HCF & LCM
HCF & LCM are acronym for words, Highest common factor and Lowest common multiple respectively.
1. H. C. F
While we all know what a multiplication is like 2 * 3 = 6. HCF is just the reverse of multiplication which is known as Factorization. Now factorization is breaking a composite number into its prime factors. Like 6 = 2 * 3, where 6 is a composite number and 2 & 3 are prime number.
“In mathematics, the Highest Common Factor (HCF) of two or more integers is the largest positive integer that divides the numbers without a remainder. For example, the HCF of 8 and 12 is 4.”
Calculation
– By Prime Factorizations
Highest Common Factor can be calculated by first determining the prime factors of the two numbers and then comparing those factors, to take out the common factors.
As in the following example: HCF (18, 42), we find the prime factors of 18 = 2 * 3 * 3 and 42 = 7 * 2 * 3 and notice the “common” of the two expressions is 2 * 3; So HCF (18, 42) = 6.
– By Division Method
In this method first divide a higher number by smaller number.
- Put the higher number in place of dividend and smaller number in place of divisor.
- Divide and get the remainder then use this remainder as divisor and earlier divisor as dividend.
- Do this until you get a zero remainder. The last divisor is the HCF.
- If there are more than two numbers then we continue this process as we divide the third lowest number by the last divisor obtained in the above steps.
First find H.C.F. of 72 and 126
72|126|1
72
54| 72|1
54
18| 54| 3
54
0
H.C.F. of 72 and 126 = 18
2. L.C.M
The Least Common Multiple of two or more integers is always divisible by all the integers it is derived from. For example, 20 is a multiple of 5 because 5 × 4 = 20, so 20 is divisible by 5 and 2. Because 10 is the smallest positive integer that is divisible by both 5 and 2, it is the least common multiple of 5 and 4.
LCM cam also be understand by this example:
Multiples of 5 are:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70 …
And the multiples of 6 are:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
Common multiples of 5 and 6 are:
30, 60, 90, 120, ….
Hence, the lowest common multiple is simply the first number in the common multiple list i.e 30.
Calculation
– By Prime Factorizations
The prime factorization theorem says that every positive integer greater than 1 can be written in only one way as a product of prime numbers.
Example: To find the value of LCM (9, 48, and 21).
First, find the factor of each number and express it as a product of prime number powers.
Like 9 = 32,
48 = 24 * 3
21 = 3 * 7
Then, write all the factors with their highest power like 32, 24, and 7. And multiply them to get their LCM.
Hence, LCM (9, 21, and 48) is 32 * 24 * 7 = 1008.
– By Division Method
Here, divide all the integers by a common number until no two numbers are further divisible. Then multiply the common divisor and the remaining number to get the LCM.
2 | 72, 240, 196
2 | 36, 120, 98
2 | 18, 60, 49
3 | 9, 30, 49
| 3, 10, 49
L.C.M. of the given numbers
= product of divisors and the remaining numbers
= 2×2×2×3×3×10×49
= 72×10×49 = 35280.
Relation between L.C.M. and H.C.F. of two natural numbers
The product of L.C.M. and H.C.F. of two natural numbers = the product of the numbers.
For Example:
LCM (8, 28) = 56 & HCF (8, 28) = 4
Now, 8 * 28 = 224 and also, 56 * 4 = 224
HCF & LCM of fractions:
Formulae for finding the HCF & LCM of a fractional number.
HCF of fraction = HCF of numerator / LCM of denominator
LCM of Fraction = LCM of Numerator / HCF of Denominator
Level-I:
1. | Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. | |||||||
|
2. | The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is: | |||||||
|
3. | Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? | |||||||
|
4. | Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is: | |||||||
|
5. | The greatest number of four digits which is divisible by 15, 25, 40 and 75 is: | |||||||
| ||||||||
6. |
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is: | |||||||
|
7. | Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is: | |||||||
|
8. | The G.C.D. of 1.08, 0.36 and 0.9 is: | |||||||
|
9. | The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is: | |||||||
|
10. | The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is: | |||||||
|
Level-II:
11. | Find the lowest common multiple of 24, 36 and 40. | |||||||
|
12. | The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is: | |||||||
|
13. |
| |||||||||||||||
|
14. | The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is: | |||||||
|
15. | A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ? | |||||||
| ||||||||
16. |
The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is: | |||||||
|
17. | What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? | |||||||
|
18. | The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is: | |||||||
|
19. | The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: | |||||||
|
20. | 252 can be expressed as a product of primes as: | |||||||
|
Answers:
Level-I:
Answer:1 Option A
Explanation:
Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)
= H.C.F. of 48, 92 and 140 = 4.
Answer:2 Option C
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
- Larger number = (23 x 14) = 322.
Answer:3 Option D
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together | 30 | + 1 = 16 times. |
2 |
Answer:4 Option A
Explanation:
N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Answer:5 Option C
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 – 399) = 9600.
Answer:6 Option C
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
Greater number = 111.
Answer:7 Option A
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Answer:8 Option C
Explanation:
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
H.C.F. of given numbers = 0.18.
Answer:9 Option B
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
Answer:10 Option D
Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Required number = (90 x 4) + 4 = 364.
Level-II:
Answer:11 Option C
Explanation:
2 | 24 – 36 – 40
——————–
2 | 12 – 18 – 20
——————–
2 | 6 – 9 – 10
——————-
3 | 3 – 9 – 5
——————-
| 1 – 3 – 5
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
Answer:12 Option C
Explanation:
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 – 37) = 23.
Answer:13 Option C
Explanation:
128352) 238368 ( 1
128352
—————
110016 ) 128352 ( 1
110016
——————
18336 ) 110016 ( 6
110016
——-
x
——-
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 ÷ 18336 7
Therefore, —— = ————– = —
238368 238368 ÷ 18336 13
Answer:14 Option B
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
Answer:15 Option D
Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Answer:16 Option C
Explanation:
Other number = |
| 11 x 7700 |
| = 308. |
275 |
Answer:17 Option B
Explanation:
L.C.M. of 12, 18, 21 30 2 | 12 – 18 – 21 – 30
—————————-
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 – 9 – 21 – 15
—————————-
Required number = (1260 ÷ 2) | 2 – 3 – 7 – 5
= 630.
Answer:18 Option D
Explanation:
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
Answer19: Option B
Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
Answer:20 Option A
Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
- TSPSC Mains Tests and Notes Program
- TSPSC Group I Prelims Exam 2024- Test Series and Notes Program
- TSPSC Prelims and Mains Tests Series and Notes Program
- TSPSC Detailed Complete Prelims Notes