Problems on H.C.F and L.C.M

1. Core Concepts & Theoretical Blueprint

HCF (Highest Common Factor, also called GCD) is the largest number that divides two or more given numbers exactly, while LCM (Lowest Common Multiple) is the smallest number that is exactly divisible by two or more given numbers. These two concepts are structural opposites — HCF looks for shared factors, LCM looks for shared multiples — and their relationship underlies nearly every question type in this chapter.

Methods of Finding HCF and LCM:

• Prime Factorization Method: HCF = product of common prime factors, each raised to the LOWEST power appearing in any of the numbers. LCM = product of ALL prime factors involved, each raised to the HIGHEST power appearing in any number.
• Division Method (Euclidean Algorithm) for HCF: Divide the larger number by the smaller; replace the larger with the remainder; repeat until remainder is 0 — the last non-zero divisor is the HCF.

The Core Product Relationship (valid ONLY for exactly two numbers):

HCF and LCM of Fractions:

Remainder-Type Formulas (extremely high-frequency in exams):

• Greatest number that divides x, y, z leaving the same remainder r in each case:
  $$\text{HCF}(x-r,\ y-r,\ z-r)$$
• Greatest number that divides x, y, z leaving different remainders $r_1, r_2, r_3$ respectively:
  $$\text{HCF}(x-r_1,\ y-r_2,\ z-r_3)$$
• Smallest number that, when divided by x, y, z, leaves the same remainder r in each case:
  $$\text{LCM}(x,y,z) + r$$
• Smallest number that, when divided by x, y, z, leaves remainders $r_1, r_2, r_3$ such that $(x-r_1)=(y-r_2)=(z-r_3)=k$ (a constant difference):
  $$\text{LCM}(x,y,z) - k$$
• Greatest number that divides N leaving remainder r: this is simply the greatest factor of $(N-r)$ that exceeds r — often just $(N-r)$ itself if asked for the greatest such divisor overall.

The Universal Trap: Four critical traps:

1. Applying HCF × LCM = product for three or more numbers — this identity is valid ONLY for exactly two numbers; for three or more, HCF and LCM must be computed independently via factorization, never via this shortcut.
2. Confusing "same remainder" and "different remainder" formulas — the same-remainder case subtracts r uniformly before taking HCF, while the different-remainder case subtracts each number's OWN specified remainder; mixing these up is the single most common scoring error in this chapter.
3. Sign error in the LCM-based remainder formulas — "leaves remainder r always" adds r to the LCM; "leaves remainders that are each (divisor − constant)" subtracts that constant from the LCM — these are structurally opposite operations easily swapped under time pressure.
4. Forgetting HCF of fractions uses LCM of denominators (not HCF), and LCM of fractions uses HCF of denominators (not LCM) — the numerator and denominator rules are deliberately crossed/inverted, and this is exploited heavily as a trap in fraction-based HCF/LCM questions.

2. Exhaustive Question Typology

                        PROBLEMS ON HCF AND LCM
                                  |
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    |            |              |               |               |          |
Type 1:       Type 2:        Type 3:        Type 4:         Type 5:     Type 6:
Basic HCF/    HCF×LCM=       HCF/LCM of     Greatest        Smallest    Bells/Traffic
LCM by        Product         Fractions      Number          Number      Lights
Factorization Relation (2                    Dividing        Divisible   Ringing
or Division   numbers only)                  x,y,z,          by x,y,z,   Together
Method                                       Leaving         Leaving     (LCM as
                                              Remainder(s)    Remainder   Time
                                                              (s)         Application)
    |            |              |
Type 7:       Type 8:        Type 9:
Least Number  HCF/LCM of     Numbers in
to be Added/  3+ Numbers,    Given Ratio
Subtracted    Successive     with Known
for Exact     Division       HCF (find
Divisibility  Method          actual numbers
                              or LCM)

Type 1 — Basic HCF/LCM computation:

• Core Scenario: "Find the HCF and LCM of 84, 120, and 140."
• Governing Equation: Prime factorization method: HCF = product of common primes at lowest powers; LCM = product of all primes at highest powers.

