Integers and the Number Line: Star problems

If all five jumps go right, the final coordinate is

Reversing a jump of length $k$ replaces a contribution of $k$ by $-k$, decreasing the endpoint by $2k$.

Since every jump length is even, each such decrease is a multiple of 4.

Thus every possible endpoint differs from 30 by a multiple of 4.

In particular, it cannot be 0: 30 is not a multiple of 4.

Also, every endpoint is even because it is a sum of signed even integers, so the endpoints 1 and $-1$ are impossible.

These observations rule out every distance less than 2.

The distance 2 is achievable.

Direct jumps 2, 4, and 8 to the right, and jumps 6 and 10 to the left.

The coordinate is

Reversing all five directions instead finishes at 2.

The order of the jumps does not affect their total.

A product is zero precisely when at least one factor is zero.

Here this happens at $n=-4$, $n=1$, or $n=5$.

These three boundary values divide the remaining integers into four regions.

If $n<-4$, all three factors are negative.

Two negative factors multiply to a positive number, and the third makes the full product negative.

These are exactly the integers $n\leq-5$.

If $-4<n<1$, the factor $n+4$ is positive and the other two are negative.

There are two negative factors, so the product is positive.

If $1<n<5$, the first two factors are positive and only $n-5$ is negative.

The product is negative, giving $n=2,3,4$.

Finally, if $n>5$, all three factors are positive.

Every integer lies either at one of the three boundaries or in one of the four regions.

The classification is therefore complete, and no large multiplication or expanded formula is needed.

The number of negative entries starts at 3.

Flipping two positive entries increases that number by 2; flipping two negative entries decreases it by 2; flipping one of each leaves it unchanged.

Therefore the number of negative entries always remains odd.

This unchanged even-or-odd status is called a parity invariant.

The absolute values total

If all entries were positive, their sum would be 28, but the invariant prevents that.

There must be at least one negative entry, whose absolute value is at least 2.

Replacing a positive 2 by a negative 2 reduces the sum by 4, so no sum can exceed $28-4=24$.

To achieve 24, flip the entries $-9,-7$, then flip $-4,2$.

The board becomes $9,7,4,-2,6$.

For a minimum, each entry is at least the negative of its absolute value, so the sum is at least $-28$.

Flip the original positive entries 2 and 6 together to reach five negative entries.

This attains $-28$ and respects the odd-negative invariant.

Neither integer can be zero.

If one were zero, both $|a|+|b|$ and $|a+b|$ would equal the absolute value of the other integer, contradicting 14 versus 6.

Nor can the two integers have the same sign: in that case adding them adds their distances from zero, again making both quantities equal.

The integers must therefore have opposite signs.

Let their larger and smaller absolute values be $L$ and $S$.

Their absolute values sum to 14, while cancellation in their sum leaves a magnitude of 6.

Thus the two lengths have total 14 and difference 6.

Remove the extra 6 from the larger length.

The remaining total is 8, made of two equal smaller lengths, so $S=4$ and $L=10$.

Either the 10 is positive and the 4 negative, or the 10 is negative and the 4 positive.

Each choice permits both orders.

These give exactly the four listed pairs.

Every pair has absolute values totaling 14 and a sum of 6 or $-6$, so all four work.

To return to 0, the total distance traveled right must equal the total traveled left.

The positive rightward totals are 5, 10, 15, and so on; the leftward totals are 3, 6, 9, 12, 15, and so on.

The first shared positive total is 15.

Reaching it requires three right moves and five left moves, or eight moves altogether.

Any larger shared total requires more moves.

Both directions must occur in a nonempty returning walk.

The eight-move route $0,5,2,7,4,1,6,3,0$ uses exactly those moves and stays nonnegative, proving that eight is attainable.

Now consider any allowed returning walk.

Its first move must reach 5.

If its second move goes right, it reaches 10 and has already visited a coordinate at least 7.

Otherwise it reaches 2.

