QuanticaPrealgebraOpen in the courseA researcher writes a fox count as \(\tfrac{84}{120}\), which is correct but bulky, while a teammate records his own count over a denominator of \(45\). Two jobs show up here. A single fraction should be written in its smallest, cleanest form. Two fractions cut into different-sized pieces need a shared piece size before you can compare them. Both jobs use the same Chapter 3 tools, just pointed in opposite directions.
Problem
The fuel reading is \(\tfrac{84}{120}\). Find \(\gcd(84,120)\), divide top and bottom by it in one move. Write \(\tfrac{84}{120}\) in simplest form as \(\tfrac{a}{b}\).
Show a hint
- It all comes down to one number, the \(\gcd\) of \(84\) and \(120\). One way to find it is to list the prime factors of each and collect every prime they both carry. \(84 = 2 \times 2 \times 3 \times 7\) and \(120 = 2 \times 2 \times 2 \times 3 \times 5\). Which primes show up in both, and how many times?
- Both numbers carry two \(2\)s and one \(3\), so \(\gcd(84,120) = 2 \times 2 \times 3 = 12\). Now make the single move, divide the top by \(12\) and the bottom by \(12\). What is \(84 \div 12\) over \(120 \div 12\)?
Show the full solution
\(84 = 2 \times 2 \times 3 \times 7\) and \(120 = 2 \times 2 \times 2 \times 3 \times 5\), so the primes they both carry are two \(2\)s and one \(3\), which makes \(\gcd(84,120) = 12\). Divide both by \(12\). Since \(84 \div 12 = 7\) and \(120 \div 12 = 10\), the fraction is $$\boxed{7/10}$$ Dividing by the gcd finishes in one move, because the biggest shared factor already contains every smaller one.
Problem
Panels generated \(168\) kWh, and the building used \(480\) kWh. Divide \(\gcd(168,480)\) into both top and bottom once. Write \(\tfrac{168}{480}\) in simplest form as \(\tfrac{a}{b}\).
Show a hint
- You only need one division to finish, but it has to be division by the right number. Ask what the largest number is that goes evenly into both \(168\) and \(480\). Find that \(\gcd\) first, then divide once.
- Break each number into prime factors. \(168 = 2^3 \cdot 3 \cdot 7\) and \(480 = 2^5 \cdot 3 \cdot 5\). The shared part is \(2^3 \cdot 3 = 24\), so the \(\gcd\) is \(24\). Now divide both \(168\) and \(480\) by \(24\) and read off the answer.
Show the full solution
\(168 = 2^3 \cdot 3 \cdot 7\) and \(480 = 2^5 \cdot 3 \cdot 5\), so the part they share is \(2^3 \cdot 3 = 24\). Divide both by \(24\). Since \(168 \div 24 = 7\) and \(480 \div 24 = 20\), the panels supplied \(\boxed{\tfrac{7}{20}}\) of the building's electricity. Nothing more cancels, since \(7\) is prime and does not divide \(20\).
Problem
Divide \(\tfrac{9}{20}\div\tfrac{3}{8}=\tfrac{9}{20}\times\tfrac{8}{3}\). Cancel before multiplying. Write the result as \(\tfrac{a}{b}\) in simplest form.
Show a hint
- Dividing by a fraction is the same as multiplying by its reciprocal, so rewrite \(\tfrac{9}{20}\div\tfrac{3}{8}\) as \(\tfrac{9}{20}\times\tfrac{8}{3}\). Once it is a multiplication, any top number can cancel with any bottom number.
- Cancel before multiplying. Replace the \(9\) and \(3\) with \(3\) and \(1\), since both lose a factor of \(3\). Replace the \(8\) and \(20\) with \(2\) and \(5\), since both lose a factor of \(4\). Now multiply the new tops and the new bottoms straight across.
Show the full solution
Flip the divisor, so \(\tfrac{9}{20}\div\tfrac{3}{8}=\tfrac{9}{20}\times\tfrac{8}{3}\). The \(9\) and the \(3\) both lose a factor of \(3\), leaving \(3\) and \(1\). The \(8\) and the \(20\) both lose a factor of \(4\), leaving \(2\) and \(5\). That gives \(\tfrac{3}{5}\times\tfrac{2}{1}\), and multiplying across gives \(\boxed{6/5}\). Once a division is rewritten as a multiplication, any top can cancel with any bottom, even across the two fractions.
