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Using the digits 1 to 9 exactly once create a set of one or two digit prime numbers.

Use those prime numbers and the math operations + - / * and parantheses to get 100.

Then use those exact same prime numbers and the above math operations to get 1000.

So your equations will have all prime numbers (showing digits 1 to 9 used exactly once) on the left and 100 or 1000 on the right.

No programming and no partial answers.

I do expect more than one solution so one with the least number of math operations will be accepted.

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How about this:

89 - 61 + 47 + (5 * (2 + 3)) = 100
(89 + 61 - 47 - 3) * (2 * 5) = 1000
Picking numbers was done much like franck vivien had suggested, there isn't much leeway.

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This does not answer the question because I am using power (^), but I think it is fun to show it, because it strictly follows the no-computers rule

Hmmm 100 and 1000. What about searching for a single and incomplete solution for 100, and using $+0$ (for example $+2+3-5$) to get 100, and $*10$ (for example $*(2+3+5)$ ) to get 1000?
Let's have a look. Keeping 2/3/5 apart, as well as any prime number containing a 2, a 3 or a 5.... the only remaining prime number with a 8 is 89
Now we must use 1, 4, 6, 7... definitely not 17 (removing 1 and 7 leaves no prime containing a 4 or a 6) , nor 11 obviously.... this leaves 41/47/61/67
OK, what about $89+61-47 = 103$?
Oh we got a solution now, using power operator:
$(89+61-47) - 2^3 +5 = 100$
$(89+61-47 -3) *(2*5) = 1000$
Note we did not stick to our original thought (using exactly the same solution $+0$ for 100 or $*10$ for 1000...

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