HCF - LCM Problems come under the category of prime factorization. There are two methods of H.C.F. and L.C.M. namely factorization method and division method.

- Resolving each given number into its prime factors.
9 = 3 × 3 = [latex]3^{2}[/latex].
15 = 3 × 5.

- The product of all the factors with highest powers.
= [latex]3^{2}[/latex] × 5 = 3 × 3 × 5 = 45.

- The required least common multiple (L.C.M) of 9 and 15 = 45.

- Resolving each given number into its prime factors.
16 = 2 x 2 x 2 x 2 = [latex]2^{4}[/latex]
24 = 2 x 2 x 2 x 3 = [latex]2^{3}[/latex] x 3

- The product of all the factors with highest powers.
= [latex]2^{4}[/latex] × 3 = 2 × 2 x 2 × 2 × 3 = 48.

- The required least common multiple (L.C.M) of 16 and 24 = 48.

Keep dividing each row till there are no common factor except 1.

Multiply the dividers with the last row.

∴ L.C.M. (12, 36, 72) = 2 × 2 × 3 × 3 × 1 × 1 × 4 = 72.

- ∴ L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.

- Factors of 12 = 2 × 2 × 3 = [latex]2^{2}[/latex] × 3
- Factors of 36 = 3 × 3 × 2 × 2 = [latex]2^{2}[/latex] × [latex]3^{2}[/latex]
- Factors of 72 = 2 × 2 × 3 × 3 × 2 = [latex]2^{3}[/latex] × [latex]3^{2}[/latex]

H.C.F. can be found by taking the product of least powers of all common factors that occur in these numbers.

∴ H.C.F. (12, 36, 72) = [latex]2^{2}[/latex] × 3 = 12.

Keep dividing each row till one of the numbers becomes 1.

Multiply the dividers.

∴ H.C.F. (12, 36, 72) = 2 × 2 × 3 = 12.

∴ H.C.F. (777, 1147) = 37.

Factors of 16: 1, 2, 4, 8, and 16

Factors of 25: 1, 5, and 25

Here, 16 and 25 have only one common factor that is 1. Hence, they are co-prime.

Factors of 21: 1, 3, 7, and 21.

Factors of 27: 1, 2, 3, 9 and 27.

Here, 21 and 27 have only two common factors that is 1 and 3. Hence, they are NOT co-prime.

Find the H.C.F. of 1.20 and 22.5.

Find the L.C.M. of 1.20 and 22.5.

There are various practical applications of H.C.F. and L.C.M. Let's take a look at a few examples to see them.

- We can interpret “greatest number that will divide” as calculating G.C.D./H.C.F. of
- Factors of 391: 1, 17, 23, 391
- Factors of 425: 1, 5, 17, 25, 425
- Factors of 527: 1, 17, 31, 527

numbers (400 - 9), (435 - 10), and (541 - 14) i.e. H.C.F.(391, 425, 527).

Clearly, H.C.F.(391, 425, 527) = 17

- We can interpret “how many seconds will they be together at the starting point” as calculating L.C.M.(18, 24, 44).
- 18 = 2 × [latex]3^{2}[/latex]
- 24 = [latex]2^{3}[/latex] × 3
- 44 = [latex]2^{2}[/latex] × 11

Clearly, L.C.M.(18, 24, 44) = [latex]2^{3}[/latex] × [latex]3^{2}[/latex] × 11 = 792

- Comparing these fractions is not a n easy task. Therefore, we use L.C.M. of the denominators to handle this.

Suppose x and y are two numbers, then.

⇒ LCM (x & y) × HCF (x & y) = x × y

We know, LCM × HCF = Product of the Numbers (First number × Second number) then we get,

18 × 1782 = 162 × Second number

18 × 1782/162 = Second number

⇒ Therefore, the second number = 198

We know, LCM × HCF = Product of the Numbers (First number × Second number) then we get,

825 × 25 = 275 × Second number

825 × 25/ 275 = Second number

⇒ Therefore, the second number = 75

26 = 2 x 13

91 = 7 x 13

HCF = 13

LCM = 2 x 7 x 13 = 182

Product of two value 26 x 91 = 2366

Product of HCF and LCM 13 x 182 = 2366

⇒ Hence Proved, LCM × HCF = Product of the two Numbers

- Since the numbers are co-prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number = [latex]\frac{551}{29}[/latex] = 19;

Third number = [latex]\frac{1073}{29}[/latex] = 37.

