Playing With Numbers — Exercise 6(D)

Find the L.C.M. of the following numbers using prime factorisation method :

18, 24

Answer

Rule :

To find the L.C.M. using prime factorization, express each number as a product of primes and then find the product of all prime factors taken the maximum number of times they appear in any single factorization.

By prime factorisation, we have:

$\begin{array}{r|r} 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

18 = 2 × 3 × 3;

24 = 2 × 2 × 2 × 3

Here, 2 appears once in the factorisation of 18 and three times in that of 24. Also, 3 appears twice in the factorisation of 18 and once in that of 24. Thus, 2 appears maximum three times and 3 appears maximum two times.

So, L.C.M. of 18 and 24 = 2 × 2 × 2 × 3 × 3 = 72

Hence, the L.C.M. of 18 and 24 is 72.

Find the L.C.M. of the following numbers using prime factorisation method :

36, 45

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

36 = 2 × 2 × 3 × 3;

45 = 3 × 3 × 5

2 appears maximum 2 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 36 and 45 = 2 × 2 × 3 × 3 × 5 = 180

Hence, the L.C.M. of 36 and 45 is 180.

Find the L.C.M. of the following numbers using prime factorisation method :

48, 72

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

48 = 2 × 2 × 2 × 2 × 3;

72 = 2 × 2 × 2 × 3 × 3

2 appears maximum 4 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations.

So, L.C.M. of 48 and 72 = 2 × 2 × 2 × 2 × 3 × 3 = 144

Hence, the L.C.M. of 48 and 72 is 144.

Find the L.C.M. of the following numbers using prime factorisation method :

63, 105

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 3 & 105 \\ \hline 5 & 35 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

63 = 3 × 3 × 7;

105 = 3 × 5 × 7

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations;

7 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 63 and 105 = 3 × 3 × 5 × 7 = 315

Hence, the L.C.M. of 63 and 105 is 315.

Find the L.C.M. of the following numbers using prime factorisation method :

27, 54, 90

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

27 = 3 × 3 × 3;

54 = 2 × 3 × 3 × 3;

90 = 2 × 3 × 3 × 5

2 appears maximum 1 time in one of the prime factorisations;

3 appears maximum 3 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 27, 54 and 90 = 2 × 3 × 3 × 3 × 5 = 270

Hence, the L.C.M. of 27, 54 and 90 is 270.

Find the L.C.M. of the following numbers using prime factorisation method :

36, 54, 81

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

36 = 2 × 2 × 3 × 3;

54 = 2 × 3 × 3 × 3;

81 = 3 × 3 × 3 × 3

2 appears maximum 2 times in one of the prime factorisations;

3 appears maximum 4 times in one of the prime factorisations.

So, L.C.M. of 36, 54 and 81 = 2 × 2 × 3 × 3 × 3 × 3 = 324

Hence, the L.C.M. of 36, 54 and 81 is 324.

Find the L.C.M. of the following numbers using prime factorisation method :

60, 75, 80, 90

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \qquad \text{and} \quad \begin{array}{r|r} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

60 = 2 × 2 × 3 × 5;

75 = 3 × 5 × 5;

80 = 2 × 2 × 2 × 2 × 5

90 = 2 × 3 × 3 × 5

2 appears maximum 4 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 2 times in one of the prime factorisations.

So, L.C.M. of 60, 75, 80 and 90 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 3600

Hence, the L.C.M. of 60, 75, 80 and 90 is 3600.

Find the L.C.M. of the following numbers using prime factorisation method :

48, 60, 72, 96

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

48 = 2 × 2 × 2 × 2 × 3;

60 = 2 × 2 × 3 × 5;

72 = 2 × 2 × 2 × 3 × 3;

96 = 2 × 2 × 2 × 2 × 2 × 3

2 appears maximum 5 times in one of the prime factorisations;

3 appears maximum 2 times in one of the prime factorisations;

5 appears maximum 1 time in one of the prime factorisations.

So, L.C.M. of 48, 60, 72 and 96 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

Hence, the L.C.M. of 48, 60, 72 and 96 is 1440.

Find the L.C.M. of the following, using common division method :

8, 10, 12, 16

Answer

By common division method, we get:

$\begin{array}{r|rrrr} 2 & 8, & 10, & 12, & 16 \\ \hline 2 & 4, & 5, & 6, & 8 \\ \hline 2 & 2, & 5, & 3, & 4 \\ \hline & 1, & 5, & 3, & 2 \\ \end{array}$

The L.C.M. of 8, 10, 12, 16 = 2 × 2 × 2 × 5 × 3 × 2 = 240

Hence, L.C.M. of 8, 10, 12, 16 = 240.

