In a contest of intelligence, three problems A, B and C were posed.
Among the contestants there were 25 who solved at least one problem each.
Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.
The number of participants who solved only problem A was one more than the number who solved problem A and at least one other problem.
Of all students who solved just one problem, half did not solve problem A.
How many students solved only problem B?
This solution is almost identical to Charlie's... barring that I narrowed it down to looking for a solution where V>X (rather than looking where all the variables are positive, as Charlie did).
They are equivalent.
_____________________________
6 students
_____________________________
Proof:
Let's identify all the possible "regions" of the problem space :
Let:
R = # of students that solved ONLY A
S = # of students that solved ONLY A and B
T = # of students that solved A, B, and C
U = # of students that solved ONLY A and C
V = # of students that solved ONLY B
W = # of students that solved ONLY B and C
X = # of students that solved ONLY C
Note: CLEARLY, all of these must be non-negative integers less than 25! This is a key constraint.
"Among the contestants there were 25 who solved at least one problem each."
R + S + T + U + V + W + X = 25
Eqn [1]
"Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.":
V + W = 2 * (W + X) ==>
W = V - 2X
Eqn [2] ==>
V = 2X + W
Eqn [2.5]
"The number of participants who solved only problem A was one more than the number who solved problem A and at least one other problem.":
R = S + T + U + 1
Eqn [3]
"Of all students who solved just one problem, half did not solve problem A. "
2 * (V + X) = R + V + X ==>
R = V + X
Eqn [4]
"How many students solved only problem B?"
means: Solve for V.
___________________________
starting with equation [1]
R + S + T + U + V + W + X = 25
add one to both sides
R + (S + T + U + 1) + V + W + X = 26
substitute R (for S+T+U+1), using equation [3]
2R + V + W + X = 26
substitute for R, using equation [4]
2(V + X) + V + W + X = 26
simplify
3V + W + 3X = 26
substitue for W, using equation [2]
3V + (V - 2X) + 3X = 26
simplify
4V + X = 26
Now we have a key relationship between V and X
Since BOTH must be non-negative integers, 4V must be between 0 and 25, meaning V must be a number from 0 to 6. Here are the possible pairings:
For V = 0, X = 26,
For V = 1, X = 22,
For V = 2, X = 18,
For V = 3, X = 14,
For V = 4, X = 10,
For V = 5, X = 6,
For V = 6, X = 2,
Remember, equation [2.5]?
V = 2X + W
This equation shows the V>=2X (since none of the values can be negative)
The last pairing is the ONLY one where V>X, so it must be the solution.
Therefore
V = 6.
solution