From 2, a left move would reach $-1$, which is forbidden, so the next move must reach 7.

Thus every allowed walk has highest coordinate at least 7, regardless of its length.

The displayed route has highest coordinate exactly 7, proving the second answer.

The two students at $-9$ and 14 are 23 units apart.

Their combined walking distance is at least 23, with equality exactly when the meeting point is between them.

Similarly, the students at $-2$ and 5 are 7 units apart, and their combined distance is at least 7, with equality exactly when the meeting point is between those two positions.

The total is therefore at least $23+7=30$.

Both bounds hold with equality precisely when $x$ lies in the smaller interval from $-2$ to 5.

There are eight integer choices, all listed in the answer.

For part (b), within that interval the outer students walk $x+9$ and $14-x$ units; the inner students cannot walk farther than both outer students.

The two outer distances add to 23, so their greater distance must be at least 12 when $x$ is an integer.

To keep both at most 12, the first requires $x\leq3$ and the second requires $x\geq2$.

Exactly $x=2$ and $x=3$ work, giving outer distances 11 and 12 in opposite orders.

For any two coordinates, the direct distance between them is no greater than the distance obtained by traveling via 0:

Equality holds when the two coordinates are on opposite sides of 0.

Apply this bound to each of the five successive legs.

Each of the four interior posts then contributes its absolute value twice, while the starting and finishing posts contribute only once.

The absolute values of all posts total

Consequently, every route has length at most 54 minus the absolute values of its two endpoint coordinates.

The smallest possible sum of two distinct endpoint absolute values is $1+2=3$.

Thus every route is at most $54-3=51$ units long.

To attain the bound, use those endpoint posts and alternate sides of 0 throughout: $-1,5,-7,8,-4,2$.

Its leg lengths are 6, 12, 15, 12, and 6, totaling 51.

Every leg crosses 0, so every individual distance bound is an equality.

The construction attains the proved upper bound.

The largest distance, 9, is the span from the leftmost to the rightmost point.

Temporarily translate all points so these extremes are 0 and 9.

Translation changes no pairwise distance.

Each interior point has two distances to the extremes adding to 9.

Among the remaining distances 2, 3, 4, 5, 7, the only pairs adding to 9 are 2 with 7 and 4 with 5.

Thus the two interiors must use these two pairs, leaving 3 as the distance between the interiors.

One interior lies at 2 or 7, and the other at 4 or 5.

The pairs {2,4} and {7,5} have gap 2 and fail; the pairs {2,5} and {7,4} have gap 3 and work.

The normalized sets are therefore {0,2,5,9} and {0,4,7,9}.

Their coordinate sums are 16 and 20.

Moving every coordinate left by 4 in the first set, or by 5 in the second, makes the sum 0.

This gives {$-4,-2,1,5$} and {$-5,-1,2,4$}.

Each has the required six distances.

The endpoint pairing exhausted the possibilities, and the sum condition permits only one translation of each normalized set.

Move the observation point from an integer $x$ to $x+1$.

A mark at or left of $x$ contributes an increase of 1 to the distance total.

Every other mark is at or right of $x+1$, since its coordinate is an integer, and contributes a decrease of 1.

If $L$ marks are at or left of $x$, the change in total is

Thus the change tells us the exact cumulative number $L$: add 5 and divide by 2.

The successive changes in the supplied totals are $-3,-3,1,1,1,3$.

They therefore imply cumulative counts of $1,1,3,3,3,4$ at or left of $-3,-2,-1,0,1,2$, respectively.

There is one mark at $-3$ and none at $-2$.

The cumulative count increases from 1 to 3 at $-1$, so there are two marks there.

There are none at 0 or 1, one at 2, and the fifth mark must be at 3.

This forces the multiset $-3,-1,-1,2,3$.

For verification, their distances from $-3$ total

The forced one-step changes then reproduce all later reports.

No alternative placement can give different cumulative counts while preserving the same changes, so the answer is unique.