What a Fraction Means

One cut at the middle makes \(2\) equal pieces, and each camper takes \(1\) of them. That share is \(\boxed{1/2}\) of the strip. The bottom number says how many equal parts the whole was cut into, and the top says how many of those parts you hold. So \(1 \div 2\) has a finished answer, and the answer is \(\tfrac{1}{2}\).

The bench is cut into \(5\) equal bands, so one band is one fifth of the bench. The tape sits on \(2\) of those bands, which is \(\boxed{2/5}\). The two numbers do different jobs. The bottom one sets the size of a piece by saying how many equal pieces make the whole, and the top one just counts how many of those pieces you have.

Cutting each of the \(3\) bars into \(4\) equal strips gives \(3 \times 4 = 12\) strips, and \(12 \div 4 = 3\) strips per friend. Each strip is \(\tfrac{1}{4}\) of a bar, so one friend's pile is $$\tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{3}{4}.$$ Each share is \(\boxed{3/4}\) of a bar. So \(3 \div 4\) never gets stuck with a leftover. The bar in \(\tfrac{3}{4}\) is the division sign, and the fraction is the finished answer.

Sharing gives \(7 \div 4 = \tfrac{7}{4}\) on each tray. Counting quarter parts gives the same amount, since \(7\) scoops cut into quarters make \(7 \times 4 = 28\) quarter parts, and \(28 \div 4 = 7\) of them land on one tray, so a tray holds $$7 \times \frac{1}{4} = \frac{7}{4}.$$ One tray holds \(\boxed{7/4}\). The two readings of a fraction always agree. And \(\tfrac{7}{4}\) is past \(1\), so each tray really does hold more than one full scoop.

Six equal hops cover the whole distance from \(0\) to \(1\), so a hop of length \(h\) satisfies \(6 \times h = 1\). The number that stacks up \(6\) times to make \(1\) is one sixth, so after a single hop the beetle sits at $$\boxed{1/6}.$$ The \(6\) underneath records that the trip to \(1\) was cut into \(6\) equal steps, and the \(1\) on top records that one of those steps has been taken.

Each step is \(\tfrac{1}{6}\) and the beetle takes \(5\) of them, so it lands at $$5 \times \tfrac{1}{6} = \tfrac{5}{6}.$$ The position is \(\boxed{5/6}\). Halfway is \(3\) steps, written \(\tfrac{3}{6}\), and \(5\) steps is more than \(3\) but fewer than the full \(6\), which matches a spot past the middle and short of \(1\).

Quarter steps land on a whole number every \(4\) steps, since \(\tfrac{4}{4}=1\) and \(\tfrac{8}{4}=2\). Seven steps clears the checkpoint at step \(4\) but falls one short of step \(8\), so \(\tfrac{7}{4}\) sits between \(1\) and \(2\). The larger of those is \(\boxed{2}\). The denominator sets the step size and the numerator counts the steps, so you can place any fraction by asking which multiples of the bottom the top falls between.

A fraction \(\frac{a}{8}\) is a whole number exactly when \(8\) divides the top. The multiples of \(8\) are \(8, 16, 24, 32, 40, 48, 56\). Of the four tops, \(8 = 8 \times 1\), \(24 = 8 \times 3\), and \(40 = 8 \times 5\) are on that list, while \(52 = 48 + 4\) leaves a remainder of \(4\). So \(\boxed{3}\) tops light the lamp. No real division is needed. A fraction is a whole number exactly when the bottom divides the top, which is last chapter's divisibility all over again.

Thirds clump into whole meters three at a time, since \(\tfrac{3}{3}=1\), so $$-\frac{7}{3} = -\frac{3}{3} - \frac{3}{3} - \frac{1}{3} = -1 - 1 - \frac{1}{3}.$$ The first six thirds carry her to exactly \(-2\), and the leftover third nudges her just past it, still short of \(-3\). She hangs between \(-2\) and \(-3\), so the deeper mark is $$\boxed{-3}.$$ Left of \(0\) the numbers get smaller as you go down, so the deeper of two marks is always the more negative one.

All three write the same number. Ava has \(-\tfrac{12}{5}\) outright. Ben's \(\tfrac{12}{-5}\) is \(12\) backward fifths, and flipping the sign of the bottom flips the sign of the quotient, so \(\tfrac{12}{-5} = -\tfrac{12}{5}\). Cleo's \(-\left(\tfrac{12}{5}\right)\) reflects \(\tfrac{12}{5}\) across \(0\), landing at \(-\tfrac{12}{5}\) too. They name \(\boxed{1}\) point. A single minus sign has three legal homes, out front, on top, or on the bottom, and none of them changes where you land.

After \(n\) hops the stick rests at \(\tfrac{n}{8}\) of a meter, which is a whole number exactly when \(n\) is a multiple of \(8\). Between \(1\) and \(40\) those are \(8, 16, 24, 32, 40\), so $$\boxed{5}$$ hops plant a flag. You can skip the listing. The count of multiples of \(8\) up to \(40\) is just \(40 \div 8 = 5\).

