Is there a mathematical solution to these problems?

[RogerDiver]

Hi. I’m trying to help my daughter with these problems but I see no mathematical way to solve them. It appears to me just to be a trial and error. How is she supposed to try 6! (360) combinations and not get frustrated by such problems? Am I missing something?

 

[Romsek]

it's more like educated trial and error.

Let's do the first one to demonstrate why it's not just random trials.

We have 4 in the denominator so it's unlikely the denominator of the addends will be 3, 5, or 6

simple ways of making 5/4 might be 1 + 1/4, 1/2 + 3/4, 2 - 3/4 etc

well we can't do 1+ 1/4 as we repeat the 1
we've got no way to get the 3/4 negative

but what we can do is 3/4 + (-1)/(-2) = 5/4

So with a little thought we knocked down that 6! to a quick check of 3 likely cases.

I suspect the remaining two will be similar.

 

[RogerDiver]

it's more like educated trial and error.

Let's do the first one to demonstrate why it's not just random trials.

We have 4 in the denominator so it's unlikely the denominator of the addends will be 3, 5, or 6

simple ways of making 5/4 might be 1 + 1/4, 1/2 + 3/4, 2 - 3/4 etc

well we can't do 1+ 1/4 as we repeat the 1
we've got no way to get the 3/4 negative

but what we can do is 3/4 + (-1)/(-2) = 5/4

So with a little thought we knocked down that 6! to a quick check of 3 likely cases.

I suspect the remaining two will be similar.

This is helpful. Thank you

 

[Steven G]

Note for exactness, 6! = 720 not 360.

 

[RogerDiver]

Note for exactness, 6! = 720 not 360.

Thanks for the correction. I used the notation incorrectly but as applied to this problem there are only 4 spots to use 6*5*4*3 combinations.