Playing With Numbers — Exercise 8(D)

Using the common multiple method, find the L.C.M. of the following:

8, 12 and 24

Answer

Multiples of 8 = 8, 16, 24, 32, 40, 48, ......

Multiples of 12 = 12, 24, 36, 48, ......

Multiples of 24 = 24, 48, 72, ......

The least common multiple = 24

Hence, L.C.M. of 8, 12 and 24 = 24

Using the common multiple method, find the L.C.M. of the following:

10, 15 and 20

Answer

Multiples of 10 = 10, 20, 30, 40, 50, 60, 70, ......

Multiples of 15 = 15, 30, 45, 60, 75, ......

Multiples of 20 = 20, 40, 60, 80, ......

The least common multiple = 60

Hence, L.C.M. of 10, 15 and 20 = 60

Using the common multiple method, find the L.C.M. of the following:

3, 6, 9 and 12

Answer

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, ......

Multiples of 6 = 6, 12, 18, 24, 30, 36, ......

Multiples of 9 = 9, 18, 27, 36, 45, ......

Multiples of 12 = 12, 24, 36, 48, ......

The least common multiple = 36

Hence, L.C.M. of 3, 6, 9 and 12 = 36

Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:

18, 24 and 96

Answer

Prime factor method :

18 = 2 × 3 × 3 = 2 × 3²

24 = 2 × 2 × 2 × 3 = 2³ × 3

96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

L.C.M. = Product of all the prime factors with the highest power of each

= 2⁵ × 3² = 32 × 9 = 288

Common division method :

$\begin{array}{l|rrr} 2 & 18, & 24, & 96 \\ \hline 2 & 9, & 12, & 48 \\ \hline 2 & 9, & 6, & 24 \\ \hline 2 & 9, & 3, & 12 \\ \hline 2 & 9, & 3, & 6 \\ \hline 3 & 9, & 3, & 3 \\ \hline 3 & 3, & 1, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$

⇒ L.C.M. of 18, 24 and 96 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:

14, 21 and 98

Answer

Prime factor method :

14 = 2 × 7

21 = 3 × 7

98 = 2 × 7 × 7 = 2 × 7²

L.C.M. = 2 × 3 × 7² = 2 × 3 × 49 = 294

Common division method :

$\begin{array}{l|rrr} 2 & 14, & 21, & 98 \\ \hline 3 & 7, & 21, & 49 \\ \hline 7 & 7, & 7, & 49 \\ \hline 7 & 1, & 1, & 7 \\ \hline & 1, & 1, & 1 \\ \end{array}$

⇒ L.C.M. of 14, 21 and 98 = 2 × 3 × 7 × 7 = 294

Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:

34, 85 and 51

Answer

Prime factor method :

34 = 2 × 17

85 = 5 × 17

51 = 3 × 17

L.C.M. = 2 × 3 × 5 × 17 = 510

Common division method :

$\begin{array}{l|rrr} 2 & 34, & 85, & 51 \\ \hline 3 & 17, & 85, & 51 \\ \hline 5 & 17, & 85, & 17 \\ \hline 17 & 17, & 17, & 17 \\ \hline & 1, & 1, & 1 \\ \end{array}$

⇒ L.C.M. of 34, 85 and 51 = 2 × 3 × 5 × 17 = 510

Find the least number which when divided by 15, 25, 40 and 50 leaves no remainder.

Answer

The least number which when divided by the given numbers leaves no remainder is their L.C.M.

$\begin{array}{l|rrrr} 2 & 15, & 25, & 40, & 50 \\ \hline 2 & 15, & 25, & 20, & 25 \\ \hline 2 & 15, & 25, & 10, & 25 \\ \hline 3 & 15, & 25, & 5, & 25 \\ \hline 5 & 5, & 25, & 5, & 25 \\ \hline 5 & 1, & 5, & 1, & 5 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$

L.C.M. = 2 × 2 × 2 × 3 × 5 × 5 = 600

Hence, the required least number is 600.