The marker \(\frac{60}{n}\) is a whole number exactly when \(n\) divides \(60\), so hunt the divisors in pairs that multiply to \(60\). $$1 \times 60, \quad 2 \times 30, \quad 3 \times 20, \quad 4 \times 15, \quad 5 \times 12, \quad 6 \times 10.$$ That gives \(1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 10,\ 12,\ 15,\ 20,\ 30,\ 60\), twelve divisors. The machine only tries \(n\) up to \(50\), and \(60\) is the single divisor above that ceiling, so \(12 - 1 = \boxed{11}\) values light the lamp. Only the largest divisor can break the ceiling here, because its partner is \(1\), the smallest divisor there is.

One coil cut into \(9\) equal pieces gives each piece \(1 \div 9\), and that division is the fraction \(\boxed{1/9}\). The top number is the one whole coil you started with, and the bottom number is how many equal pieces the clamps made.

Sharing \(5\) sheets equally among \(8\) frames is \(5 \div 8\), so each frame receives \(\boxed{5/8}\) of a sheet. That is less than one full sheet, which fits, since \(5\) sheets cannot give \(8\) frames a whole sheet each.

The hose is cut into \(7\) equal sections, so each section is \(\tfrac{1}{7}\) of it. The sticker covers \(4\) sections, which is \(4\) copies of \(\tfrac{1}{7}\), or \(\boxed{4/7}\) of the hose. The bottom of the fraction is how many equal pieces make the whole, and the top is how many of them you take.

A step of \(\tfrac{1}{12}\) is one piece out of \(12\) equal pieces that fill a whole, so twelve steps cover \(12 \times \tfrac{1}{12} = \tfrac{12}{12} = 1\). The beetle takes \(\boxed{12}\) steps. The step count matches the denominator because the denominator was telling you how many equal pieces make one whole all along.

Five steps of \(\tfrac{1}{5}\) make one unit, so whole numbers sit at \(0, 5, 10, 15, 20\) steps out, which are the addresses \(0, 1, 2, 3, 4\). The token is at \(17\) steps, between \(15\) and \(20\), so \(\tfrac{17}{5}\) falls between \(3\) and \(4\). The smaller is \(\boxed{3}\). Checking directly, \(3 = \tfrac{15}{5}\) and \(4 = \tfrac{20}{5}\), and \(15 < 17 < 20\).

Whole meters come in groups of three thirds, since \(\frac{3}{3}=1\). Three whole meters use up \(\frac{9}{3}\), leaving \(\frac{11}{3}-\frac{9}{3}=\frac{2}{3}\), so \(\frac{11}{3}=3+\frac{2}{3}\). That is past the \(3\) mark and short of \(4\), so the larger whole number is \(\boxed{4}\).

Since \(\tfrac{9}{4} = 2\tfrac14\), the sensor sits at \(-2\tfrac14\), a quarter meter below the \(-2\) mark and not yet down to \(-3\). The deeper of the two marks is $$\boxed{-3}$$ Deeper means more negative, so the lower mark is the one farther from the surface.

Counting up in sevens gives \(7,14,21,28,35,42,49,56,63,70,77,84\), and \(84\) lands right on the twelfth step. So \(7\) divides \(84\) with nothing left over, and \(\tfrac{84}{7} = \boxed{12}\) bins.

The stretch from \(0\) to \(4\) measures \(4\) units, and sharing it among \(3\) friends means \(4 \div 3\), so one piece is \(\tfrac{4}{3}\) long. The first piece starts at \(0\), so it ends at address $$\boxed{4/3}$$ That is a little past \(1\), which makes sense, since \(4\) split three ways gives each person a bit more than one whole ribbon.

Read each card as one signed number. The first is already \(-\tfrac{8}{3}\). A positive over a negative is negative, so \(\tfrac{8}{-3} = -\tfrac{8}{3}\). And \(-\left(\tfrac{8}{3}\right)\) is the opposite of \(\tfrac{8}{3}\), again \(-\tfrac{8}{3}\). All three land on one spot, so they name \(\boxed{1}\) point. A minus sign on top, on the bottom, or out front all do the same job, they flip the sign once.

The top is always \(80\), so an address is a whole number exactly when its bottom divides \(80\). Checking each one, \(\frac{80}{5}=16\), \(\frac{80}{16}=5\), and \(\frac{80}{10}=8\) come out even, while \(6\) and \(3\) both leave a remainder. That is \(\boxed{3}\) real lockers.

\(\tfrac{72}{n}\) is a whole number exactly when \(n\) divides \(72\), so count the divisors of \(72\). Hunting in pairs that multiply to \(72\) gives \(1\times 72\), \(2\times 36\), \(3\times 24\), \(4\times 18\), \(6\times 12\), and \(8\times 9\), which lists \(1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72\). That is \(\boxed{12}\) values of \(n\). Once the two partners in a pair meet in the middle, like \(8\) and \(9\) here, you have caught every divisor.

Split the top. $$\frac{n+7}{n} = \frac{n}{n} + \frac{7}{n} = 1 + \frac{7}{n}.$$ The \(1\) is already whole, so the whole expression is a whole number exactly when \(\frac{7}{n}\) is. Since \(7\) is prime, only \(n=1\) and \(n=7\) divide it, giving \(\frac{8}{1}=8\) and \(\frac{14}{7}=2\). That is \(\boxed{2}\) values of \(n\).