∴ Required sum = (19 + 29 + 37) = 85.

- Well, consider a = 12 and b = 25. They are not prime but coprime (have no common dividers except 1).

a=[latex]2^{2}[/latex] x 3, b=[latex]5^{2}[/latex].

Their product is ab = 12 x 25 = 300. Their HCF is 1 as for any coprime numbers. Their LCM must include all their prime factors with their degrees, so LCM = [latex]2^{2}[/latex] x 3 x [latex]5^{2}[/latex]=300.

And yes, ab = HCF (a, b) x LCM (a, b) for these a and b.

We have: 6 = 2 x 3,

72 = [latex]2^{3}[/latex] x [latex]3^{2}[/latex],

120 = [latex]2^{3}[/latex] x 3 x 5

Here, [latex]2^{1}[/latex] and [latex]3^{1}[/latex] are the smallest powers of the common factors 2 and 3 respectively.

So, HCF (6, 72, 120) = [latex]2^{1}[/latex] X [latex]3^{1}[/latex] = 2 X 3 = 6

[latex]2^{3}[/latex], [latex]3^{2}[/latex] and [latex]5^{1}[/latex] are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.

So, LCM (6, 72, 120) = [latex]2^{3}[/latex] x [latex]3^{2}[/latex] x [latex]5^{1}[/latex] = 360.

- The first watch ticks every [latex]\frac{95}{90}[/latex] seconds.

They will tick together after (LCM. Of [latex]\frac{95}{90}[/latex] & [latex]\frac{323}{315}[/latex]) Seconds.

Now, LCM of [latex]\frac{95}{90}[/latex] and [latex]\frac{323}{315}[/latex] = [latex]\frac{LCM \: of \: 95, 323}{HCF \: of \: 90, 315}[/latex] = [latex]\frac{19 \times 5 \times 17}{45}[/latex]

The number of times they will tick in the first 3600 seconds = [latex]3600 \div \frac{19 \times 5 \times 17}{45}[/latex] = [latex]\frac{3600 \times 45}{19 \times 5 \times 17}[/latex] = 100 [latex]\frac{100}{323}[/latex]

Once they have already ticked in the beginning.

So in 1 hour they will tick 100 + 1 = 101 times.

- We know, LCM of fractions = [latex]\frac{LCM \:of \:numerators}{HCF \:of \:denominators}[/latex]

LCM ([latex]\frac{1}{2}[/latex], [latex]\frac{2}{3}[/latex], [latex]\frac{3}{7}[/latex]) = [latex]\frac{LCM \:(1,\: 2,\: 3)}{HCF \:(2,\: 3,\: 7)}[/latex]

LCM (1, 2, 3) = 1 x 2 x 3 = 6

Now in order to find out HCF, use the prime factorization method i.e;

Here 2, 3 and 7 are prime numbers, so they are visible by "1" and itself only.

so, 2 = 2 x 1 3 = 3 x 1 7 = 7 x 1

In prime factorization method in case of HCF, we have to consider all the common factors with greatest index, in this case the common factor is "1"

So, HCF(2, 3, 7) = 1

∴ LCM of fractions = [latex]\frac{6}{1}[/latex] = 6

Now, Let's find out HCF of fraction

We know, HCF of fractions = [latex]\frac{HCF \:of \:numerators}{LCM \:of \:denominators}[/latex]

HCF ([latex]\frac{1}{2}[/latex], [latex]\frac{2}{3}[/latex], [latex]\frac{3}{7}[/latex]) = [latex]\frac{HCF \:(1,\: 2,\: 3)}{LCM \:(2,\: 3,\: 7)}[/latex]

As we know in order to find out HCF, use the prime factorization method i.e;

so, 1 = 1 x 1 2 = 2 x 1 3 = 3 x 1

In prime factorization method in case of HCF, we have to consider all the common factors with greatest index, in this case the common factor is "1"

So, HCF(1, 2, 3) = 1

Now let's find LCM(2, 3, 7) using divison method.