Find the smallest number which when divided by 18, 36 and 48, leaves a remainder of 7.

Answer

First we find the L.C.M. of 18, 36 and 48.

$\begin{array}{l|rrr} 2 & 18, & 36, & 48 \\ \hline 2 & 9, & 18, & 24 \\ \hline 2 & 9, & 9, & 12 \\ \hline 2 & 9, & 9, & 6 \\ \hline 3 & 9, & 9, & 3 \\ \hline 3 & 3, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$

L.C.M. = 2 × 2 × 2 × 2 × 3 × 3 = 144

The smallest number divisible by 18, 36 and 48 is 144.

∴ The smallest number which leaves a remainder of 7 = 144 + 7 = 151

Hence, the required number is 151.

Find the smallest number which when increased by 11 is completely divisible by 16, 24, 40 and 45.

Answer

First we find the L.C.M. of 16, 24, 40 and 45.

$\begin{array}{l|rrrr} 2 & 16, & 24, & 40, & 45 \\ \hline 2 & 8, & 12, & 20, & 45 \\ \hline 2 & 4, & 6, & 10, & 45 \\ \hline 2 & 2, & 3, & 5, & 45 \\ \hline 3 & 1, & 3, & 5, & 45 \\ \hline 3 & 1, & 1, & 5, & 15 \\ \hline 5 & 1, & 1, & 5, & 5 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$

L.C.M. = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

The smallest number completely divisible by 16, 24, 40 and 45 is 720.

∴ The required number = 720 - 11 = 709

Hence, the required number is 709.

Find the L.C.M. and H.C.F. of 24 and 30. Divide the L.C.M. obtained by the H.C.F. Is L.C.M. completely divisible by H.C.F.?

Answer

Splitting each number into its prime factors :

24 = 2 × 2 × 2 × 3

30 = 2 × 3 × 5

H.C.F. = product of common prime factors = 2 × 3 = 6

L.C.M. = 2 × 2 × 2 × 3 × 5 = 120

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

$\dfrac{\text{L.C.M.}}{\text{H.C.F.}} = \dfrac{120}{6} = 20$

Since, the remainder is 0,

Hence, L.C.M. = 120, H.C.F. = 6 and the L.C.M. is completely divisible by the H.C.F.

The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.

Answer

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ First number × Second number = H.C.F. × L.C.M.

⇒ 150 × Second number = 50 × 300

⇒ Second number = $\dfrac{50 \times 300}{150}$

⇒ Second number = $\dfrac{15000}{150}$

⇒ Second number = 100

Hence, the other number is 100.

The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.

Answer

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ 432 = H.C.F. × 72

⇒ H.C.F. = $\dfrac{432}{72}$

⇒ H.C.F. = 6

Hence, the H.C.F. of the two numbers is 6.

The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.

Answer

We know that,

Product of two numbers = H.C.F. × L.C.M.

⇒ 19200 = 40 × L.C.M.

⇒ L.C.M. = $\dfrac{19200}{40}$

⇒ L.C.M. = 480

Hence, the L.C.M. of the two numbers is 480.

Find the smallest number which when divided by 12, 15, 18, 24 and 36 leaves no remainder.

Answer

The smallest number which when divided by the given numbers leaves no remainder is their L.C.M.

$\begin{array}{l|rrrrr} 2 & 12, & 15, & 18, & 24, & 36 \\ \hline 2 & 6, & 15, & 9, & 12, & 18 \\ \hline 2 & 3, & 15, & 9, & 6, & 9 \\ \hline 3 & 3, & 15, & 9, & 3, & 9 \\ \hline 3 & 1, & 5, & 3, & 1, & 3 \\ \hline 5 & 1, & 5, & 1, & 1, & 1 \\ \hline & 1, & 1, & 1, & 1, & 1 \\ \end{array}$

L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 = 360

Hence, the required smallest number is 360.