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Summing to 100
By Yatir Halevi on Sunday, January 20, 2002 - 12:01 pm:

You have the digits 1-9 at your disposal, you can only use each one once, and you have 7 operation (minus and\or plus) at your disposal.
From this arrive to the sum of 100.
This is quite easy.

I saw this riddle in a newspaper.
Now, the writer also wrote that there is only ONE way to solve this, How is that proven?

Yatir

By Kerwin Hui on Sunday, January 20, 2002 - 05:45 pm:

Yatir, I think you mean that you must use all the digits 1-9 exactly once, and there are exactly 7 +/- to use. We are allow to use numbers such as 45 but that would count as using digit 4 and 5.

The question as you quoted has more than 1 solutions. I think you need an extra condition that the digits must occur in the order 1 2 3 4 5 6 7 8 9. Then it is quite easy to show that we must have 78 or 89 as one of our term (to get the sum of all 8 numbers high enough). Now by considering parity, we reject 89. We also have 1+2+3+4+5+6+78+9=108, so we just change the sign of 4. If we have a minus sign in front of 1, we must have 2 double-digit number, but it is rather easy to check this cannot happen.

Kerwin

By Yatir Halevi on Sunday, January 20, 2002 - 08:56 pm:

You're right kerwin.We have to use all of the +/- signs, therefore there must be a 2-digit number somewhere. I don't know about the fact that they have to be in some kind of order, it didn't say.
My solution was: 98+7+6-5-4-3+2-1
So if your extra condition is really correct, then it is proven. And also our 2-digit number MUST be even...

Yatir