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 Happy New Year (Posted on 2011-01-01)

2011=670*3+1+2+4+8-9-5

We have just welcomed the New Year, using all 10 digits and 4 basic arithmetic signs(+, -, /, and *).

Try to get 2011 with ALL 10 digits, significantly fewer signs and no brackets.

Hint: My solution uses 10 distinct digits and 3 signs only, but maybe a better solution exists.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 best I can do | Comment 1 of 9
Using two distinct signs (although 5 total), you can form 2011 with:

978 * 2 + 4 + 5 + 10 + 36 = 1956 + 55 = 2011

Using two distinct signs, and only 4 signs total, there are many solutions of the form:

a * b + c * d + e = 2011, my favorite being:

2 * 47 + 3 * 89 + 1650 = 2011, where a, b, c, and d are all prime

Using two distinct signs, and only 3 signs total, there are again many solutions of the form:

a * b + c + d = 2011, my favorite being:

3 * 287 + 509 + 641, where a, b, and d are all prime.

Finally, after testing numerous combinations of signs, I moved on to form a - b - c = 2011, and after finding that there were indeed solutions of this form, I decided to minimize the sum of a, b, and c to .

So, using one distinct sign, and only 2 signs total while minimizing (a + b + c):

2495 - 106 - 378 = 2011

That's the "best" solution that I can seem to come up with. Happy New Year everyone!

 Posted by Justin on 2011-01-01 13:29:41

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