1. Standard Representation of Two Numbers
The most useful idea in advanced HCF-LCM problems is to remove the HCF from both numbers. This converts the original pair into a pair of co-prime integers and makes many apparently difficult questions much simpler.
Fundamental Representation:
If HCF of two positive integers is h, write:
First Number = hm
Second Number = hn
where HCF(m,n)=1.
Then:
HCF = h
LCM = hmn
Product of Numbers = h2mn
Example:
If HCF=12 and the reduced co-prime factors are 5 and 7, then:
Numbers=12×5 and 12×7
=60 and 84.
LCM=12×5×7=420.
Exam Insight:
The reduced factors m and n must be co-prime. This condition is essential in advanced pair-finding questions.
2. Numbers Given in a Ratio with HCF
If two numbers are in a ratio and their HCF is given, first reduce the ratio to its lowest terms.
If the reduced ratio is m:n and HCF=h:
Numbers = hm and hn
LCM = hmn
Example:
Ratio of two numbers=3:5 and HCF=8.
Numbers:
24 and 40.
LCM=8×3×5=120.
Important:
The ratio parts must first be co-prime.
For example, if the ratio is 6:10, reduce it to 3:5 before using the HCF as the common multiplier.
Example:
Ratio=6:10 and HCF=14.
Reduced ratio=3:5.
Numbers=42 and 70.
LCM=14×3×5=210.
3. Numbers Given in a Ratio with LCM
If the reduced ratio m:n and the LCM are given, use the fact that LCM=hmn.
Formula:
HCF = LCM ÷ (m×n)
Example:
Two numbers are in the ratio 4:7 and their LCM is 420.
Since 4 and 7 are co-prime:
HCF=420÷(4×7)
=420÷28
=15.
Numbers=60 and 105.
Validity Check:
For a valid reduced ratio m:n, the given LCM must be divisible by m×n.
4. Finding Number Pairs from HCF and LCM
If both HCF and LCM are given, the possible number pairs can be generated systematically.
Let:
HCF=h
LCM=L
Numbers=hm and hn, where HCF(m,n)=1.
Key Relation:
mn = L/h
Method:
1. Calculate N=L/h.
2. Find factor pairs of N.
3. Keep only the pairs whose two factors are co-prime.
4. Multiply both factors by h.
Example:
HCF=12 and LCM=420.
N=420/12=35.
Co-prime factor pairs of 35:
(1,35) and (5,7).
Therefore possible unordered number pairs are:
(12,420)
(60,84).
Common Mistake:
Every factor pair of L/H does not necessarily work. The reduced factors must be co-prime.
5. Counting the Number of Possible Pairs
Questions sometimes ask only how many pairs can have a given HCF and LCM.
Basic Method:
Let N=LCM/HCF.
Count the unordered co-prime factor pairs (m,n) satisfying mn=N.
Example:
HCF=18 and LCM=900.
N=900/18=50.
Factor pairs:
(1,50), (2,25), (5,10).
Co-prime pairs:
(1,50), (2,25).
Therefore number of valid unordered pairs=2.
Advanced Shortcut:
If N>1 has k distinct prime factors, the number of unordered co-prime factor pairs is:
2k−1.
Example:
Suppose LCM/HCF=105.
105=3×5×7 contains 3 distinct prime factors.
Number of possible unordered pairs:
23−1=4.
Why the Shortcut Works:
Each complete prime-power block must go entirely to one reduced factor or the other; otherwise the reduced factors would not remain co-prime.
Special Case:
If LCM=HCF, then the only possible pair is the equal pair (HCF,HCF).
6. Product Given with HCF or LCM
The product relation is especially useful when product, HCF or LCM is known.
For two numbers:
Product = HCF × LCM
If Product P and HCF h are given:
LCM=P/h.
If Product P and LCM L are given:
HCF=P/L.
For Pair Construction when Product and HCF are given:
If numbers=hm and hn, then:
mn=P/h2.
Example:
Product=6760 and HCF=13.
mn=6760/132
=6760/169
=40.
Co-prime factor pairs of 40:
(1,40) and (5,8).
Hence two unordered number pairs are possible.
Validity Check:
P must be divisible by h2 if h is truly the HCF of both numbers.
7. Sum or Difference Given with HCF
If HCF=h and numbers are hm and hn, then sum and difference conditions become conditions on m and n.
If Sum=S:
m+n = S/h
If Difference=D:
|m−n| = D/h
Necessary Condition:
S or D must be divisible by the HCF whenever the corresponding equation is expected to produce integer reduced factors.
Example:
Two numbers have HCF=12 and sum=96.
m+n=96/12=8.
Possible positive co-prime pairs with sum 8 include:
(1,7) and (3,5).
Hence possible number pairs include:
(12,84) and (36,60).
Note:
Sum or difference with HCF alone may not uniquely determine the two numbers. Another condition such as LCM, product or ratio may be required.
