# Possible solution to "The Haybaler Problem"

#### Fandoms Fangirl

##### New member
[In reference to this thread from 2011.]

I'm doing the same problem. Here is the way I thought about it.

a+b=80
d+e=91

So I said

a=39, b=41, d=44, e=47

Then you need a number that works with all these, so it has to be less than 44 and more than 41. So it can be 42 or 43.

But it can't be 42, because then you would get 81 and 89 which are unneeded. So c=43.

And the answer doesn't have multiple solutions, because you would have to change all of the numbers, and you can't equal sidedly change them, like in algebra.

Can someone tell me if I'm right? I am trying to do the same problem!

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#### stapel

##### Super Moderator
Staff member
[In reference to this thread from 2011.]

I'm doing the same problem. Here is the way I thought about it.

a+b=80
d+e=91
How did you arrive at the above?

Please be complete, showing all of your work and reasoning, specifying how it differs from what was provided in the earlier thread. Thank you! 