LCM(2, 3, 7) = 2 x 3 x 7 = 42.

∴ HCF of fractions = [latex]\frac{1}{42}[/latex]

- Here the required HCF is,

18 = 3 x 3 x 2

12 = 3 x 2 x 2

21 = 3 x 7

9 = 3 x 3

In prime factorization method in case of HCF, we have to consider all the common factors with greatest index, in this case the common factor is "3".

So, HCF(18, 12, 21, 9) = 3

Now let's find LCM of,

35 = 3 x 7

25 = 5 x 5

40 = 2 x 2 x 2 x 5

10 = 2 x 5

Highest factors of all these numbers is 2 x 2 x 2 x 5 x 5 x 7 = 1400.

∴ H.C.F. of [latex]\frac{18}{35}[/latex], [latex]\frac{12}{25}[/latex], [latex]\frac{21}{40}[/latex], [latex]\frac{9}{10}[/latex] = [latex]\frac{3}{1400}[/latex]

2. [latex]Two \ numbers \ are \ said \ to \ be \ co-primes \ if \ their \ H.C.F. \ is \ 1[/latex]

3. [latex]H.C.F. = \frac{H.C.F. \ of \ numerators}{L.C.M \ of \ denominators}[/latex]

4. [latex]L.C.M. = \frac{L.C.M. \ of \ numerators}{H.C.F. \ of \ denominators}[/latex]

Given numbers are 144, 288, 368

First factorize the given numbers, i.e.

[latex] 144 = 2^4 × 3^2[/latex];

[latex]288 = 2^5 × 3^2[/latex];

[latex]368 = 2^5 × 3^2[/latex]

Therefore H.C.F. = [latex]2^4 × 3^2[/latex] = 144.

Given fraction is [latex] \frac{108}{360}[/latex]

[latex]108 = 2^2 × 3^3[/latex]

[latex]360 = 2^3 × 3^2 × 5[/latex]

H.C.F. of 108 and 360 is [latex]2^2 × 3^2[/latex] = 36

Now, By dividing the numerator and denominator by 36,

⇒[latex]\frac{108}{360} = \frac{108}{360} × \frac{360}{360}[/latex]

⇒[latex]\frac{3}{10}[/latex]

Therefore [latex]\frac{108}{360} = \frac{3}{10}[/latex]

Given that

H.C.F. of two numbers = 42

L.C.M. of two numbers = 1260

One of the number = 210

[latex]Other number = \frac{ H.C.F. of two numbers × L.C.M. of two numbers}{one of the number}[/latex]

substitute the given values in the above fromula,

[latex]Other number = \frac{42 × 1260}{210}[/latex] = 252

Hence, other number = 252.

Here the difference between any divisor and the corresponding remainder is 17 i.e.

(35 - 18) = 17

(45 - 28) = 17

(55 - 38) = 17

Therefore Required number = (L.C.M. of 35,45,55) -17

L.c.m of 35, 45, 55 is 3465

⇒ Required number = 3465 - 17 = 3448

Hence, least number = 3448.

Given that

Ratio of two numbers = 13 : 15

Their L.C.M. is 39780

Assume the two numbers as [latex]13x and 15x [/latex]

Now 13[latex]x × 15x [/latex] = 39780

⇒[latex] x [/latex] × 195 = 39780

⇒[latex] x = \frac{39780}{195}[/latex]

⇒[latex] x [/latex]= 204

Therefore, substitute the value of [latex] x [/latex] i.e.

13 x [latex] x [/latex] = 13 x 204 = 2652

15 x [latex] x [/latex] = 15 x 204 = 3060

Hence, the two numbers are 2652 and 3060.