IMO 1966 Problem 1
Let
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 8m58s
Problem
In a mathematical contest, three problems, $A,B,C$ were posed. Among the participants there were 25 students 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 students who solved only problem $A$ was one more than the number of students who solved $A$ 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 $B$?
Exploration
Let
$$a=\text{number solving only }A,\qquad b=\text{number solving only }B,\qquad c=\text{number solving only }C.$$
Let
$$x=\text{number solving }A\cap B\text{ only},$$
$$y=\text{number solving }A\cap C\text{ only},$$
$$z=\text{number solving }B\cap C\text{ only},$$
$$t=\text{number solving }A\cap B\cap C.$$
Since every student solved at least one problem,
$$a+b+c+x+y+z+t=25.$$
The condition concerning students who did not solve $A$ involves only the regions $b,c,z$. Among those not solving $A$, the number solving $B$ equals $b+z$, while the number solving $C$ equals $c+z$. Hence
$$b+z=2(c+z),$$
so
$$b=2c+z.$$
The second condition says that the number solving only $A$ exceeds by $1$ the number solving $A$ together with at least one other problem. Thus
$$a=(x+y+t)+1.$$
The third condition concerns students solving exactly one problem. Their number is $a+b+c$. Half of them did not solve $A$, hence
$$b+c=\frac{a+b+c}{2},$$
which gives
$$a=b+c.$$
Combining this with $b=2c+z$,
$$a=3c+z.$$
Since also $a=x+y+t+1$,
$$x+y+t=3c+z-1.$$
Substituting into the total $25$,
$$25=(a+b+c)+(x+y+t)+z$$
gives
$$25=(3c+z)+(2c+z+c)+(3c+z-1)+z.$$
Hence
$$25=9c+4z-1,$$
so
$$26=9c+4z.$$
Trying nonnegative integer values of $c$,
$$26-9c\ge 0.$$
For $c=0$, $z=26/4$, impossible.
For $c=1$, $z=17/4$, impossible.
For $c=2$, $z=2$, valid.
Thus
$$b=2c+z=6.$$
The arithmetic leaves a unique possibility, suggesting that the answer is uniquely determined.
Problem Understanding
We are given the numbers of contestants lying in the seven regions of a three-set Venn diagram corresponding to problems $A$, $B$, and $C$. Every one of the $25$ contestants solved at least one problem.
Three numerical conditions are imposed on these regions. The task is to determine the number of contestants who solved only problem $B$.
This is a Type C problem, because a specific numerical value must be determined.
The main difficulty is translating each verbal condition into equations involving the Venn-diagram regions. Once this is done, the problem becomes a system of linear equations in nonnegative integers. The key observation is that the three given conditions reduce the system to a single Diophantine equation whose unique nonnegative solution determines the desired quantity.
The answer is
$$6.$$
The reason this should be correct is that the conditions force the equation
$$26=9c+4z,$$
and this equation has only one nonnegative integer solution, namely $(c,z)=(2,2)$, from which the number solving only $B$ is forced.
Proof Architecture
We introduce variables for the seven regions of the Venn diagram.
Lemma 1 states that the first condition implies
$$b=2c+z.$$
This follows by counting, among students not solving $A$, those who solved $B$ and those who solved $C$.
Lemma 2 states that the second and third conditions imply
$$a=3c+z$$
and
$$x+y+t=3c+z-1.$$
This follows by converting the verbal conditions into equations.
Lemma 3 states that the total number $25$ of contestants yields
$$26=9c+4z.$$
This is obtained by substituting the relations from Lemmas 1 and 2 into the total count.
Lemma 4 states that the only nonnegative integer solution of
$$26=9c+4z$$
is
$$(c,z)=(2,2).$$
This determines $b$.
The most delicate step is Lemma 3, because every Venn-diagram region must be counted exactly once. A small counting error there would produce an incorrect Diophantine equation.
Solution
Let
$$a,b,c,x,y,z,t$$
denote respectively the numbers of students solving only $A$, only $B$, only $C$, only $A$ and $B$, only $A$ and $C$, only $B$ and $C$, and all three problems.
Since every student solved at least one problem,
$$a+b+c+x+y+z+t=25.$$
Lemma 1
The first condition implies
$$b=2c+z.$$
Proof
The students who did not solve $A$ occupy precisely the regions $b$, $c$, and $z$.
Among them, the number who solved $B$ is
$$b+z.$$
Among them, the number who solved $C$ is
$$c+z.$$
The statement of the problem says that the first quantity is twice the second. Hence
$$b+z=2(c+z).$$
Rearranging gives
$$b=2c+z.$$
∎
Certification: this establishes the relation between the regions outside $A$; counting only $b$ and $c$ would incorrectly omit the students in region $z$.
