3 men went fishing they would divide all fish evenly. 1st guy get them divided with 1 left over, next guy divides without prior knowledge again throws the extra over finally the 3rd one gets up and divides equally with one lft over.
What is the smallest number of fish caught.
To find the smallest number of fish caught in this scenario, we need to work backwards and determine what numbers, when divided by each person in turn, leave one fish remaining.
Let's start with the third person. We know that after they divide the remaining fish equally, one fish is left over. This means that the number of fish caught was (2 * X) + 1, where X represents the number of fish after the second person divided them.
Now, let's move to the second person. We know that after their division, one fish is left over. This means that the number of fish divided by the second person was (2 * Y) + 1, where Y represents the number of fish after the first person divided them.
Finally, we consider the first person. We know that after their division, one fish is left over. This means that the number of fish divided by the first person was (2 * Z) + 1, where Z represents the total number of fish caught.
To find the smallest possible value for Z, we need to find the smallest integers for X, Y, and Z that satisfy all the conditions above.
Let's find the pattern by finding the smallest possible values for X, Y, and Z through trial and error:
If we assume Z = 1, then X must be (2 * 1) + 1 = 3.
If X = 3, then Y must be (2 * 3) + 1 = 7.
So, the smallest possible number of fish caught is Z = (2 * 7) + 1 = 15.
Therefore, the smallest number of fish caught in this scenario is 15.