Type 2 — HCF × LCM = Product relation (two numbers only):

• Core Scenario: "The HCF and LCM of two numbers are 12 and 336 respectively. If one number is 48, find the other."
• Governing Equation: $\text{Other number} = \dfrac{\text{HCF}\times\text{LCM}}{\text{Given number}}$

Type 3 — HCF/LCM of fractions:

• Core Scenario: "Find the HCF and LCM of $\frac{2}{3}, \frac{4}{9}, \frac{6}{15}$."
• Governing Equation: HCF $=\dfrac{\text{HCF(numerators)}}{\text{LCM(denominators)}}$; LCM $=\dfrac{\text{LCM(numerators)}}{\text{HCF(denominators)}}$

Type 4 — Greatest number dividing x, y, z leaving remainder(s):

• Core Scenario: "Find the greatest number that divides 285 and 1249, leaving remainders 9 and 7 respectively."
• Governing Equation: $\text{HCF}(x-r_1, y-r_2)$ for different remainders; $\text{HCF}(x-r,y-r)$ for same remainder.

Type 5 — Smallest number divisible by x, y, z leaving remainder(s):

• Core Scenario: "Find the smallest number which when divided by 4, 5, and 6 leaves remainder 3 in each case."
• Governing Equation: $\text{LCM}(x,y,z)+r$ for same remainder in each case.

Type 6 — Bells/traffic lights ringing/changing together (LCM as a time-synchronization tool):

• Core Scenario: "Three bells ring at intervals of 12, 15, and 20 minutes respectively. If they ring together at 8:00 AM, when will they next ring together?"
• Governing Equation: $\text{Time of next simultaneous event} = \text{LCM}(\text{all intervals})$, added to the starting time.

Type 7 — Least number to be added/subtracted for exact divisibility:

• Core Scenario: "Find the least number that must be added to 1056 to make it exactly divisible by 23," or "subtracted to make it divisible."
• Governing Equation: For addition: find remainder $r$ when dividing by the divisor; number to add $= (\text{divisor}-r)$ (or 0 if $r=0$). For subtraction: the number to subtract $= r$ itself.

Type 8 — HCF/LCM of 3+ numbers via successive division:

• Core Scenario: "Find the HCF of 391, 425, and 527 using the successive/continued division method."
• Governing Equation: Apply Euclidean algorithm pairwise: $\text{HCF}(a,b,c)=\text{HCF}(\text{HCF}(a,b),c)$

Type 9 — Numbers in a given ratio with known HCF (find actual numbers/LCM):

• Core Scenario: "Two numbers are in the ratio 3:4, and their HCF is 4. Find their LCM," or "find the numbers."
• Governing Equation: If numbers are in ratio $a:b$ (in lowest terms) with HCF = h, actual numbers are $ah$ and $bh$; since a, b are coprime, $\text{LCM}=a\times b\times h$.

3. Type-wise Practice MCQs with Full Solutions

Type 1 — Basic HCF/LCM Computation

MCQ 1. Find the HCF of 84 and 144. (A) 12 (B) 24 (C) 6 (D) 36

Correct Answer: (A) Solution: $84=2^2\times3\times7$; $144=2^4\times3^2$. Common primes at lowest power: $2^2\times3=12$.

MCQ 2. Find the LCM of 15, 20, and 30. (A) 60 (B) 90 (C) 120 (D) 150

Correct Answer: (A) Solution: $15=3\times5$; $20=2^2\times5$; $30=2\times3\times5$. LCM = product of all primes at highest power $=2^2\times3\times5=60$.

MCQ 3. Find the HCF of 391, 425, and 527 using successive division. (A) 17 (B) 13 (C) 19 (D) 23

Correct Answer: (A) Solution: HCF(391,425): $425=391\times1+34$; $391=34\times11+17$; $34=17\times2+0$ → HCF=17. Now HCF(17,527): $527\div17=31$ exactly → HCF=17. So overall HCF = 17.

Type 2 — HCF × LCM = Product Relation

MCQ 1. The HCF and LCM of two numbers are 9 and 693 respectively. If one of the numbers is 63, find the other. (A) 99 (B) 89 (C) 109 (D) 91

Correct Answer: (A) Solution: Other number $=\dfrac{\text{HCF}\times\text{LCM}}{\text{Given number}}=\dfrac{9\times693}{63}=\dfrac{6237}{63}=99$.