Problem
Evaluate \(\tfrac{3}{4}\times\tfrac{4}{5}\times\cdots\times\tfrac{8}{9}\). Every interior bottom is the next top and cancels. Write the collapsed result as \(\tfrac{a}{b}\) in simplest form.
Show a hint
- You do not have to multiply this out. Write the whole product as one giant fraction, with all six tops on top and all six bottoms underneath. Now hunt for matching numbers. Does anything on top also appear on the bottom?
- Every middle number shows up once on top and once on the bottom, so each of those pairs cancels to \(1\). Cross out the \(4\), \(5\), \(6\), \(7\), and \(8\) wherever they appear. The only top with no matching bottom is the first \(3\), and the only bottom with no matching top is the last \(9\). What single fraction is left?
Show the full solution
Stack all the tops over all the bottoms. $$\frac{3\times4\times5\times6\times7\times8}{4\times5\times6\times7\times8\times9}$$ Every number from \(4\) through \(8\) sits once on top and once on the bottom, so all of those pairs cancel. That leaves \(\tfrac{3}{9}\), which reduces to \(\boxed{\dfrac{1}{3}}\). In a chain like this only the first top and the last bottom ever survive, so there is nothing to multiply out.
Problem
\(1{,}485=3^3\times 5\times 11\) and \(1{,}890=2\times 3^3\times 5\times 7\). Cancel shared primes, read what remains. Write \(\tfrac{1{,}485}{1{,}890}\) in simplest form as \(\tfrac{a}{b}\).
Show a hint
- Line the two prime lists up factor by factor. Both the top and the bottom carry three \(3\)s and one \(5\). Those are exactly the primes you are allowed to cross off, one from the top for each one you cross off the bottom.
- Strike the shared \(3\times 3\times 3\times 5\) from both floors. On top you are left with just \(11\). On the bottom you are left with \(2\times 7\). Multiply each leftover pile back up and write the fraction.
Show the full solution
Line the prime lists up. The top is \(1{,}485 = 3\times 3\times 3\times 5\times 11\) and the bottom is \(1{,}890 = 2\times 3\times 3\times 3\times 5\times 7\), so the shared primes are three \(3\)s and one \(5\). Cancel each one against its twin. $$\frac{1{,}485}{1{,}890}=\frac{\cancel{3}\times \cancel{3}\times \cancel{3}\times \cancel{5}\times 11}{2\times \cancel{3}\times \cancel{3}\times \cancel{3}\times \cancel{5}\times 7}=\frac{11}{2\times 7}$$ What survives is \(\boxed{\dfrac{11}{14}}\). The cancelled primes multiply to \(135\), which is the gcd, so factoring hands it to you without any searching.
Problem
Simplify \(\dfrac{k^4\times 10}{k^2\times 15}\): cancel two \(k\)s and the shared factor \(5\). Result is \(\dfrac{2k^2}{3}\). Now set \(k=3\). What single whole number does the fraction equal?
Show a hint
- You already simplified to \(\frac{2k^2}{3}\). Now \(k^2\) just means \(k\times k\), so with \(k=3\) that piece is \(3\times 3\). Build the top first, then divide by \(3\).
- With \(k=3\), \(k^2=9\). The top is \(2\times 9=18\). Then \(\frac{18}{3}\) is the final number. Divide it out.
Show the full solution
There are four \(k\)s on top and two on the bottom, so two pairs cancel and \(k^2\) is left upstairs. The numbers give \(\frac{10}{15}=\frac{2}{3}\), so the expression is \(\frac{2k^2}{3}\). Now put in \(k=3\). Then \(k^2=9\), the top is \(2\times 9=18\), and \(18\div 3\) gives $$\boxed{6}$$ Cancelling before substituting keeps the numbers small, and it works on letters because a letter is just a factor you have not filled in yet.
Simplest form gives a single fraction its smallest name. But \(\tfrac{5}{6}\) and \(\tfrac{7}{8}\) are both already in simplest form, and you still cannot say which is larger, because sixths and eighths are different-sized pieces and counting one kind against the other tells you nothing. So this time we go the other way. Instead of shrinking the denominators, we grow them until both fractions are cut into pieces of the same size.
Problem
The lcd of \(6\) and \(8\) is \(24\). Eighths \(\times 3\): \(\tfrac{7}{8}=\tfrac{21}{24}\). Do the same for \(\tfrac{5}{6}\). What does \(\tfrac{5}{6}\) become over \(24\)?
