LCM — prime factorization method, relationship HCF × LCM = product: Exercises

Two symbols we use constantly, defined once so nothing is a mystery:

Level 1 — Recognition

Goal: read a recipe, pick the right power.

Q1. $20 = 2^{2} \cdot 5$ and $50 = 2 \cdot 5^{2}$ . Which power of $5$ goes into the LCM, and which into the HCF?

Recall Solution — Q1

Prime $5$ appears as $5^{1}$ in 20 and $5^{2}$ in 50.

LCM takes the max: $max (1, 2) = 2 \Rightarrow 5^{2}$ .
HCF takes the min: $min (1, 2) = 1 \Rightarrow 5^{1}$ . Why? To be a multiple of both you need at least $5^{2}$ (the greedier requirement wins); to divide both you can only lean on $5^{1}$ (the poorer number caps you).

Q2. True or false: for $9 = 3^{2}$ and $27 = 3^{3}$ , the LCM is $27$ and the HCF is $9$ .

Recall Solution — Q2

Only prime is $3$ . $LCM = 3^{m a x (2, 3)} = 3^{3} = 27$ ; $HCF = 3^{m i n (2, 3)} = 3^{2} = 9$ . True. Sanity: when one number divides the other, the bigger is the LCM and the smaller is the HCF.

Q3. $14 = 2 \cdot 7$ and $15 = 3 \cdot 5$ share no prime. What is their HCF? What is their LCM?

Recall Solution — Q3

Every prime is present in only one number, so each min is $0$ . $HCF = 2^{0} \cdot 3^{0} \cdot 5^{0} \cdot 7^{0} = 1$ . $LCM = 2 \cdot 3 \cdot 5 \cdot 7 = 210 = 14 \times 15$ . Numbers with HCF $= 1$ are called coprime; their LCM is just the product.

Level 2 — Application

Goal: run the full 4-step procedure.

Q4. Find the LCM and HCF of $24$ and $36$ .

Recall Solution — Q4

Step 1 (factorize): $24 = 2^{3} \cdot 3$ , $36 = 2^{2} \cdot 3^{2}$ . Step 2 (list primes): $2$ and $3$ . Step 3 (pick powers):

$HCF$ : $2^{m i n (3, 2)} \cdot 3^{m i n (1, 2)} = 2^{2} \cdot 3^{1} = 12$ .
$LCM$ : $2^{m a x (3, 2)} \cdot 3^{m a x (1, 2)} = 2^{3} \cdot 3^{2} = 72$ . Step 4 (check): $HCF \times LCM = 12 \times 72 = 864 = 24 \times 36$ ✓.

Q5. Find the LCM of $45$ , using HCF as a shortcut, of $45$ and $75$ . (HCF first, then $LCM = \frac{ab}{HCF}$ .)

Recall Solution — Q5

$45 = 3^{2} \cdot 5$ , $75 = 3 \cdot 5^{2}$ . $HCF = 3^{m i n (2, 1)} \cdot 5^{m i n (1, 2)} = 3 \cdot 5 = 15$ . $LCM = \frac{45 \times 75}{15} = \frac{3375}{15} = 225.$ Cross-check by max powers: $3^{m a x (2, 1)} \cdot 5^{m a x (1, 2)} = 3^{2} \cdot 5^{2} = 9 \cdot 25 = 225$ ✓.

Q6. Find the LCM of $8$ , $12$ and $20$ (three numbers).

Recall Solution — Q6

$8 = 2^{3}$ , $12 = 2^{2} \cdot 3$ , $20 = 2^{2} \cdot 5$ . Take the max exponent of each prime across all three:

$2$ : $max (3, 2, 2) = 3$ .
$3$ : $max (0, 1, 0) = 1$ .
$5$ : $max (0, 0, 1) = 1$ . $LCM = 2^{3} \cdot 3 \cdot 5 = 8 \cdot 15 = 120$ .

Level 3 — Analysis

Goal: reason backwards from HCF/LCM to a missing number.

Q7. $HCF (a, b) = 6$ , $LCM (a, b) = 72$ , and $a = 24$ . Find $b$ .

Recall Solution — Q7

Two numbers, so $HCF \times LCM = a \times b$ . $b = \frac{HCF \times LCM}{a} = \frac{6 \times 72}{24} = \frac{432}{24} = 18.$ Verify: $24 = 2^{3} \cdot 3$ , $18 = 2 \cdot 3^{2}$ . $HCF = 2 \cdot 3 = 6$ ✓, $LCM = 2^{3} \cdot 3^{2} = 72$ ✓.

Q8. Can two numbers have $HCF = 6$ and $LCM = 90$ ? Justify.

Recall Solution — Q8

A necessary test: the HCF must divide the LCM (the shared part is a slice of the common multiple). Check: $90/6 = 15$ , a whole number, so it passes the divisibility test. But we need real numbers. Both must be multiples of the HCF $6$ and divisors of the LCM $90$ . Write $a = 6 m$ , $b = 6 n$ with $m, n$ coprime. Then $LCM = 6 mn = 90 \Rightarrow mn = 15$ . Coprime pairs with $mn = 15$ : $(1, 15)$ or $(3, 5)$ .

$(1, 15) \to a = 6, b = 90$ : $HCF (6, 90) = 6$ ✓, $LCM = 90$ ✓.
$(3, 5) \to a = 18, b = 30$ : $HCF (18, 30) = 6$ ✓, $LCM = 90$ ✓. Yes — and in fact two different pairs work.

Q9. Is it possible to have $HCF = 4$ and $LCM = 90$ ?