MCQ 2. Two numbers are in the ratio 4:5. If their HCF is 8, find their LCM. (A) 160 (B) 140 (C) 180 (D) 200

Correct Answer: (A) Solution: Numbers are $4\times8=32$ and $5\times8=40$. LCM $=a\times b\times h=4\times5\times8=160$ (using coprime-ratio shortcut).

MCQ 3. The product of two numbers is 2028, and their HCF is 13. Find the number of such possible pairs. (A) 2 (B) 3 (C) 4 (D) 1

Correct Answer: (A) Solution: If HCF=13, numbers can be written as $13a$ and $13b$ where $\gcd(a,b)=1$. Product $=169ab=2028\Rightarrow ab=12$. Coprime pairs (a,b) with product 12: (1,12) and (3,4) — both coprime pairs. So there are 2 such pairs.

Type 3 — HCF/LCM of Fractions

MCQ 1. Find the HCF of $\dfrac{4}{9}, \dfrac{10}{21}, \dfrac{2}{3}$. (A) $\dfrac{2}{63}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{4}{63}$ (D) $\dfrac{2}{21}$

Correct Answer: (A) Solution: HCF of numerators (4,10,2) $=2$. LCM of denominators (9,21,3) $=63$. HCF of fractions $=\dfrac{2}{63}$.

MCQ 2. Find the LCM of $\dfrac{2}{3}, \dfrac{4}{9}, \dfrac{6}{15}$. (A) 12 (B) $\dfrac{12}{1}$ (C) 4 (D) 24

Correct Answer: (A) Solution: LCM of numerators (2,4,6) $=12$. HCF of denominators (3,9,15) $=3$. LCM of fractions $=\dfrac{12}{3}=4$. (Recheck arithmetic — this gives 4, not 12; correcting the marked option.)

MCQ 2 (verified). Find the LCM of $\dfrac{2}{3}, \dfrac{4}{9}, \dfrac{6}{15}$. (A) 4 (B) 12 (C) 8 (D) 24

Correct Answer: (A) Solution: As derived: LCM(numerators)=12, HCF(denominators)=3, LCM of fractions $=\dfrac{12}{3}=4$.

MCQ 3. Find the HCF of $2\dfrac{1}{4}, 3\dfrac{3}{8}, 5\dfrac{5}{6}$ (convert to improper fractions first). (A) $\dfrac{1}{24}$ (B) $\dfrac{1}{12}$ (C) $\dfrac{3}{24}$ (D) $\dfrac{1}{8}$

Correct Answer: (A) Solution: Convert: $2\frac14=\frac94$, $3\frac38=\frac{27}{8}$, $5\frac56=\frac{35}{6}$. HCF of numerators (9,27,35): HCF(9,27)=9, HCF(9,35)=1 → overall HCF=1. LCM of denominators (4,8,6): LCM=24. HCF of fractions $=\dfrac{1}{24}$.

Type 4 — Greatest Number Dividing with Remainder(s)

MCQ 1. Find the greatest number that divides 615 and 963, leaving remainder 6 in each case. (A) 87 (B) 78 (C) 93 (D) 69

Correct Answer: (A) Solution: Required number $=\text{HCF}(615-6, 963-6)=\text{HCF}(609,957)$. Using Euclidean algorithm: $957=609\times1+348$; $609=348\times1+261$; $348=261\times1+87$; $261=87\times3+0$ → HCF=87.

MCQ 2. Find the greatest number that will divide 285 and 1249, leaving remainders 9 and 7 respectively. (A) 23 (B) 46 (C) 69 (D) 92

Correct Answer: (A) Solution: Required number $=\text{HCF}(285-9,1249-7)=\text{HCF}(276,1242)$. $1242=276\times4+138$; $276=138\times2+0$ → HCF=138. (Recheck: this gives 138, not 23; let's re-verify factor: $1242/138=9$ exactly, $276/138=2$ exactly, confirming HCF=138. Correcting the option set below.)

MCQ 2 (verified). Find the greatest number that will divide 285 and 1249, leaving remainders 9 and 7 respectively. (A) 138 (B) 46 (C) 69 (D) 92

Correct Answer: (A) Solution: As derived: HCF(276, 1242) = 138 via the Euclidean algorithm.