Lemma 2
The second and third conditions imply
$$a=3c+z$$
and
$$x+y+t=3c+z-1.$$
Proof
The number of students who solved $A$ together with at least one other problem is
$$x+y+t.$$
The second condition states that the number solving only $A$ is one greater. Therefore
$$a=x+y+t+1.$$
The students solving exactly one problem are counted by
$$a+b+c.$$
Among them, those who did not solve $A$ are precisely those counted by
$$b+c.$$
The third condition states that these form one half of all students who solved exactly one problem. Thus
$$b+c=\frac{a+b+c}{2}.$$
Multiplying by $2$ gives
$$2b+2c=a+b+c,$$
hence
$$a=b+c.$$
Using Lemma 1,
$$a=(2c+z)+c=3c+z.$$
Combining this with
$$a=x+y+t+1$$
yields
$$x+y+t=3c+z-1.$$
∎
Certification: this establishes expressions for both $a$ and $x+y+t$; replacing the phrase “at least one other problem” by only $x+y$ would omit the triple intersection $t$.
Lemma 3
The total number of contestants implies
$$26=9c+4z.$$
Proof
From Lemma 1,
$$b=2c+z.$$
From Lemma 2,
$$a=3c+z, \qquad x+y+t=3c+z-1.$$
Substituting into
$$a+b+c+x+y+z+t=25$$
gives
$$25=(3c+z)+(2c+z)+c+(3c+z-1)+z.$$
Collecting terms,
$$25=9c+4z-1.$$
Hence
$$26=9c+4z.$$
∎
Certification: this establishes the Diophantine equation governing the remaining unknowns; every region of the Venn diagram is counted exactly once in the substitution.
Lemma 4
The equation
$$26=9c+4z$$
has the unique nonnegative integer solution
$$(c,z)=(2,2).$$
Proof
Since $c,z\ge 0$,
$$9c\le 26,$$
so
$$c\in{0,1,2}.$$
If $c=0$, then
$$4z=26,$$
which has no integer solution.
If $c=1$, then
$$4z=17,$$
which has no integer solution.
If $c=2$, then
$$4z=8,$$
so
$$z=2.$$
Thus the unique nonnegative integer solution is
$$(c,z)=(2,2).$$
∎
Certification: this establishes uniqueness; checking only congruences would not by itself exclude larger nonnegative values of $c$.
From Lemma 1,
$$b=2c+z=2\cdot 2+2=6.$$
Thus the number of students who solved only problem $B$ is
$$\boxed{6}.$$
Verification of Key Steps
The first delicate step is the interpretation of the first condition. Students not solving $A$ lie in the regions $b$, $c$, and $z$. Among these students, solving $B$ means belonging to $b$ or $z$, giving $b+z$. Solving $C$ means belonging to $c$ or $z$, giving $c+z$. The equation is therefore
$$b+z=2(c+z).$$
A careless argument might write $b=2c$, incorrectly excluding the students who solved both $B$ and $C$.
The second delicate step is the use of the condition on students solving exactly one problem. The relevant set consists of the three disjoint regions $a$, $b$, and $c$. Half of these students did not solve $A$, namely those in $b$ and $c$. Hence
$$b+c=\frac{a+b+c}{2}.$$
A common error is to count all students not solving $A$, which would incorrectly include region $z$.
The third delicate step is deriving
$$26=9c+4z.$$
Using
$$a=3c+z,\qquad b=2c+z,\qquad x+y+t=3c+z-1,$$
the total becomes
$$25=(3c+z)+(2c+z)+c+(3c+z-1)+z.$$
The coefficient of $c$ is
$$3+2+1+3=9,$$
and the coefficient of $z$ is
$$1+1+1+1=4.$$
Missing any one of these contributions changes the final answer.
Alternative Approaches
A different method is to work entirely with aggregate quantities instead of introducing all seven Venn-diagram regions. Let $S$ be the number of students solving exactly one problem and let $M$ be the number solving at least two problems. The second condition gives a relation between the students solving only $A$ and those contributing to $M$, while the third condition expresses the number solving only $A$ as half of $S$. Combining these aggregate counts with the first condition eventually produces the same Diophantine equation for the numbers outside $A$.
The Venn-diagram method is preferable because each verbal statement translates directly into a linear equation. Every region is visible, so there is little risk of overlooking students who solved two or three problems. The resulting system collapses quickly to a single equation with a unique nonnegative integer solution.