8. Combining HCF, LCM and Sum
If HCF, LCM and sum are given, the reduced factors can often be found directly.
Let:
Numbers=hm and hn.
Then:
mn=L/h
m+n=S/h
Useful Equation:
m and n are roots of:
t2 − (S/h)t + (L/h) = 0
Example:
HCF=15, LCM=180 and sum=105.
mn=180/15=12
m+n=105/15=7.
The co-prime pair is 3 and 4.
Numbers:
45 and 60.
Difference=15.
9. Combining HCF, LCM and Difference
If Difference=D:
mn=L/h
|m−n|=D/h
Example:
HCF=10, LCM=120 and difference=10.
mn=12
|m−n|=1.
The co-prime factor pair of 12 with difference 1 is 3 and 4.
Numbers=30 and 40.
Shortcut:
When L/H is small, listing its co-prime factor pairs is often faster than forming equations.
10. Maximum and Minimum Pair Behaviour
For fixed HCF h and LCM L, different valid pairs may be possible.
Always Possible Pair:
(h,L)
Reason:
Since h divides L:
HCF(h,L)=h and LCM(h,L)=L.
For Minimum Sum or Difference:
Among valid co-prime factor pairs of L/h, look for the pair whose factors are closest to each other.
For Maximum Difference:
The extreme reduced pair (1,L/h) gives the pair (h,L), which produces the greatest separation among valid positive pairs.
11. Derived HCF-LCM Relationships
Mean Proportional of Two Numbers:
If the numbers are a and b:
Mean proportional=√(ab)
=√(HCF×LCM)
Example:
HCF=2 and LCM=12.
ab=24.
Mean proportional=√24=2√6.
Sum of Reciprocals:
If sum a+b=S, then:
1/a + 1/b
=S/(ab)
=S/(HCF×LCM)
Example:
Sum=18, HCF=3 and LCM=54.
Sum of reciprocals:
18/(3×54)=1/9.
12. Validity Checks for Advanced Questions
Check 1:
HCF must divide LCM.
Check 2:
Every valid number must be a multiple of the HCF.
Check 3:
For fixed HCF and LCM, every valid number must divide the LCM.
Check 4:
Product of the two numbers must equal HCF×LCM.
Check 5:
After dividing both numbers by the HCF, the reduced parts must be co-prime.
Exam Trap:
A pair may have the correct product but still be invalid because its actual HCF is larger than the stated HCF.
Example:
Suppose HCF=18 and LCM=900.
Reduced product=900/18=50.
The factor pair (5,10) cannot be used because HCF(5,10)=5, not 1.
Using it would make the actual HCF 18×5=90.
13. Fast Exam Strategy
Ratio + HCF:
Reduce ratio → multiply ratio parts by HCF.
Ratio + LCM:
Reduce ratio → HCF=LCM/(product of ratio parts).
HCF + LCM:
Compute LCM/HCF → find co-prime factor pairs.
Product + HCF:
Compute Product/HCF² → find co-prime factor pairs.
Sum + HCF:
Compute Sum/HCF → search for co-prime reduced factors satisfying the sum.
HCF + LCM + Sum/Difference:
Use mn=LCM/HCF together with m+n or |m−n|.
14. Common Exam Traps
Trap 1:
Using an unreduced ratio directly with the given HCF.
Trap 2:
Ignoring the co-prime condition after removing the HCF.
Trap 3:
Counting every factor pair of LCM/HCF as a valid pair.
Trap 4:
Counting (a,b) and (b,a) separately when the question asks for unordered pairs.
Trap 5:
Assuming HCF and LCM always determine a unique pair.
Trap 6:
Forgetting that product=HCF×LCM applies to exactly two numbers.
Trap 7:
Accepting a proposed pair without checking its actual HCF and LCM.
Trap 8:
Using the pair-count shortcut without first factorising LCM/HCF into distinct prime factors.
15. Quick Revision
Remember:
• If HCF=h, write numbers as hm and hn where HCF(m,n)=1.
• Then LCM=hmn.
• Ratio must be reduced before using HCF directly.
• Reduced ratio m:n with HCF h gives numbers hm and hn.
• Reduced ratio m:n with LCM L gives HCF=L/(mn).
• If HCF=h and LCM=L, then mn=L/h.
• Valid reduced factor pairs must be co-prime.
• Product of two numbers=h×L.
• With product P and HCF h, mn=P/h².
• With sum S and HCF h, m+n=S/h.
• With difference D and HCF h, |m−n|=D/h.
• For pair counting, count co-prime factor pairs of L/h.
• If L/h has k distinct prime factors, unordered pair count=2^(k−1), for L>h.
• Mean proportional=√(HCF×LCM).
• Sum of reciprocals=(sum of numbers)/(HCF×LCM).
• Always verify HCF, LCM, product and co-prime conditions.