MCQ 3. Find the greatest number that divides 43, 91, and 183, leaving the same remainder in each case. (A) 4 (B) 8 (C) 12 (D) 16

Correct Answer: (A) Solution: Since the remainder is the same but unknown, use pairwise differences: $\text{HCF}(91-43, 183-91, 183-43)=\text{HCF}(48,92,140)$. $\text{HCF}(48,92)$: $92=48\times1+44$; $48=44\times1+4$; $44=4\times11+0$→HCF=4. Check HCF(4,140)=4. So required number = 4.

Type 5 — Smallest Number Divisible with Remainder(s)

MCQ 1. Find the smallest number which, when divided by 6, 9, and 12, leaves remainder 5 in each case. (A) 41 (B) 36 (C) 45 (D) 40

Correct Answer: (A) Solution: $\text{LCM}(6,9,12)=36$. Required number $=36+5=41$.

MCQ 2. Find the least number which, when divided by 5, 6, 7, and 8, leaves remainder 3 in each case, but is exactly divisible by 9. (A) 723 (B) 843 (C) 963 (D) 723 + something

Correct Answer: (A) Solution: $\text{LCM}(5,6,7,8)=840$. Numbers satisfying the remainder condition: $840k+3$. We need this divisible by 9. Test k=1: $840+3=843$; $843\div9=93.67$ (not exact). Test k=0: $3$, not divisible by 9 and too small. Systematically: $840\equiv840-93\times9=840-837=3\pmod9$, so $840k+3\equiv3k+3\pmod9$. Need $3k+3\equiv0\pmod9\Rightarrow3(k+1)\equiv0\pmod9\Rightarrow k+1\equiv0\pmod3\Rightarrow k\equiv2\pmod3$. Smallest such k=2: $840\times2+3=1683$. Check $1683\div9=187$, exact. So the answer is 1683, not matching initial guess — correct the option set.

MCQ 2 (verified). Find the least number which, when divided by 5, 6, 7, and 8, leaves remainder 3 in each case, but is exactly divisible by 9. (A) 1683 (B) 843 (C) 963 (D) 1443

Correct Answer: (A) Solution: As derived: the general form is $840k+3$, and the smallest k satisfying divisibility by 9 is $k=2$, giving $840\times2+3=1683$.

MCQ 3. Find the smallest number which, when divided by 7, 11, and 13, leaves remainders 6, 10, and 12 respectively. (A) 1000 (B) 1001 (C) 999 (D) 1002

Correct Answer: (C) Solution: Notice $(7-6)=(11-10)=(13-12)=1$, a constant difference k=1. Required number $=\text{LCM}(7,11,13)-1=1001-1=1000$. (Recheck: LCM(7,11,13)=1001 since all are pairwise coprime; formula gives 1001−1=1000. Correcting marked answer to (A).)

MCQ 3 (verified). Correct Answer: (A) 1000 Solution: As derived: LCM(7,11,13)=1001; constant difference k=1; required number = 1001−1=1000.

Type 6 — Bells/Traffic Lights Ringing Together

MCQ 1. Three bells ring at intervals of 6, 8, and 12 minutes respectively. If they all ring together at 9:00 AM, at what time will they next ring together? (A) 9:24 AM (B) 9:30 AM (C) 9:20 AM (D) 9:36 AM

Correct Answer: (A) Solution: $\text{LCM}(6,8,12)=24$. Next simultaneous ring = 9:00 AM + 24 minutes = 9:24 AM.

MCQ 2. Four bells toll at intervals of 5, 6, 8, and 12 seconds respectively. In 30 minutes, how many times do they toll together (including the start)? (A) 16 (B) 15 (C) 18 (D) 20

Correct Answer: (A) Solution: $\text{LCM}(5,6,8,12)=120$ seconds = 2 minutes. In 30 minutes, they toll together every 2 minutes, so number of simultaneous tolls (including the starting toll at t=0) $=\dfrac{30}{2}+1=15+1=16$.

MCQ 3. Three traffic lights change at intervals of 48, 72, and 108 seconds. If they change simultaneously at 10:00:00 AM, find the next time they change together. (A) 10:07:12 AM (B) 10:05:00 AM (C) 10:06:00 AM (D) 10:08:00 AM

Correct Answer: (A) Solution: $\text{LCM}(48,72,108)$: $48=2^4\times3$; $72=2^3\times3^2$; $108=2^2\times3^3$. LCM $=2^4\times3^3=16\times27=432$ seconds $=7$ minutes 12 seconds. Next simultaneous change = 10:00:00 + 7:12 = 10:07:12 AM.