Show a hint
- You want to rebuild \(\tfrac{5}{6}\) so the bottom reads \(24\) instead of \(6\). Ask yourself what you multiply \(6\) by to land on \(24\), then remember the equivalent fraction rule says the top has to get the exact same treatment.
- Since \(6\times 4=24\), the multiplier is \(4\). Multiply the top by \(4\) as well, so \(5\times 4=20\), and the rebuilt fraction is \(\tfrac{20}{24}\).
Show the full solution
Since \(6\times 4=24\), the bottom was multiplied by \(4\), so the top gets the same \(4\). That gives \(5\times 4=20\) on top. $$\tfrac{5}{6}=\boxed{20/24}$$ Same ribbon, smaller pieces. Each sixth was cut into \(4\) parts, which is why the count went from \(5\) to \(20\).
Problem
To add \(\tfrac{1}{6}+\tfrac{4}{9}\) find a shared denominator. \(6\times 9=54\) works but overshoots, and \(6\) and \(9\) share \(3\). What is \(\operatorname{lcm}(6,9)\)?
Show a hint
- You are hunting for the smallest number that both \(6\) and \(9\) divide into evenly. Try walking up the multiples of the bigger one, \(9, 18, 27, \dots\), and stop at the first that \(6\) also divides.
- List the multiples of \(9\): \(9, 18, 27\). Is \(9\) a multiple of \(6\)? No. Is \(18\) a multiple of \(6\)? Yes, \(18=6\times 3\). That first match is your answer.
Show the full solution
Break each denominator into primes, \(6=2\times 3\) and \(9=3\times 3\). Covering both takes one \(2\) and two \(3\)s, so \(\operatorname{lcm}(6,9)=2\times 3\times 3=\boxed{18}\). The product \(6\times 9=54\) also works as a common denominator, but it counts the shared \(3\) twice, so it comes out \(3\) times bigger than it needs to be.
Problem
\(\operatorname{lcm}(4,6,9)\): primes are \(2^2\) and \(3^2\). Multiply the highest powers. What is the lcd of \(\tfrac{1}{4}\), \(\tfrac{5}{6}\), \(\tfrac{7}{9}\)?
Show a hint
- Line up the prime factorizations, \(4 = 2^2\), \(6 = 2 \times 3\), \(9 = 3^2\). The lcd has to be divisible by every one of them, so it needs enough \(2\)s for the \(4\) and enough \(3\)s for the \(9\) at the same time.
- Take the highest power of each prime seen anywhere. The biggest stack of \(2\)s is \(2^2\) and the biggest stack of \(3\)s is \(3^2\). The lcd is \(2^2 \times 3^2 = 4 \times 9\). Just multiply.
Show the full solution
Write each denominator in primes, \(4 = 2^2\), \(6 = 2 \times 3\), and \(9 = 3^2\). Take the highest power of each prime, \(2^2\) from the \(4\) and \(3^2\) from the \(9\), then multiply, \(2^2 \times 3^2 = 4 \times 9 = 36\). The lcd is \(\boxed{36}\). It checks out, since \(36 \div 4 = 9\), \(36 \div 6 = 6\), and \(36 \div 9 = 4\), and dropping any factor would leave one denominator unable to divide in.
Problem
lcd is \(36\). \(\tfrac{1}{4}\to\tfrac{9}{36}\). \(\tfrac{7}{9}\to\tfrac{28}{36}\). What does \(\tfrac{5}{6}\) become over \(36\)?
Show a hint
- Ask the bottom first. What times \(6\) lands on \(36\)? Whatever that number is, the top has to ride along with the exact same multiplier.
- The bottom \(6\) needs \(\times 6\) to become \(36\), so multiply the top by \(6\) too. That is \(5\times 6\) over \(6\times 6\).
Show the full solution
The bottom needs \(6\times 6 = 36\), so the multiplier is \(6\) and the top takes the same \(6\). That gives \(5\times 6 = 30\), so \(\tfrac{5}{6} = \tfrac{5\times 6}{6\times 6} = \boxed{30/36}\). Scaling both parts by the same number renames a fraction without changing its value, and now all three read over \(36\), ready to compare or add.
