HCF and LCM problems questions with answers

Questions

1. Question 1Largest equal groups or pieces (HCF)

   Two ribbons are $18$ cm and $24$ cm long. Mei Ling cuts both into pieces that are all the same length, with nothing left over. Find the greatest possible length of each piece.

   Show the worked solution

   1. Step 1: Short division of 18 by primes, smallest first, until you reach 1; then 18 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;18\;} \\ 3 & \underline{\;9\;} \\ 3 & \underline{\;3\;} \\ & \;1 \end{array}$$18 = 2 \times 3^{2}$
   2. Step 2: Short division of 24 by primes, smallest first, until you reach 1; then 24 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;24\;} \\ 2 & \underline{\;12\;} \\ 2 & \underline{\;6\;} \\ 3 & \underline{\;3\;} \\ & \;1 \end{array}$$24 = 2^{3} \times 3$
   3. Step 3: Decide what the question needs: the largest number that divides every amount exactly, or the smallest number that every amount divides intoThe largest number that divides every amount exactly: the HCF.
   4. Step 4: Work it out: HCF uses the lowest power of each common prime; LCM uses the highest power of every primeHCF $= 2 \times 3$$= 6$
2. Question 2Next time together, smallest number (LCM)

   Two bus services leave an interchange together at 7.00 a.m. One leaves every $9$ minutes and the other every $12$ minutes. Find the number of minutes before they next leave together.

   Show the worked solution

   1. Step 1: Short division of 9 by primes, smallest first, until you reach 1; then 9 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 3 & \underline{\;9\;} \\ 3 & \underline{\;3\;} \\ & \;1 \end{array}$$9 = 3^{2}$
   2. Step 2: Short division of 12 by primes, smallest first, until you reach 1; then 12 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;12\;} \\ 2 & \underline{\;6\;} \\ 3 & \underline{\;3\;} \\ & \;1 \end{array}$$12 = 2^{2} \times 3$
   3. Step 3: Decide what the question needs: the largest number that divides every amount exactly, or the smallest number that every amount divides intoThe smallest number that every amount divides into: the LCM.
   4. Step 4: Work it out: HCF uses the lowest power of each common prime; LCM uses the highest power of every primeLCM $= 2^{2} \times 3^{2}$$= 36$
3. Question 3Two steps: HCF or LCM, then use it

   A rectangular floor measures $225$ cm by $315$ cm. It is covered with the largest possible identical square tiles, without cutting any tile. Find the number of tiles needed.

   Show the worked solution

   1. Step 1: Short division of 225 by primes, smallest first, until you reach 1; then 225 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 3 & \underline{\;225\;} \\ 3 & \underline{\;75\;} \\ 5 & \underline{\;25\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}$$225 = 3^{2} \times 5^{2}$
   2. Step 2: Short division of 315 by primes, smallest first, until you reach 1; then 315 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 3 & \underline{\;315\;} \\ 3 & \underline{\;105\;} \\ 5 & \underline{\;35\;} \\ 7 & \underline{\;7\;} \\ & \;1 \end{array}$$315 = 3^{2} \times 5 \times 7$
   3. Step 3: Decide what the question needs: the largest number that divides every amount exactly, or the smallest number that every amount divides intoThe largest number that divides every amount exactly: the HCF.
   4. Step 4: Work it out: HCF uses the lowest power of each common prime; LCM uses the highest power of every primeHCF $= 3^{2} \times 5$$= 45$
   5. Step 5: Count the tiles along each side, then multiplyTiles $= (225 \div 45) \times (315 \div 45)$$= 5 \times 7$$= 35$
4. Question 4Two numbers from their HCF and LCM

   Two whole numbers have a highest common factor of $9$ and a lowest common multiple of $126$. Neither number is equal to $9$. Find the two numbers.

   Show the worked solution

   1. Step 1: Each number is the HCF times a factor; the two factors have no common factor and multiply to LCM ÷ HCF$\dfrac{126}{9} = 14$$14 = 2 \times 7$
   2. Step 2: Split 14 into two factors with no common factor, neither of them 1 (so neither number is the HCF)$14 = 2 \times 7$
   3. Step 3: Multiply each factor by the HCFThe numbers are $9 \times 2 = 18$ and $9 \times 7 = 63$.
5. Question 5Cubes in a box, blocks into a cube

   A box in the shape of a cuboid measures $45$ cm by $60$ cm by $75$ cm. It is completely filled with identical cubes, each as large as possible, with no gaps. Find the number of cubes in the box.

   Show the worked solution

   1. Step 1: Short division of 45 by primes, smallest first, until you reach 1$\def\arraystretch{1.8}\begin{array}{r|l} 3 & \underline{\;45\;} \\ 3 & \underline{\;15\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}$
   2. Step 2: Short division of 60 by primes, smallest first, until you reach 1$\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;60\;} \\ 2 & \underline{\;30\;} \\ 3 & \underline{\;15\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}$
   3. Step 3: Short division of 75 by primes, smallest first, until you reach 1$\def\arraystretch{1.8}\begin{array}{r|l} 3 & \underline{\;75\;} \\ 5 & \underline{\;25\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}$
   4. Step 4: Count how many times each prime divides: the numbers in index notation$45 = 3^{2} \times 5$$60 = 2^{2} \times 3 \times 5$$75 = 3 \times 5^{2}$
   5. Step 5: The largest cube that fits each edge exactly: its edge is the HCFHCF $= 3 \times 5$$= 15$
   6. Step 6: Count the cubes along each edge of the box, then multiply$\dfrac{45}{15} \times \dfrac{60}{15} \times \dfrac{75}{15} = 3 \times 4 \times 5$$= 60$
6. Question 6Edges of a cuboid from its volume

   A cuboid has a volume of $570$ cm$^3$. Its length, breadth and height are whole numbers of centimetres, all different, and each is greater than $3$ cm. Find its length, breadth and height.

   Show the worked solution

   1. Step 1: Short division of 570 by primes, smallest first, until you reach 1; then 570 in index notation$\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;570\;} \\ 3 & \underline{\;285\;} \\ 5 & \underline{\;95\;} \\ 19 & \underline{\;19\;} \\ & \;1 \end{array}$$570 = 2 \times 3 \times 5 \times 19$
   2. Step 2: Group the primes into three different whole numbers, each greater than 3: try the smallest ones first$570 = 5 \times 6 \times 19$The cuboid measures $5$ cm by $6$ cm by $19$ cm (the only way with every edge greater than $3$ cm and all different).
7. Question 7Together again: the time of day (LCM)

   Three ferries leave a jetty together at 1405. They leave every $10$ minutes, every $15$ minutes and every $18$ minutes. Find the time at which all three next leave together. Give your answer using the 24-hour clock.

   Show the worked solution

   1. Step 1: They are together again after a time that every interval divides into: the LCM$10 = 2 \times 5$, $15 = 3 \times 5$, $18 = 2 \times 3^{2}$$\text{LCM} = 2 \times 3^{2} \times 5$$\phantom{\text{LCM}} = 90$ minutes
   2. Step 2: Change the minutes to hours and minutes (60 minutes in an hour)$90$ min $= 1$ h $30$ min
   3. Step 3: Add this to the start time, 1405$1405 + 1$ h $30$ min $= 1535$