Type 7 — Least Number to Add/Subtract for Exact Divisibility

MCQ 1. Find the least number that must be added to 1056 to make it exactly divisible by 23. (A) 2 (B) 3 (C) 1 (D) 4

Correct Answer: (A) Solution: $1056\div23=45$ remainder $21$ (since $23\times45=1035$, $1056-1035=21$). Number to add $=23-21=2$.

MCQ 2. Find the least number that must be subtracted from 9999 to make it exactly divisible by 73. (A) 43 (B) 30 (C) 45 (D) 28

Correct Answer: (A) Solution: $9999\div73$: $73\times137=10001$, too large; $73\times136=9928$; $9999-9928=71$. Number to subtract $=71$. (Recheck: recompute $73\times136$: $73\times136=73\times100+73\times36=7300+2628=9928$; $9999-9928=71$. This gives 71, not matching option (A). Correcting the option set.)

MCQ 2 (verified). Find the least number that must be subtracted from 9999 to make it exactly divisible by 73. (A) 71 (B) 30 (C) 45 (D) 28

Correct Answer: (A) Solution: As derived: remainder on dividing 9999 by 73 is 71, so 71 must be subtracted to leave a number exactly divisible by 73.

MCQ 3. Find the smallest 4-digit number exactly divisible by 12, 15, and 20. (A) 1020 (B) 1000 (C) 1080 (D) 1200

Correct Answer: (A) Solution: $\text{LCM}(12,15,20)=60$. Smallest 4-digit number $=1000$. $1000\div60=16$ remainder $40$. Number to add $=60-40=20$. Required number $=1000+20=1020$.

Type 8 — HCF/LCM of 3+ Numbers via Successive Division

MCQ 1. Find the HCF of 96, 240, and 336. (A) 48 (B) 24 (C) 16 (D) 32

Correct Answer: (A) Solution: HCF(96,240): $240=96\times2+48$; $96=48\times2+0$→HCF=48. Check HCF(48,336): $336\div48=7$ exact → HCF=48. Overall HCF = 48.

MCQ 2. Find the LCM of 24, 36, and 40. (A) 360 (B) 240 (C) 480 (D) 720

Correct Answer: (A) Solution: $24=2^3\times3$; $36=2^2\times3^2$; $40=2^3\times5$. LCM $=2^3\times3^2\times5=8\times9\times5=360$.

MCQ 3. Find the HCF of 1152 and 1664 using the division method. (A) 128 (B) 64 (C) 96 (D) 32

Correct Answer: (A) Solution: $1664=1152\times1+512$; $1152=512\times2+128$; $512=128\times4+0$ → HCF=128.

Type 9 — Numbers in Given Ratio with Known HCF

MCQ 1. Two numbers are in the ratio 5:6, and their HCF is 4. Find the numbers. (A) 20, 24 (B) 15, 18 (C) 10, 12 (D) 25, 30

Correct Answer: (A) Solution: Since 5 and 6 are coprime, numbers = $5\times4=20$ and $6\times4=24$.

MCQ 2. Two numbers are in the ratio 3:4. Their LCM is 84. Find their HCF. (A) 7 (B) 12 (C) 6 (D) 21

Correct Answer: (A) Solution: Since 3 and 4 are coprime, LCM $=3\times4\times h=12h=84\Rightarrow h=7$.

MCQ 3. Two numbers are in the ratio 2:3. If their LCM is 48, find the sum of the numbers. (A) 40 (B) 45 (C) 36 (D) 50

Correct Answer: (A) Solution: LCM $=2\times3\times h=6h=48\Rightarrow h=8$. Numbers $=2\times8=16$ and $3\times8=24$. Sum $=16+24=40$.

4. High-Yield Speed Tricks & Shortcut Mental Models

Shortcut 1 — The Coprime-Ratio LCM Shortcut

• Application: Any Type 9 problem where two numbers are given in a ratio and either HCF or LCM is known.
• Mental Model: If numbers are in ratio $a:b$ with $a,b$ coprime (always true once a ratio is expressed in lowest terms) and HCF = h, the actual numbers are simply $ah$ and $bh$, and since $a,b$ share no common factor, their LCM is trivially $a\times b\times h$ — no need to separately factorize the actual numbers at all.

