CF 233B - Non-square Equation

Rating: 1400
Tags: binary search, brute force, math
Solve time: 49s
Verified: no

Solution

Problem-Type Check

This is a Type B - “Prove that / determine a value” problem. The task is to deduce a uniquely determined numerical quantity from the hypotheses.

The proof must therefore show that the given conditions force a unique value for the number of students who solved only problem $B$, and that the derived value is consistent with all conditions.

The proposed solution does this: it derives a system of equations from the hypotheses, solves the resulting Diophantine equation, and obtains a unique value $b=6$. No additional existence construction is strictly required because the argument derives the value directly from the hypotheses and the intermediate variables remain nonnegative integers.

Step-by-Step Verification

Step 1: Definition of variables $a,b,c,d,e,f,g$ for the seven Venn-diagram regions - VALID

The partition of the contestants into the seven mutually exclusive regions is correct and exhaustive.

Step 2: Total count

$$a+b+c+d+e+f+g=25$$

VALID

This follows directly from the partition and the statement that every student solved at least one problem.

Step 3: Translation of the “did not solve $A$” condition into

$$b+f=2(c+f)$$

and hence

$$b=2c+f$$

VALID

Among students outside $A$, the $B$-solvers are exactly the regions $b,f$, while the $C$-solvers are exactly $c,f$. The algebraic rearrangement is correct.

As a quick parity/divisibility sanity test:

If $c=1,f=0$, then $b=2$, and indeed $b+f=2$, $c+f=1$, satisfying the condition.

If $c=2,f=2$, then $b=6$, and indeed $b+f=8$, $c+f=4$.

Step 4: Translation of the “half of the one-problem solvers did not solve $A$” condition into

$$b+c=\frac{a+b+c}{2}$$

and hence

$$a=b+c$$

VALID

The one-problem solvers are exactly $a,b,c$, and among them those not solving $A$ are precisely $b,c$. The algebra is correct:

$$2(b+c)=a+b+c \implies a=b+c.$$

Check with small values:

If $a=4,b=3,c=1$, then exactly half of the one-problem solvers avoid $A$:

$$b+c=4,\quad a+b+c=8.$$

Step 5: Translation of the “only $A$” condition into

$$a=d+e+g+1$$

VALID

Students solving $A$ together with at least one other problem are exactly the regions $d,e,g$.

Step 6: Deduction

$$d+e+g=b+c-1$$

from Steps 4 and 5 - VALID

Substituting $a=b+c$ into $a=d+e+g+1$ gives the claimed relation.

Step 7: Substitution into the total count to derive

$$26=3(b+c)+f$$

VALID

The substitution is:

$$25=(b+c)+b+c+(b+c-1)+f.$$

Combining terms:

$$25=3(b+c)+f-1,$$

hence

$$26=3(b+c)+f.$$

The arithmetic is correct.

Step 8: Substituting $b=2c+f$ into the previous equation to obtain

$$26=9c+4f$$

VALID

Indeed,

$$26=3((2c+f)+c)+f=3(3c+f)+f=9c+4f.$$

Step 9: Solving the Diophantine equation by congruences:

$$9c+4f=26$$

implies

$$4f\equiv 8 \pmod 9$$

and therefore

$$f\equiv 2\pmod 9$$

VALID

Since $4^{-1}\equiv 7 \pmod 9$,

$$f\equiv 7\cdot 8=56\equiv 2\pmod 9.$$

Step 10: Using $0\le f\le 6$ to conclude $f=2$, then $c=2$, then $b=6$ - VALID

From $9c+4f=26$,

$$4f\le 26 \implies f\le 6.$$

Among integers $0,\dots,6$, only $2$ is congruent to $2\pmod 9$.

Then:

$$9c+8=26 \implies 9c=18 \implies c=2,$$

and

$$b=2c+f=4+2=6.$$

Step 11: Implicit consistency of the obtained values with all constraints - VALID

From the derived relations:

$$a=b+c=8,$$

$$d+e+g=a-1=7.$$

Since only the sum $d+e+g$ matters, one may choose for example

$$(d,e,g)=(7,0,0),$$

which keeps all variables nonnegative.

Checking all conditions:

Total:

$$8+6+2+7+0+2+0=25.$$

Outside $A$:

$$b+f=6+2=8,\quad c+f=2+2=4.$$

Only-$A$ versus mixed-$A$:

$$8=7+1.$$

Exactly one problem:

$$a+b+c=16,\quad b+c=8.$$

Everything is consistent.

Completeness Check

The proof handles all variables and all hypotheses.

No cases are omitted. The integrality condition is properly used. The final conclusion follows directly from the proved system.

The exploratory section contains an intentionally incorrect intermediate computation producing $25/3$, but the author explicitly identifies and corrects the algebraic mistake. Since the final proof section is logically independent and correct, this does not invalidate the solution.

No hidden assumptions remain unjustified.

Summary

The proof is logically correct and complete. Every condition is translated accurately into algebraic relations, the eliminations are valid, and the Diophantine equation is solved rigorously. The obtained value $b=6$ is uniquely forced and consistent with the original hypotheses.

VERDICT: PASS - the proof is complete and all steps are correctly justified.