Playing With Numbers — Exercise 8(F)

Find:

the smallest number that is completely divisible by 28 and 42.

Answer

The smallest number that is completely divisible by 28 and 42 is their L.C.M.

$\begin{array}{l|rr} 2 & 28, & 42 \\ \hline 2 & 14, & 21 \\ \hline 3 & 7, & 21 \\ \hline 7 & 7, & 7 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 × 2 × 3 × 7 = 84

Hence, the smallest number completely divisible by 28 and 42 is 84.

Find:

the largest number that can divide 28 and 42 completely.

Answer

The largest number that can divide 28 and 42 completely is their H.C.F.

28 = 2 × 2 × 7

42 = 2 × 3 × 7

The prime factors common to both numbers are 2 and 7.

⇒ H.C.F. = 2 × 7 = 14

Hence, the largest number that can divide 28 and 42 completely is 14.

Take two numbers each of which is divisible by 8.

(i) Add the numbers taken. Is the sum so obtained also divisible by 8?

(ii) Subtract the smaller number from the bigger number. Is the difference so obtained also divisible by 8?

Answer

Let the two numbers each divisible by 8 be 16 and 24.

(i) Sum of the numbers = 16 + 24 = 40

40 ÷ 8 = 5

Since, 40 is completely divisible by 8, the sum is also divisible by 8.

Yes, the sum is divisible by 8.

(ii) Difference of the numbers = 24 - 16 = 8

8 ÷ 8 = 1

Since, 8 is completely divisible by 8, the difference is also divisible by 8.

Yes, the difference is divisible by 8.

What is the H.C.F. of two consecutive

(i) numbers?

(ii) even numbers?

(iii) odd numbers?

Answer

(i) The H.C.F. of two consecutive numbers is 1. For example, H.C.F. of 7 and 8 = 1.

(ii) The H.C.F. of two consecutive even numbers is 2. For example, H.C.F. of 8 and 10 = 2.

(iii) The H.C.F. of two consecutive odd numbers is 1. For example, H.C.F. of 7 and 9 = 1.

Find the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Answer

First we find the L.C.M. of 6, 8 and 12.

$\begin{array}{l|rrr} 2 & 6, & 8, & 12 \\ \hline 2 & 3, & 4, & 6 \\ \hline 2 & 3, & 2, & 3 \\ \hline 3 & 3, & 1, & 3 \\ \hline & 1, & 1, & 1 \\ \end{array}$

L.C.M. = 2 × 2 × 2 × 3 = 24

The multiples of 24 are 24, 48, 72, 96, 120, ......

The smallest 3-digit multiple of 24 is 120.

Hence, the smallest 3-digit number exactly divisible by 6, 8 and 12 is 120.

Find the greatest 3-digit number which is exactly divisible by 6, 8 and 12.

Answer

First we find the L.C.M. of 6, 8 and 12.

$\begin{array}{l|rrr} 2 & 6, & 8, & 12 \\ \hline 2 & 3, & 4, & 6 \\ \hline 2 & 3, & 2, & 3 \\ \hline 3 & 3, & 1, & 3 \\ \hline & 1, & 1, & 1 \\ \end{array}$

The L.C.M. of 6, 8 and 12 = 24

The greatest 3-digit number is 999.

On dividing 999 by 24, we get :

$\begin{array}{l} 24\overline{\smash{\big)}999\smash{\big(}} 41 \\ \phantom{)}\phantom{)}\underline{-96} \\ \phantom{++} 39 \\ \phantom{+)}\underline{-24} \\ \phantom{++} 15 \\ \end{array}$

Remainder = 15

∴ The greatest 3-digit number divisible by 24 = 999 - 15 = 984

Hence, the greatest 3-digit number exactly divisible by 6, 8 and 12 is 984.

Find the L.C.M. of 140 and 168. Use the L.C.M. obtained to find the H.C.F. of the given numbers.

Answer

$\begin{array}{l|rr} 2 & 140, & 168 \\ \hline 2 & 70, & 84 \\ \hline 2 & 35, & 42 \\ \hline 3 & 35, & 21 \\ \hline 5 & 35, & 7 \\ \hline 7 & 7, & 7 \\ \hline & 1, & 1 \\ \end{array}$

L.C.M. = 2 × 2 × 2 × 3 × 5 × 7 = 840

We know that,

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

⇒ 140 × 168 = H.C.F. × 840

⇒ H.C.F. = $\dfrac{140 \times 168}{840}$

⇒ H.C.F. = $\dfrac{23520}{840}$

⇒ H.C.F. = 28

Hence, L.C.M. = 840 and H.C.F. = 28.

Find the H.C.F. of 108 and 450 then use the H.C.F. obtained to find the L.C.M. of the given numbers.

Answer

Splitting each number into its prime factors :

108 = 2 × 2 × 3 × 3 × 3

450 = 2 × 3 × 3 × 5 × 5

The prime factors common to both numbers are 2, 3 and 3.

⇒ H.C.F. = 2 × 3 × 3 = 18

We know that,

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

⇒ 108 × 450 = 18 × L.C.M.

⇒ L.C.M. = $\dfrac{108 \times 450}{18}$

⇒ L.C.M. = $\dfrac{48600}{18}$

⇒ L.C.M. = 2700

Hence, H.C.F. = 18 and L.C.M. = 2700.

Take any two numbers. Find their H.C.F. and L.C.M. Show that L.C.M. obtained is completely divisible by the H.C.F.

Answer

Let the two numbers be 12 and 18.

Splitting each number into its prime factors :

12 = 2 × 2 × 3

18 = 2 × 3 × 3

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

L.C.M. = 2 × 2 × 3 × 3 = 36

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

$\dfrac{\text{L.C.M.}}{\text{H.C.F.}} = \dfrac{36}{6} = 6$

Since, the remainder is 0,

Hence, the L.C.M. (36) is completely divisible by the H.C.F. (6).

Find L.C.M. and H.C.F. of numbers 16, 32 and 96. Show that the L.C.M. of 16, 32 and 96 is divisible by their H.C.F.

Answer

Splitting each number into its prime factors :

16 = 2 × 2 × 2 × 2

32 = 2 × 2 × 2 × 2 × 2

96 = 2 × 2 × 2 × 2 × 2 × 3

H.C.F. = product of common prime factors = 2 × 2 × 2 × 2 = 16

L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 = 96

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

$\dfrac{\text{L.C.M.}}{\text{H.C.F.}} = \dfrac{96}{16} = 6$

Since, the remainder is 0,

Hence, L.C.M. = 96, H.C.F. = 16 and the L.C.M. of 16, 32 and 96 is divisible by their H.C.F.