Problem Statement
We hay bales in pairs to find equals of 80, 82, 83 ,84 ,85, 86, 87, 88, 90, and 91.
Is there more than one answer to this problem.
Process
So far I was able to have bales like “bales 2 and 3” I was able to get 10 different kinds. Also since there are 10 different numbers two it would make sense. The different bales were 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3, 2 and 4, 2 and 5, 3 and 4, 3 and
5, and for last 4 and 5. I was able to get this because at the beginning it showed the those numbers to 2 and 4. Then it said so on. I also wondered why they stopped at five in the second number. So I did keep going on and found ten groups of numbers. Which were going with the total numbers that were ten.
The numbers were 80, 82, 83, 84, 85, 86, 87, 88, 90,91. In total there were ten …show more content…
Just like those groups of numbers.
I haven’t found the total of the groups with the total numbers but I know they have a connection in some way.
Like I talked about in the beginning. Those groups of numbers the first one the 1 didn’t go with another one. So it would do for the other ones like it showed for number 2.
“2 and 3” It started with the twos not 2 and 1.
I also made a little chart on my paper to see what will go with what.
82= A and C
83= A and D
84= B and C
85= A and D
86= A and