HackerRank HCF & LCM Quiz- 1

Let their ages be 7p and 9p

LCM will be 9 x 7 x p = 378

378 = 63p

p = \frac{378}{63} = 6

Difference in age = 9p - 7p = 2p = 2 x 6 = 12

LCM of 5, 7 and 8

= LCM(5, 7, 8)

= 2 × 2 × 2 × 5 × 7

= 280

Let the number 280x + 4 is divisible by 6

Least possible number for x is 2

So the number is 280 x 2 + 2 =564

Required capacity = HCF of 960, 540 and 120

The factors of 120 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

The factors of 540 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 27, 30, 36, 45, 54, 60, 90, 108, 135, 180, 270, 540

 

The factors of 960 are: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 64, 80, 96, 120, 160, 192, 240, 320, 480, 960

Then the greatest common factor is 60.

So he must by cup of 60ml

Let the numbers be 2x, 4x and 7x

LCM of 2, 4 and 7 is 28

So LCM of 2x 3x and 7x = 28x

28x= 140

x =  \frac{140}{28} = 5

So numbers are 10, 20 and 35

The factors of 10 are: 1, 2, 5, 10

The factors of 20 are: 1, 2, 4, 5, 10, 20

The factors of 35 are: 1, 5, 7, 35

Then the HCF is 5.

Let the numbers be 12a and 12b where a and b are coprimes

12a x 12b = 144ab = 720

ab = \frac{720}{144} = 5

Coprimes with product 5 = (1,5)

So 12a = 12 x 1 = 12

12b = 12 x 5 = 60

Greater number is 60

L.C.M = \frac{ L.C.M of numerators}{ H.C.F of denominators}

LCM of 1 and 2 = 2

The factors of 3 are: 1, 3

The factors of 4 are: 1, 2, 4

The factors of 5 are: 1, 5

Then theH.C.F of denominators is 1.

L.C.M of fraction = \frac{2}{ 1}

= 2

We need to find the LCM of 40, 30 and 55

Prime factorization of the numbers:

30 = 2 × 3 × 5

40 = 2 × 2 × 2 × 5

55 = 5 × 11

LCM(30, 40, 55)

= 2 × 2 × 2 × 3 × 5 × 11

= 1320

Prime factorization of the numbers:

6 = 2 × 3

8 = 2 × 2 × 2

10 = 2 × 5

12 = 2 × 2 × 3

LCM(6, 8, 10)

= 2 × 2 × 2 × 3 × 5

= 120

LCM(6, 8, 12)

= 2 × 2 × 2 × 3

= 24

 

LCM(6, 10, 12)

= 2 × 2 × 3 × 5

= 60

LCM(8, 10, 12)

= 2 × 2 × 2 × 3 × 5

= 120

LCM(6, 8, 12) is minimum

Let the numbers be 4a and 4b where a and b are coprime

4a x 4b = 16ab = 880

ab =  \frac{880}{16} = 55

a and b can be (1,55) and (5,11)

When a =1 and b = 55

Numbers are 4 and  220

When a =5 and b = 11

Numbers are 20 and 44

So from options,

We need to find LCM of 3, 5 and 12

Prime factorization of the numbers:

3 = 3

5 = 5

12 = 2 × 2 × 3

LCM(3, 5, 12)

= 2 × 2 × 3 × 5

= 60

So bulb glows together again after 60 minutes

i.e. at 12am