + 68 = 1341 - divisible by 9

every possible sum is divisible by 9.

indeed:

601 + 243 + 98 + 57 = 999 - divisible by 9

the above proof says this is true whatever sum we construct, so one which sums to 1000 must also be divisible by 9, this is impossible so no such sum can exist.

[Magic1]

Sellotape 59  Thanks for your time and effort.  It seems that this problem cannot be solved.

Magic 1