Lcm & Hcf
Key Concepts & Formulas
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Prime-Factor Method | Break every number into primes; LCM = product of highest powers of all primes, HCF = product of lowest powers of common primes. |
| 2 | LCM × HCF = Product | For any two positive integers a & b: LCM(a,b) × HCF(a,b) = a × b. |
| 3 | Division Method | Repeatedly divide the larger number by the smaller; remainder becomes new divisor until 0; last non-zero remainder = HCF. |
| 4 | Coprime Numbers | Two numbers whose HCF is 1; their LCM is simply their product. |
| 5 | LCM of Fractions | LCM = LCM(numerators) ÷ HCF(denominators); HCF of fractions = HCF(numerators) ÷ LCM(denominators). |
| 6 | Remainder Consistency | If N leaves same remainder r when divided by a, b, c… then N = k × LCM(a,b,c…) + r. |
10 Practice MCQs
1. The HCF of two numbers is 12 and their product is 2592. Find their LCM.
**Answer:** 216 **Solution:** LCM × HCF = Product ⇒ LCM = 2592 / 12 = 216 **Shortcut:** Direct division once HCF known. **Tag:** LCM-HCF-product relation2. Find the smallest 3-digit number exactly divisible by 12, 15 and 18.
**Answer:** 180 **Solution:** LCM(12,15,18)=180; smallest 3-digit multiple is 180 itself. **Shortcut:** LCM first, then first multiple in range. **Tag:** Smallest number divisible3. The LCM of two coprime numbers is 255. If one number is 15, the other is
**Answer:** 17 **Solution:** For coprime numbers, LCM = product ⇒ 255 = 15 × x ⇒ x = 17 **Shortcut:** Product = LCM when HCF = 1. **Tag:** Coprime property4. Three bells toll at intervals 8, 12 and 18 minutes. If they toll together at 8:00 am, when next together?
**Answer:** 9:12 am **Solution:** LCM(8,12,18)=72 min ⇒ 8:00 + 72 min = 9:12 am **Shortcut:** LCM of intervals gives simultaneous period. **Tag:** Real-life LCM5. Find HCF of 1.5, 2.5 and 3.5.
**Answer:** 0.5 **Solution:** Make integers: 15, 25, 35 → HCF = 5 → divide by 10 ⇒ 0.5 **Shortcut:** Remove decimal, find HCF, restore decimal. **Tag:** Decimal HCF6. Two numbers are in ratio 3:4 and their HCF is 8. Their sum is
**Answer:** 56 **Solution:** Numbers = 3×8 = 24 & 4×8 = 32; sum = 56 **Shortcut:** Multiply ratio terms by HCF. **Tag:** Ratio & HCF7. The largest number that divides 403, 434 and 465 leaving remainder 3 in each case is
**Answer:** 31 **Solution:** Subtract 3: 400, 431, 462 → HCF = 31 **Shortcut:** Remainder adjustment → HCF of reduced numbers. **Tag:** Common remainder8. LCM of fractions 2/3, 4/5, 5/6 is
**Answer:** 20 **Solution:** LCM(num)=2×4×5=40; HCF(den)=1; LCM(frac)=40/1=40 → but simplified form 20/1=20 **Shortcut:** LCM(num)/HCF(den). **Tag:** Fraction LCM9. Find the least square number divisible by 8, 12 and 15.
**Answer:** 3600 **Solution:** LCM(8,12,15)=120; make perfect square ⇒ 120×2×3×5=3600 **Shortcut:** LCM first, then multiply missing prime pairs to square. **Tag:** Perfect square & LCM10. If HCF of 408 and 544 is 136, their LCM is
**Answer:** 1632 **Solution:** Product = 408×544 = 221952; LCM = 221952/136 = 1632 **Shortcut:** Use product formula. **Tag:** Reverse calculation5 Previous Year Questions
[RRB NTPC 2021] The product of two numbers is 2160 and their HCF is 12. Find their LCM.
**Answer:** 180 **Solution:** LCM = 2160 / 12 = 180 **Tag:** Product formula[RRB Group-D 2019] Find the greatest number that divides 1657 and 2037 leaving remainder 7.
**Answer:** 127 **Solution:** 1657-7=1650; 2037-7=2030; HCF(1650,2030)=127 **Tag:** Remainder type[RRB NTPC 2016] Three traffic lights change every 25, 40 and 60 seconds. If they change together at 7:00 am, next together?
**Answer:** 7:05 am **Solution:** LCM(25,40,60)=600 s = 10 min **Tag:** Real-life LCM[RRB ALP 2018] HCF of 1.75, 2.25 and 3.5 is
**Answer:** 0.25 **Solution:** 175, 225, 350 → HCF = 25 → 25/100 = 0.25 **Tag:** Decimal HCF[RRB NTPC 2020] Two numbers are in ratio 5:7 and their LCM is 315. Find their HCF.
**Answer:** 9 **Solution:** Let numbers = 5x, 7x; LCM = 35x = 315 ⇒ x = 9 = HCF **Tag:** Ratio & LCMSpeed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Same remainder r | Subtract r from each, find HCF | HCF(123−3, 237−3)=HCF(120,234)=6 |
| Coprime check | HCF must be 1 | HCF(15,22)=1 → coprime |
| Decimal HCF | Multiply by 100, find HCF, divide back | HCF(1.2,1.5)=HCF(12,15)/10=3/10=0.3 |
| LCM of (a, 2a, 3a) | Simply 6a | LCM(7,14,21)=42 |
| Quick 2-number LCM | Use LCM = (a×b)/HCF | a=18, b=24, HCF=6 → LCM=72 |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Ignoring remainder | Directly taking HCF of original numbers | Always adjust by subtracting remainder first |
| Decimal misplacement | Forgetting to divide back after removing decimal | Restore same number of decimal places |
| Fraction LCM upside-down | Using HCF of denominators instead of LCM | Remember: LCM(frac)=LCM(num)/HCF(den) |
| Assuming numbers coprime | Seeing small numbers | Always verify HCF=1 before treating as coprime |
Quick Revision Flashcards
| Front | Back |
|---|---|
| LCM × HCF for two numbers equals | their product |
| HCF of coprime numbers | 1 |
| Smallest number divisible by a,b,c | LCM(a,b,c) |
| Largest number dividing a,b,c leaving remainder r | HCF(a−r, b−r, c−r) |
| LCM of fractions formula | LCM(num)/HCF(den) |
| HCF of decimals trick | Remove decimal, find HCF, restore |
| If ratio a:b and HCF = h, numbers are | ah and bh |
| Three numbers in ratio 2:3:4 with HCF 5 are | 10, 15, 20 |
| Perfect square divisible by given numbers | LCM first, then pair up all primes |
| Remainder 5 when divided by 6,8,10 → number form | N = k·LCM(6,8,10)+5 = 120k+5 |
BETA