Problem
Simplify \(\dfrac{a^4 b^2}{a^2 b^5}\): cancel \(a\)s and \(b\)s in pairs. Then set \(a=3\) and \(b=2\). What single fraction in simplest form does the whole expression equal?
Show a hint
- The two letters never mix, so deal with them separately. There are four \(a\)s on top and two on the bottom, so two \(a\)s upstairs survive. There are two \(b\)s on top and five on the bottom, so three \(b\)s downstairs survive. Write what is left as a tidy fraction before you put numbers in.
- After cancelling, you are left with \(\dfrac{a\times a}{b\times b\times b}\), which is \(\dfrac{a^2}{b^3}\). Now \(a=3\) gives \(a^2=3\times 3=9\), and \(b=2\) gives \(b^3=2\times 2\times 2=8\). Put the top over the bottom.
Show the full solution
Four \(a\)s on top against two on the bottom leaves \(a^2\) upstairs. Two \(b\)s on top against five on the bottom leaves \(b^3\) downstairs, so the expression is \(\dfrac{a^2}{b^3}\). With \(a=3\) the top is \(3\times 3=9\), and with \(b=2\) the bottom is \(2\times 2\times 2=8\). $$\boxed{9/8}$$ Each letter cancels only against itself, so count the \(a\)s and the \(b\)s as two separate piles.
Problem
\(\tfrac{5}{9}\) vs \(\tfrac{7}{12}\). Which is larger? Write it as \(\tfrac{a}{b}\).
Show a hint
- The two fractions are hard to compare because they count different-sized pieces. Rewrite each one over \(36\). What do you multiply the top and bottom of \(\tfrac{5}{9}\) by to land on \(36\)? What about \(\tfrac{7}{12}\)?
- Since \(9 \times 4 = 36\), \(\tfrac{5}{9} = \tfrac{20}{36}\). Since \(12 \times 3 = 36\), \(\tfrac{7}{12} = \tfrac{21}{36}\). Now both count thirty-sixths, so the one with more of them wins. Compare \(20\) and \(21\).
Show the full solution
The least common denominator of \(9\) and \(12\) is \(36\). Grow both, \(\tfrac{5}{9} = \tfrac{5 \times 4}{9 \times 4} = \tfrac{20}{36}\) and \(\tfrac{7}{12} = \tfrac{7 \times 3}{12 \times 3} = \tfrac{21}{36}\). Both count thirty-sixths now, and \(21 > 20\), so the larger fraction is \(\boxed{\tfrac{7}{12}}\). Numerators only race fairly once the denominators match.
Practice these ideas
Practice
A contact sheet has \(24\) thumbnails, and \(18\) are keepers. Reduce \(\tfrac{18}{24}\) by dividing by \(\gcd(18,24)\). Write the answer as \(\tfrac{a}{b}\).
Show the solution
The divisors of \(18\) are \(1, 2, 3, 6, 9, 18\) and the divisors of \(24\) are \(1, 2, 3, 4, 6, 8, 12, 24\), so \(\gcd(18,24)=6\). Divide top and bottom by \(6\). $$\frac{18}{24}=\frac{18\div 6}{24\div 6}=\boxed{3/4}$$ Three of every four pictures made the cut. Going straight to the gcd does in one step what dividing by \(2\) and then by \(3\) does in two.
Practice
A big batch uses \(360\) g of one ingredient per \(504\) g of another. Simplify \(\tfrac{360}{504}\) to simplest form and write as \(\tfrac{a}{b}\).
Show the solution
Chase the difference to get the gcd. Since \(504-360=144\), \(\gcd(360,504)=\gcd(360,144)\), and \(360-2\cdot 144=72\), so that equals \(\gcd(144,72)=72\). Divide both by \(72\), where \(360\div 72 = 5\) and \(504\div 72 = 7\), so \(\tfrac{360}{504}=\boxed{5/7}\). Subtracting the smaller number from the larger never changes the gcd, which is what makes that loop safe.
Practice
Compute \(\tfrac{8}{21}\times\tfrac{7}{12}\). Cancel shared factors across tops and bottoms before multiplying. Write the product as \(\tfrac{a}{b}\) in simplest form.
Show the solution
Write it as one fraction, \(\tfrac{8 \times 7}{21 \times 12}\), and cancel before multiplying. The \(7\) and the \(21\) share a \(7\), leaving \(1\) and \(3\). The \(8\) and the \(12\) share a \(4\), leaving \(2\) and \(3\). What is left is \(\tfrac{2 \times 1}{3 \times 3} = \boxed{2/9}\). Cancelling diagonally is allowed because every top is about to pile onto one top anyway.
