Systems of Linear Equations: Star problems

Adding all three equations gives $3x+3y+3z=36$, so $x+y+z=12$.

Subtract this total equation from $x+2y=11$ to obtain $y-z=-1$.

Subtract it from $y+2z=14$ to obtain $z-x=2$.

Consequently $z=x+2$ and $y=z-1=x+1$.

Substitution into the total gives $x+(x+1)+(x+2)=12$, hence $3x+3=12$ and $x=3$.

It follows that $y=4$ and $z=5$.

The original left sides are $3+2\cdot4=11$, $4+2\cdot5=14$, and $5+2\cdot3=11$, so the triple works.

Every solution had to satisfy the total and both difference relations, which forced these values.

Thus there cannot be a second solution.

The sum of all three readings is

Independently of their order, their algebraic sum is

Thus $a+b=11$, and its label is recovered immediately.

The remaining readings are $(a+b)+a=11+a$ and

Subtracting $11$ from $16$ and $17$ shows that the two numbers are $5$ and $6$.

Their order is not determined, so the possible ordered pairs are $(5,6)$ and $(6,5)$.

For the first pair, the three expressions are $11,16,17$ in the stated order; for the second they are $11,17,16$.

Both are positive and valid.

The total forced the first label, and the remaining differences forced the two entries, so the list is complete.

Adding the original equations gives $(a+3)(x+y)=6$, and subtracting the second from the first gives

At $a=-3$, the sum equation reads $0=6$, so the original system is inconsistent.

If $a\ne-3,1$, the difference equation forces $x=y$.

The sum equation then gives $2x=6/(a+3)$, hence

Substitution into either original equation gives $(a+3)\cdot3/(a+3)=3$, so the forced pair works.

At $a=1$, both original equations become $2x+2y=3$.

Their solutions are all pairs with $x+y=3/2$, equivalently $(x,y)=(t,3/2-t)$ for any real $t$.

These cases exhaust the real parameter values.

The two exceptional numbers have different effects: one makes the equations contradictory, while the other makes one equation repeat the information in the other.

For any three real numbers,

The first four reported numbers satisfy this identity, since

If exactly one of these four reports were wrong, the other three would be correct.

The identity would then determine the fourth value uniquely; because the displayed numbers already obey it, that fourth report would also be correct, a contradiction.

The promise of at most one error therefore forces all four of the first reports to be correct.

Subtracting their pair sums from the total gives $x=14-11=3$, $y=14-10=4$, and $z=14-7=7$.

These values give the first three sums $7,11,10$ as required.

The fifth quantity is $3+2\cdot4+3\cdot7=32$, not $31$.

Thus the fifth instrument is the unique incorrect one.

The consistency argument is essential: choosing three equations and solving them without using the error promise would not by itself justify trusting those equations.

Adding the first, third, and fifth equations gives

Adding the second, fourth, and sixth gives the same left side and right side $16+k$.

Therefore $k=4$ is necessary; any other value makes the system inconsistent.

Let $a=t$.

Successive substitution yields

The closing sum is then $f+a=4$, regardless of $t$.

Thus when $k=4$, it supplies no new restriction.

Positivity and integrality require $t$ to be an integer with $t\ge1$ and $4-t\ge1$, so $t=1,2,3$.

The remaining expressions are positive for each of these values.

Substituting the three choices gives exactly the listed sextuples, and their construction verifies every neighboring sum.

Any solution had to arise from its first entry $t$, so no additional sextuples are possible.

Set $u=x/y$ and $v=y/x$.

Addition and subtraction of the equations give $u=(5+2k)/4$ and

However, because the original variables are nonzero, these new quantities must also satisfy $uv=1$.

Substituting the candidate values into that condition gives $(25-4k^2)/16=1$, so $k^2=9/4$.

Thus $k=3/2$ or $k=-3/2$.

This step uses only the two real numbers whose square is $9/4$.

If $k=3/2$, then $u=2,v=1/2$, so all original pairs have the form $(x,y)=(2t,t)$ with $t\ne0$.

If $k=-3/2$, then $u=1/2,v=2$, giving $(x,y)=(t,2t)$ with $t\ne0$.

Each family satisfies both original equations.

The substitution made a linear system, but it also introduced a relation between its new variables.

Checking that relation is what eliminates the other parameter values.

Scaling both original numbers together explains why each admissible case has infinitely many pairs.

The three clues give

Subtracting the first equation from the second yields

Subtracting the second from the third and dividing by $2$ yields

Their difference is $c+3d=27$.

Let $d=t$.

Then $c=27-3t$, the first difference equation gives $b=3t-16$, and the count equation gives $a=11-t$.

Because each count is a positive integer, $b\ge1$ forces $t\ge6$, while $c\ge1$ forces $t\le8$.

The conditions on $a,d$ add no further restriction for $t=6,7,8$.

These three values produce the listed quadruples.

They satisfy the count equation and both difference equations, which reconstruct all three original equations.

Conversely, every possible bag must have exactly this parameterized form and meet the positivity bounds.

Thus the list is complete, even though three equations alone did not determine four real unknowns uniquely.

Subtracting the first given equation from the second gives $y+2z+3w=0$.

Hence $y=-2z-3w$ and $x=z+2w$.

The values of $z,w$ are free: substituting these formulas verifies both original equations for every real choice of them.

The proposed conclusion therefore becomes

For this to hold for every real $z,w$, both coefficients must vanish.

Taking $(z,w)=(1,0)$ and $(0,1)$ proves this necessity directly.

Thus $C=2B-A$ and $D=3B-2A$.

Set $r=A$ and $s=B-A$ to obtain

Conversely, coefficients of this form give

which is zero by the hypotheses.

Thus every listed coefficient pattern is guaranteed, and the free-variable argument proves that there are no others.

Let $u,v,w$ be the move counts in a returning sequence.

Balancing each box gives $-2u+6w=0$, $u-3v=0$, and $v-w=0$.

Thus $u=3w$ and $v=w$.

A nonempty sequence requires $w\ge1$, so its length is $5w\ge5$.

Each move preserves $V=x+2y+6z$.

Every returning sequence includes a $C$ move, which requires a counter in $Z$, so at that moment $V\ge6$.

Therefore $V\ge6$ initially is necessary.

Now assume $V\ge6$.

If $z\ge1$, the five moves $C,A,A,A,B$ are legal and restore all counts.

If $y\ge3$, the sequence $B,C,A,A,A$ does the same.

The remaining case is $z=0$ and $y\in\{0,1,2\}$.

Here $x+2y\ge6$, so

Perform $A$ exactly $3-y$ times, producing three counters in $Y$; then perform $B$, then $C$, then $A$ exactly $y$ times.

After $C$, box $X$ contains $x+2y$ counters, enough for these final moves.

The sequence restores $(x,y,0)$ and uses exactly five moves.

Thus the condition is sufficient, and the move-count lower bound proves that the constructed five-move return is shortest.