Shortcut 2 — Same-Remainder vs Different-Remainder Instant Recognition

• Application: Every Type 4/5 remainder-based problem — the fastest way to avoid the chapter's #1 trap.
• Mental Model: Before writing anything, explicitly check: does the problem state ONE remainder that applies to all numbers, or DIFFERENT remainders per number? If one remainder r for all: subtract r uniformly, then HCF (for "greatest divisor" type) or add r to LCM (for "smallest dividend" type). If different remainders with a CONSTANT gap $(x-r_1)=(y-r_2)$: this signals the LCM-minus-constant formula instead. Naming which sub-case applies out loud (even mentally) before computing prevents the most common error in the entire chapter.

Shortcut 3 — Euclidean Algorithm Over Full Factorization for Large Numbers

• Application: Any HCF computation involving numbers too large to factorize quickly (3+ digit numbers, especially primes-heavy ones).
• Mental Model: Never attempt full prime factorization of large, awkward numbers under time pressure. The Euclidean algorithm (repeated "divide larger by smaller, replace with remainder") reaches the HCF in a small, bounded number of steps regardless of how large or factorization-resistant the numbers are — always faster than searching for prime factors of an unfamiliar large number.

5. Deep-Dive: Most Frequently Asked Questions

Problem 1 (SSC/RRB Standard): Find the greatest number that will divide 445, 572, and 699, leaving remainders 4, 5, and 6 respectively.

Traditional Method (Slow): Subtract respective remainders: $445-4=441$; $572-5=567$; $699-6=693$. Find HCF(441,567): $567=441\times1+126$; $441=126\times3+63$; $126=63\times2+0$→HCF=63. Verify HCF(63,693): $693\div63=11$, exact → HCF=63. (Requires three separate subtractions, then a full pairwise Euclidean algorithm across three numbers — ~50-55 seconds.)

Exam Shortcut (Fast): Recognize immediately this is a "different remainders" case (4,5,6 are all different) → apply $\text{HCF}(x-r_1,y-r_2,z-r_3)$ directly without hesitation, then run the Euclidean algorithm efficiently in one pass by finding HCF of the two smaller adjusted numbers first (441,567→63), then a single quick division check against the third (693÷63=11 exact) — same total work as above but executed with zero hesitation about which formula applies, saving the 10-15 seconds typically lost second-guessing "same or different remainder." Answer: 63.

Problem 2 (UPSC/Banking Advanced): Three pieces of timber measuring 42 m, 49 m, and 63 m in length are to be cut into planks of equal length, with no wood wasted. Find the greatest possible length of each plank, and the total number of planks obtained.

Step-by-Step Breakdown:

1. Since each piece must be cut into planks of EQUAL length with none wasted, the plank length must exactly divide all three timber lengths — and to maximize plank length (minimizing the number of cuts/waste), we need the GREATEST such length, i.e., the HCF of 42, 49, and 63.
2. Prime factorize: $42=2\times3\times7$; $49=7^2$; $63=3^2\times7$.
3. Common prime factor across all three: only 7 (2 and 3 do not appear in 49; 7² doesn't divide 42 or 63 fully). HCF $=7$.
4. Greatest possible plank length = 7 m.
5. Number of planks from each piece: $\dfrac{42}{7}=6$; $\dfrac{49}{7}=7$; $\dfrac{63}{7}=9$.
6. Total number of planks $=6+7+9=22$.
7. Answer: The greatest possible length of each plank is 7 m, yielding a total of 22 planks. This is a classic "maximize the common measuring unit" application of HCF, structurally identical to tiling/flooring problems that ask for the largest square tile that can exactly cover a rectangular floor.

6. Chapter Checklist for Students

• I only apply HCF × LCM = Product for exactly TWO numbers, never for three or more.
• I explicitly identify whether a remainder problem uses the SAME remainder for all numbers or DIFFERENT remainders before selecting the corresponding HCF/LCM formula.
• I correctly invert numerator/denominator operations for fraction HCF/LCM: HCF of fractions uses LCM of denominators; LCM of fractions uses HCF of denominators.
• I use the Euclidean division algorithm instead of full prime factorization whenever numbers are large or factorization-resistant.
• I recognize "equal-length cutting" and "largest tile covering a floor" word problems as disguised HCF applications, and "bells/lights ringing together" or "events recurring simultaneously" as disguised LCM applications.