Practice
Divide \(\tfrac{15}{16}\div\tfrac{25}{8}\). Flip the divisor, cancel, then multiply. Write the quotient as \(\tfrac{a}{b}\) in simplest form.
Show the solution
Flip the divisor, so \(\tfrac{15}{16}\div\tfrac{25}{8}=\tfrac{15}{16}\times\tfrac{8}{25}\). The \(8\) and the \(16\) share an \(8\), leaving \(1\) and \(2\). The \(15\) and the \(25\) share a \(5\), leaving \(3\) and \(5\). So the product is \(\tfrac{3}{2}\times\tfrac{1}{5}=\boxed{\tfrac{3}{10}}\). Cancelling first means you never have to write \(120\) over \(400\) and reduce it after.
Practice
\(\tfrac{2310}{2730}\): top is \(2\times 3\times 5\times 7\times 11\), bottom is \(2\times 3\times 5\times 7\times 13\). Cancel shared primes. Write the reduced fraction as \(\tfrac{a}{b}\).
Show the solution
Both lists carry \(2\times 3\times 5\times 7\), so those four primes cancel and only \(11\) is left on top with \(13\) on the bottom. $$\frac{2310}{2730}=\frac{2\times 3\times 5\times 7\times 11}{2\times 3\times 5\times 7\times 13}=\boxed{\frac{11}{13}}$$ That shared chunk is \(210\), the gcd. Two different primes share no factor, so this is already in lowest terms.
Practice
\(\tfrac{2}{3}\times\tfrac{3}{4}\times\tfrac{4}{5}\times\tfrac{5}{6}\times\tfrac{6}{7}\). Cancel matching pairs. Write the result as \(\tfrac{a}{b}\) in simplest form.
Show the solution
Multiply across with the factors lined up. $$\frac{2\times 3\times 4\times 5\times 6}{3\times 4\times 5\times 6\times 7}$$ The \(3, 4, 5, 6\) appear on both rows and cancel to \(1\), leaving \(\boxed{2/7}\). In a chain like this only the first top and the last bottom ever go unpartnered, and \(2\) and \(7\) share nothing.
Practice
Simplify \(\dfrac{m^5}{m^2\times 4}\): cancel \(m\)s, leaving \(\dfrac{m^3}{4}\). Set \(m=2\). What single number does the expression equal?
Show the solution
There are five \(m\)s on top and two on the bottom, so two pairs cancel and \(m^3\) is left over the \(4\). Set \(m=2\). The top becomes \(2\times 2\times 2=8\), so the fraction is \(\dfrac{8}{4}=\boxed{2}\). Letters cancel exactly like numbers, so simplify before you substitute and the arithmetic stays small.
Practice
A measuring strip must mark both \(\tfrac{3}{8}\) in and \(\tfrac{1}{10}\) in tick marks. Find the smallest shared bottom: \(\operatorname{lcm}(8,10)\). What is that least common denominator?
Show the solution
Break each bottom into primes, \(8 = 2\times 2\times 2\) and \(10 = 2\times 5\). The lcm takes the most of each prime that shows up in either one, so \(\operatorname{lcm}(8,10) = 2\times 2\times 2\times 5 = \boxed{40}\). A strip ruled into \(40\)ths carries both marks exactly, since \(\tfrac{3}{8} = \tfrac{15}{40}\) and \(\tfrac{1}{10} = \tfrac{4}{40}\).
Practice
Three rhythm loops land every \(4\), \(10\), and \(15\) beats. The next beat where all three align is \(\operatorname{lcm}(4,10,15)\). What is that number?
Show the solution
In primes, \(4=2^2\), \(10=2\cdot 5\), and \(15=3\cdot 5\). Keep the highest power of each prime, two \(2\)s from the \(4\), one \(3\) from the \(15\), and one \(5\), so the lcm is \(2^2\cdot 3\cdot 5 = 4\cdot 3\cdot 5 = \boxed{60}\). Check the fits, \(60\div 4=15\), \(60\div 10=6\), and \(60\div 15=4\), all whole.
Practice
Rewrite \(\tfrac{7}{12}\) as an equivalent fraction over \(60\). Find what \(12\) is multiplied by to reach \(60\), then scale the top. What is the new numerator?