Find the L.C.M. of the following, using common division method :

12, 15, 18, 24

Answer

By common division method, we get:

$\begin{array}{r|rrrr} 2 & 12, & 15, & 18, & 24 \\ \hline 2 & 6, & 15, & 9, & 12 \\ \hline 3 & 3, & 15, & 9, & 6 \\ \hline & 1, & 5, & 3, & 2 \\ \end{array}$

The L.C.M. of 12, 15, 18, 24 = 2 × 2 × 3 × 5 × 3 × 2 = 360

Hence, L.C.M. of 12, 15, 18, 24 = 360.

Find the L.C.M. of the following, using common division method :

15, 18, 20, 27

Answer

By common division method, we get:

$\begin{array}{r|rrrr} 2 & 15, & 18, & 20, & 27 \\ \hline 3 & 15, & 9, & 10, & 27 \\ \hline 3 & 5, & 3, & 10, & 9 \\ \hline 5 & 5, & 1, & 10, & 3 \\ \hline & 1, & 1, & 2, & 3 \\ \end{array}$

The L.C.M. of 15, 18, 20, 27 = 2 × 3 × 3 × 5 × 2 × 3 = 540

Hence, L.C.M. of 15, 18, 20, 27 = 540.

Find the L.C.M. of the following, using common division method :

20, 25, 30, 45

Answer

By common division method, we get:

$\begin{array}{r|rrrr} 2 & 20, & 25, & 30, & 45 \\ \hline 3 & 10, & 25, & 15, & 45 \\ \hline 5 & 10, & 25, & 5, & 15 \\ \hline & 2, & 5, & 1, & 3 \\ \end{array}$

The L.C.M. of 20, 25, 30, 45 = 2 × 3 × 5 × 2 × 5 × 3 = 900

Hence, L.C.M. of 20, 25, 30, 45 = 900.

Find the L.C.M. of the following, using common division method :

21, 45, 63, 81

Answer

By common division method, we get:

$\begin{array}{r|rrrr} 3 & 21, & 45, & 63, & 81 \\ \hline 3 & 7, & 15, & 21, & 27 \\ \hline 7 & 7, & 5, & 7, & 9 \\ \hline & 1, & 5, & 1, & 9 \\ \end{array}$

The L.C.M. of 21, 45, 63, 81 = 3 × 3 × 7 × 5 × 9 = 2835

Hence, L.C.M. of 21, 45, 63, 81 = 2835.

Find the L.C.M. of the following, using common division method :

16, 18, 24, 32, 36

Answer

By common division method, we get:

$\begin{array}{r|rrrrr} 2 & 16, & 18, & 24, & 32, & 36 \\ \hline 2 & 8, & 9, & 12, & 16, & 18 \\ \hline 2 & 4, & 9, & 6, & 8, & 9 \\ \hline 2 & 2, & 9, & 3, & 4, & 9 \\ \hline 3 & 1, & 9, & 3, & 2, & 9 \\ \hline 3 & 1, & 3, & 1, & 2, & 3 \\ \hline & 1, & 1, & 1, & 2, & 1 \\ \end{array}$

The L.C.M. of 16, 18, 24, 32, 36 = 2 × 2 × 2 × 2 × 3 × 3 × 2 = 288

Hence, L.C.M. of 16, 18, 24, 32, 36 = 288.

Find the L.C.M. of the following, using common division method :

120, 210, 225

Answer

By common division method, we get:

$\begin{array}{r|rrr} 2 & 120, & 210, & 225 \\ \hline 3 & 60, & 105, & 225 \\ \hline 5 & 20, & 35, & 75 \\ \hline & 4, & 7, & 15 \\ \end{array}$

The L.C.M. of 120, 210, 225 = 2 × 3 × 5 × 4 × 7 × 15 = 12600

Hence, L.C.M. of 120, 210, 225 = 12600.

Find the L.C.M. of the following, using common division method :

175, 182, 350

Answer

By common division method, we get:

$\begin{array}{r|rrr} 2 & 175, & 182, & 350 \\ \hline 5 & 175, & 91, & 175 \\ \hline 5 & 35, & 91, & 35 \\ \hline 7 & 7, & 91, & 7 \\ \hline & 1, & 13, & 1 \\ \end{array}$

The L.C.M. of 175, 182, 350 = 2 × 5 × 5 × 7 × 13 = 4550

Hence, L.C.M. of 175, 182, 350 = 4550.

The H.C.F. of two numbers is 23 and their L.C.M. is 276. If one of the numbers is 92, find the other.

Answer

It is given that,

H.C.F. = 23

L.C.M. = 276

Given number = 92

We know that,

The other number = $\dfrac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$

= $\dfrac{23 \times 276}{92} = \dfrac{6348}{92}$ = 69

Hence, the other number is 69.