16. Verified Previous-Year Questions
SSC GD PYQ
Ratio & HCF
1 December 2021 · Shift II
Q1. The ratio of two numbers is 3:5 and their HCF is 6. Find their LCM.
A. 30
B. 90
C. 80
D. 60
Correct Answer: B. 90
3 and 5 are co-prime.
Numbers:
6×3=18
6×5=30.
LCM=6×3×5
=90.
SSC CGL PYQ
Product Relation
1 December 2022 · Tier-I · Shift II
Q2. The LCM of two numbers is 48 and their product is 384. What is the ratio of their HCF to their LCM?
A. 1:4
B. 1:6
C. 1:3
D. 2:5
Correct Answer: B. 1:6
Product=HCF×LCM.
384=HCF×48.
HCF=8.
Required ratio:
8:48=1:6.
RRB NTPC PYQ
Possible Number Pair
23 July 2021 · CBT-I · Shift II
Q3. If the HCF of two numbers is 18 and their LCM is 378, which of the following can be the pair of numbers?
A. 54 and 252
B. 18 and 252
C. 54 and 126
D. 27 and 252
Correct Answer: C. 54 and 126
LCM/HCF=378/18=21.
For 54 and126:
54=18×3
126=18×7.
3 and7 are co-prime and 3×7=21.
Therefore:
HCF=18 and LCM=18×3×7=378.
SSC MTS PYQ
Mixed HCF-LCM Relation
14 July 2022 · Shift I
Q4. The ratio of the LCM of two numbers to their sum is 12:7. If their HCF is 4, find the product of the two numbers.
A. 192
B. 172
C. 196
D. 169
Correct Answer: A. 192
Let the numbers be 4m and4n, where m and n are co-prime.
LCM=4mn
Sum=4(m+n).
Therefore:
mn/(m+n)=12/7.
The co-prime pair satisfying this is m=3 and n=4:
mn=12 and m+n=7.
Numbers=12 and16.
Product=192.
17. Practice MCQs
Practice MCQ
Q1. Two numbers are in the ratio 4:9 and their HCF is 7. Find their LCM.
A. 126
B. 196
C. 252
D. 504
Correct Answer: C. 252
4 and9 are co-prime. LCM=7×4×9=252.
Practice MCQ
Q2. Two numbers are in the ratio 6:10 and their HCF is 14. Find the larger number.
A. 70
B. 84
C. 98
D. 140
Correct Answer: A. 70
Reduce 6:10 to3:5. Numbers=14×3 and14×5=42 and70. Larger number=70.
Practice MCQ
Q3. Two numbers are in the ratio 4:7 and their LCM is 420. Find their HCF.
A. 10
B. 12
C. 15
D. 20
Correct Answer: C. 15
HCF=420/(4×7)=420/28=15.
Practice MCQ
Q4. The HCF and LCM of two numbers are 12 and 420. How many unordered pairs are possible?
A. 1
B. 2
C. 3
D. 4
Correct Answer: B. 2
LCM/HCF=35. Co-prime factor pairs are (1,35) and (5,7). Therefore 2 unordered pairs are possible.
Practice MCQ
Q5. The HCF of two numbers is 18 and their LCM is 1260. How many unordered pairs are possible?
A. 2
B. 3
C. 4
D. 6
Correct Answer: C. 4
LCM/HCF=70=2×5×7. It has 3 distinct prime factors, so unordered pair count=23−1=4.
Practice MCQ
Q6. The product of two numbers is 6760 and their HCF is 13. How many unordered pairs are possible?
A. 1
B. 2
C. 3
D. 4
Correct Answer: B. 2
Reduced product=6760/13²=40. Co-prime factor pairs are (1,40) and (5,8). Hence 2 unordered pairs.
Practice MCQ
Q7. Two numbers have HCF 15, LCM 180 and sum 105. Find their difference.
A. 10
B. 15
C. 20
D. 30
Correct Answer: B. 15
mn=180/15=12 and m+n=105/15=7. Thus m,n=3,4. Numbers=45,60. Difference=15.
Practice MCQ
Q8. The HCF and LCM of two numbers are 8 and 360. Which pair is possible?
A. 24 and 120
B. 40 and 72
C. 48 and 60
D. 16 and 180
Correct Answer: B. 40 and 72
40=8×5 and72=8×9. Since 5 and9 are co-prime, HCF=8 and LCM=8×5×9=360.
Practice MCQ
Q9. The HCF and LCM of two positive numbers are 6 and 150. Find their mean proportional.
A. 20
B. 25
C. 30
D. 36
Correct Answer: C. 30
Product=6×150=900. Mean proportional=√900=30.
Practice MCQ
Q10. The sum of two numbers is 66, their HCF is 6 and their LCM is 180. Find the larger number.
A. 30
B. 36
C. 42
D. 60
Correct Answer: B. 36
mn=180/6=30 and m+n=66/6=11. The co-prime factor pair satisfying both is5 and6. Therefore numbers=30 and36, and the larger number is 36.