Show the solution
Start with the bottoms. Since \(60 \div 12 = 5\), the denominator was multiplied by \(5\), so the numerator takes the same \(5\). That gives \(7 \times 5 = 35\), so \(\tfrac{7}{12} = \tfrac{35}{60}\) and the new numerator is \(\boxed{35}\). Scaling top and bottom by the same number renames the fraction without changing its value.
Practice
A track is \(\tfrac{4}{9}\) done. Rewrite over the common denominator \(45\). Find what \(9\) is multiplied by to reach \(45\), then scale the top. What is the new numerator over \(45\)?
Show the solution
The denominator went from \(9\) to \(45\), and \(9 \times 5 = 45\), so the multiplier is \(5\). The top grows by the same \(5\), giving \(4 \times 5 = 20\), so \(\tfrac{4}{9} = \tfrac{20}{45}\) and the new numerator is \(\boxed{20}\). Dividing \(\tfrac{20}{45}\) back by \(5\) returns \(\tfrac{4}{9}\), so the value never moved.
Practice
Maya finishes \(\tfrac{4}{7}\) of a loop, and Theo finishes \(\tfrac{5}{9}\). Grow both to the lcd of \(7\) and \(9\), compare the tops. Which is larger? Write the answer as \(\tfrac{a}{b}\).
Show the solution
Since \(7\) and \(9\) share no common factor, the least common denominator is \(7 \times 9 = 63\). Grow both, \(\tfrac{4}{7} = \tfrac{4 \times 9}{7 \times 9} = \tfrac{36}{63}\) and \(\tfrac{5}{9} = \tfrac{5 \times 7}{9 \times 7} = \tfrac{35}{63}\). Since \(36 > 35\), Maya covered more, so the larger fraction is \(\boxed{\tfrac{4}{7}}\). When two denominators share no factor, their lcm is simply the product.
Practice
A turbine produced \(280\) MWh out of \(630\) regional MWh. Break both into primes, cancel shared factors, and write \(\tfrac{280}{630}\) in simplest form.
Show the solution
Break both into primes, \(280 = 2\times 2\times 2\times 5\times 7\) and \(630 = 2\times 3\times 3\times 5\times 7\). They both carry one \(2\), one \(5\), and one \(7\), so \(\gcd(280,630) = 2\times 5\times 7 = 70\). Divide top and bottom by \(70\), where \(280 \div 70 = 4\) and \(630 \div 70 = 9\), giving \(\boxed{4/9}\). Only shared primes cancel, and \(4 = 2\times 2\) and \(9 = 3\times 3\) have none left in common.
Practice
Rope 1 reaches \(\tfrac{11}{15}\) of a wall, and Rope 2 reaches \(\tfrac{7}{9}\). The lcd is \(45\). Rename both over \(45\) and compare. Which fraction is larger? Write it as \(\tfrac{a}{b}\).
Show the solution
The least common denominator of \(15\) and \(9\) is \(45\), since \(45 = 15 \times 3 = 9 \times 5\). Grow both, \(\tfrac{11}{15} = \tfrac{11 \times 3}{15 \times 3} = \tfrac{33}{45}\) and \(\tfrac{7}{9} = \tfrac{7 \times 5}{9 \times 5} = \tfrac{35}{45}\). Since \(35 > 33\), Rope 2 reaches farther, so the larger fraction is \(\boxed{7/9}\). Once the pieces are the same size, the comparison is just the numerators.
Practice
Evaluate \(\tfrac{12}{25}\times\tfrac{35}{18}\div\tfrac{7}{15}\). Flip the last fraction, then cancel every shared factor across all tops and bottoms. What single number unlocks the box?
Show the solution
Flip the last fraction, so the expression becomes $$\frac{12}{25}\times\frac{35}{18}\times\frac{15}{7}.$$ The \(35\) and the \(7\) share a \(7\), leaving \(5\) on top, and that \(5\) cuts the \(25\) down to \(5\). Then the \(15\) over that \(5\) leaves \(3\) on top. The \(12\) and the \(18\) share a \(6\), becoming \(2\) and \(3\), and that bottom \(3\) cancels the \(3\) above. Only \(\tfrac{2}{1}\) is left, so the box opens with \(\boxed{2}\). Cancelling across all three fractions at once keeps a messy product down to single digits.