Answer:
a = 45
b = 45
c = 73
The order doesn't matter as long as we have two copies of 45 and one copy of 73.
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Step-by-step explanation:
If a,b,c are all different values, then there's no way to have a mode since the mode is the most frequent item. We need 45 to repeat itself more than once.
This means either a,b,c are the same value (45) or exactly two of those variables are equal (either a = b, b = c or a = c) to 45.
Let's assume that a = b = c = 45. If so, then
{41,46,a,b,c}
would update to
{41,46,45,45,45}
Adding those items up and dividing by 5 gets us
(41+46+45+45+45)/5 = 44.4
We don't get the mean of 50 we want. So all three a,b,c values cannot be 45.
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Instead, we have the case that exactly two of a,b,c are 45 and the third is something else.
Without loss of generality, let's say a = b = 45 and c is something else.
The set {41,46,a,b,c} updates to {41,46,45,45,c}
Add those values up, divide by 5, and set the result equal to the target goal of 50. The idea is to solve for c.
(41+46+45+45+c)/5 = 50
(177+c)/5 = 50
177+c = 5*50
177+c = 250
c = 250-177
c = 73
As a check,
(41+46+45+45+73)/5 = 50
which works out.
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So that's how we arrive to a = 45, b = 45, c = 73
The order of the letters doesn't matter. We could easily have a = 45, c = 73 and b = 45; or we could have a = 73, b = 45, c = 45.
All that matters is exactly two of the a,b,c are 45 and the third variable is 73.