The H.C.F. of two numbers is 144 and their L.C.M. is 2880. If one of the numbers is 576, find the other.

Answer

It is given that,

H.C.F. = 144

L.C.M. = 2880

Given number = 576

We know that,

The other number = $\dfrac{\text{H.C.F.} \times \text{L.C.M.}}{\text{given number}}$

= $\dfrac{144 \times 2880}{576} = \dfrac{414720}{576}$ = 720

Hence, the other number is 720.

The product of two numbers is 2925 and their H.C.F. is 15. Find their L.C.M.

Answer

It is given that,

H.C.F. = 15

Product of two numbers = 2925

We know that,

L.C.M. = $\dfrac{\text{Product of the two numbers}}{\text{H.C.F.}}$

= $\dfrac{2925}{15}$

= 195

Hence, the L.C.M. is 195.

The product of two numbers is 3750 and their L.C.M. is 150. Find their H.C.F.

Answer

It is given that,

L.C.M. = 150

Product of two numbers = 3750

We know that,

H.C.F. = $\dfrac{\text{Product of the two numbers}}{\text{L.C.M.}}$

= $\dfrac{3750}{150}$

= 25

Hence, the H.C.F. is 25.

Can there be two numbers having H.C.F. 12 and L.C.M. 64? Give reasons in support of your answer.

Answer

The H.C.F. of two numbers perfectly divides their L.C.M. without leaving any remainder.

We are given that, H.C.F. is 12 and L.C.M. is 64.

On dividing the L.C.M. by the H.C.F.,

$\begin{array}{l} \phantom{22 ))}{5} \\ 12\overline{\smash{\big)}64} \\ \phantom{22}\phantom{)}\mathllap{-,}\underline{60} \\ \phantom{{22 )}}04 \\ \end{array}$

The remainder is 4.

Thus, 12 does not divide 64 exactly.

Hence, there cannot be two numbers having an H.C.F. of 12 and an L.C.M. of 64.

Find the least number which is exactly divisible by each of the numbers 15, 18 and 24.

Answer

The least number exactly divisible by the given numbers will be their L.C.M. So, we need to find the L.C.M. of 15, 18 and 24.

$\begin{array}{r|rrr} 2 & 15, & 18, & 24 \\ \hline 3 & 15, & 9, & 12 \\ \hline & 5, & 3, & 4 \\ \end{array}$

The L.C.M. of 15, 18, 24 = 2 × 3 × 5 × 3 × 4 = 360

Hence, the least number exactly divisible by 15, 18, and 24 is 360.

Find the smallest number which when divided by 15, 20, 25 and 30 leaves 5 as remainder in each case.

Answer

First, we find the L.C.M. of 15, 20, 25, and 30. By common division method, we get:

$\begin{array}{r|rrrr} 2 & 15, & 20, & 25, & 30 \\ \hline 3 & 15, & 10, & 25, & 15 \\ \hline 5 & 5, & 10, & 25, & 5 \\ \hline & 1, & 2, & 5, & 1 \\ \end{array}$

The L.C.M. of 15, 20, 25 and 30 = 2 × 3 × 5 × 2 × 5 = 300

Required number = L.C.M. of (15, 20, 25, 30) + 5 = 300 + 5 = 305

Hence, the required smallest number is 305.

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 minutes respectively. After what interval of time will they toll together again?

Answer

The bells will toll together again after an interval of time equal to the L.C.M. of their individual tolling intervals (2, 4, 6, 8, 10, and 12 minutes).

By common division method, we get:

$\begin{array}{r|rrrrrr} 2 & 2, & 4, & 6, & 8, & 10, & 12 \\ \hline 2 & 1, & 2, & 3, & 4, & 5, & 6 \\ \hline 3 & 1, & 1, & 3, & 2, & 5, & 3 \\ \hline & 1, & 1, & 1, & 2, & 5, & 1 \\ \end{array}$

The L.C.M. of 2, 4, 6, 8, 10, and 12 = 2 × 2 × 3 × 2 × 5 = 120 minutes = 2 hours

Hence, the bells will toll together again after 2 hours.

An electronic device makes a beep after every 15 minutes. Another device makes a beep after every 20 minutes. They beeped together at 10 a.m. At what time will they make the next beep together?

Answer

The devices will beep together again after an interval of time equal to the L.C.M. of 15 and 20.

By common division method, we get:

$\begin{array}{r|rr} 5 & 15, & 20 \\ \hline & 3, & 4 \\ \end{array}$

The L.C.M. of 15 and 20 = 5 × 3 × 4 = 60

Therefore, the devices will beep together again after 60 minutes. Since 60 minutes equals 1 hour, we add 1 hour to their initial beep time.

Next beep time = 10 a.m. + 1 hour = 11 a.m.

Hence, the devices will make the next beep together at 11 a.m.