Recall Solution — Q9

Apply the divisibility test: does $HCF$ divide $LCM$ ? $90/4 = 22.5$ — not a whole number. Structurally: $HCF = 4 = 2^{2}$ demands both numbers carry $2^{2}$ , so their LCM must carry at least $2^{2}$ . But $90 = 2 \cdot 3^{2} \cdot 5$ carries only $2^{1}$ . Contradiction. Impossible.

Level 4 — Synthesis

Goal: combine several ideas at once.

Q10. Find the smallest number that is divisible by every integer from $1$ to $10$ .

Recall Solution — Q10

"Divisible by all of $1..10$ " means "a common multiple of $1, 2, \dots, 10$ " — we want the least one, i.e. $LCM (1, 2, \dots, 10)$ . Take the highest power of each prime $\leq 10$ :

$2$ : highest power $\leq 10$ is $2^{3} = 8$ (since $8$ appears via the number $8$ ).
$3$ : highest is $3^{2} = 9$ (from $9$ ).
$5$ : $5^{1}$ (from $5, 10$ ).
$7$ : $7^{1}$ (from $7$ ). $LCM = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7 = 8 \cdot 9 \cdot 5 \cdot 7 = 2520.$ Every number $1..10$ divides $2520$ , and nothing smaller does.

Q11. Two numbers are in the ratio $3 : 4$ and their HCF is $5$ . Find the numbers and their LCM.

Recall Solution — Q11

Ratio $3 : 4$ with the parts already coprime means the numbers are $3 k$ and $4 k$ where $k$ is exactly the HCF. So $k = 5$ : $a = 3 \times 5 = 15, b = 4 \times 5 = 20.$ $LCM = \frac{a \cdot b}{HCF} = \frac{15 \times 20}{5} = 60$ . Check via primes: $15 = 3 \cdot 5$ , $20 = 2^{2} \cdot 5$ , $LCM = 2^{2} \cdot 3 \cdot 5 = 60$ ✓.

Level 5 — Mastery

Goal: full real-world multi-step, no scaffolding.

Q12 (bells). Three bells ring at intervals of $9$ , $12$ and $15$ minutes. They all ring together at 10:00 a.m. When do they next all ring together, and how many times in the following $6$ hours (including 10:00) do all three coincide?

Recall Solution — Q12

"Ring together" = a common multiple of $9, 12, 15$ ; "next" = the least one, so we need $LCM (9, 12, 15)$ . $9 = 3^{2}$ , $12 = 2^{2} \cdot 3$ , $15 = 3 \cdot 5$ . $LCM = 2^{2} \cdot 3^{2} \cdot 5 = 4 \cdot 9 \cdot 5 = 180$ minutes $= 3$ hours. So they next coincide at 1:00 p.m. In a $6$ -hour window starting at 10:00, coincidences fall at $t = 0, 180, 360$ minutes → 10:00, 1:00, 4:00. That is 3 times (including the 10:00 start).

Figure — LCM — prime factorization method, relationship HCF × LCM = product

Look at the three chalk number-lines: each bell's marks are spaced by its own interval, and the yellow columns are the only times all three marks line up — exactly every

180

minutes.

Q13 (tiling). A rectangular floor is $f a r m - sc a l e :$ 6 $mb y$ 8 $m$ 40 $c m . W ai t — c l e an l y : a f l oor i s$ 6,\text{m} = 600,\text{cm} $b y$ 8.4,\text{m} = 840,\text{cm}$. Find the side of the largest square tile that tiles it exactly (no cutting), and the number of tiles.

Recall Solution — Q13

"Largest square that fits both sides exactly" = the greatest common divisor of the side lengths, so we need $HCF (600, 840)$ . $600 = 2^{3} \cdot 3 \cdot 5^{2}$ , $840 = 2^{3} \cdot 3 \cdot 5 \cdot 7$ . $HCF = 2^{m i n (3, 3)} \cdot 3^{m i n (1, 1)} \cdot 5^{m i n (2, 1)} \cdot 7^{m i n (0, 1)} = 2^{3} \cdot 3 \cdot 5 = 120$ . So the largest tile is $120 cm \times 120 cm$ . Number of tiles $= \frac{600}{120} \times \frac{840}{120} = 5 \times 7 = 35$ tiles.

The grid shows

5

tiles along the

600

side and

7

along the

840

side — the biggest square that divides both lengths evenly.

Q14 (syncing cycles). Two runners circle a track. Runner A completes a lap every $126$ s, runner B every $154$ s. They start together. After how long are they together at the start line again, and how many laps has each run by then?

Recall Solution — Q14

They meet at the start when the elapsed time is a common multiple of both lap times; "again, soonest" = $LCM (126, 154)$ . $126 = 2 \cdot 3^{2} \cdot 7$ , $154 = 2 \cdot 7 \cdot 11$ . $HCF = 2 \cdot 7 = 14$ , so use the shortcut: $LCM = \frac{126 \times 154}{14} = \frac{19404}{14} = 1386 s .$ $1386 s = 23 min 6 s$ . Laps: A runs $1386/126 = 11$ laps; B runs $1386/154 = 9$ laps. Notice $11$ and $9$ are coprime — that's the signature of an LCM meeting: each finishes a whole number of laps with no common factor left over.

Recall Self-check summary

Which key equation rescued Q7? ::: HCF × LCM = a × b (two numbers). What test killed Q9 instantly? ::: HCF must divide LCM; 90/4 is not an integer. Why is the tiling answer an HCF, not an LCM? ::: We want the largest square that divides both side lengths — a common divisor.

Connections

LCM — prime factorization method, relationship HCF × LCM = product
HCF — prime factorization and Euclid's algorithm
Fundamental Theorem of Arithmetic
Prime Numbers and Factorization
Divisibility Rules
Applications — bells ringing, tiling, syncing cycles