Playing With Numbers — Multiple Choice Questions

Which of the following numbers is divisible by 8 ?

1. 536146

2. 789214

3. 955626

4. 813256

Answer

A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

536146 $\longrightarrow$ Last three digits are 146. But 146 is not divisible by 8.

789214 $\longrightarrow$ Last three digits are 214. But 214 is not divisible by 8.

955626 $\longrightarrow$ Last three digits are 626. But 626 is not divisible by 8.

813256 $\longrightarrow$ Last three digits are 256. 256 is divisible by 8.

Since the number formed by the last three digits of 813256 is divisible by 8, the number 813256 is also divisible by 8.

Hence, Option 4 is the correct option.

Which of the following numbers is divisible by 9 ?

1. 314526

2. 578142

3. 964315

4. 876421

Answer

A number is divisible by 9 if the sum of its digits is divisible by 9.

We need to find the sum of the digits of each given number.

314526 $\longrightarrow$ 3 + 1 + 4 + 5 + 2 + 6 = 21

578142 $\longrightarrow$ 5 + 7 + 8 + 1 + 4 + 2 = 27

964315 $\longrightarrow$ 9 + 6 + 4 + 3 + 1 + 5 = 28

876421 $\longrightarrow$ 8 + 7 + 6 + 4 + 2 + 1 = 28

Among the sums, 27 is divisible by 9. Therefore 578142 is also divisible by 9.

Hence, Option 2 is the correct option.

Which of the following numbers is divisible by 11 ?

1. 3333333

2. 66666

3. 999999

4. none of these

Answer

A number is divisible by 11 if the difference between the sum of the digits at odd positions and the sum of the digits at even positions is either 0 or a multiple of 11.

Calculating the difference for each number,

3333333 $\longrightarrow$ (3 + 3 + 3 + 3) - (3 + 3 + 3) = 12 - 9 = 3

66666 $\longrightarrow$ (6 + 6 + 6) - (6 + 6) = 18 - 12 = 6

999999 $\longrightarrow$ (9 + 9 + 9) - (9 + 9 + 9) = 27 - 27 = 0

Since the difference for 999999 is 0, the number 999999 is divisible by 11.

Hence, Option 3 is the correct option.

Which of the following numbers is divisible by 6 ?

1. 345672

2. 154636

3. 236794

4. none of these

Answer

A number is divisible by 6 if it is divisible by both 2 and 3.

345672 $\longrightarrow$ Last digit is 2 (divisible by 2) and the sum of digits 3 + 4 + 5 + 6 + 7 + 2 = 27 (divisible by 3).

154636 $\longrightarrow$ Last digit is 6 (divisible by 2) but the sum of digits 1 + 5 + 4 + 6 + 3 + 6 = 25 (not divisible by 3).

236794 $\longrightarrow$ Last digit is 4 (divisible by 2) but the sum of digits 2 + 3 + 6 + 7 + 9 + 4 = 31 (not divisible by 3).

Since 345672 is divisible by both 2 and 3, it is divisible by 6.

Hence, Option 1 is the correct option.

Which of the following is a prime number ?

1. 81

2. 87

3. 91

4. 97

Answer

A prime number is a number that has only two factors; 1 and itself.

81 $\longrightarrow$ Factors are 1, 3, 9, 27 and 81. It has more than two factors.

87 $\longrightarrow$ Factors are 1, 3, 29 and 87. It has more than two factors.

91 $\longrightarrow$ Factors are 1, 7, 13 and 91. It has more than two factors.

97 $\longrightarrow$ Factors are 1 and 97. It has exactly two factors.

Since 97 has exactly two factors, it is a prime number.

Hence, Option 4 is the correct option.

Which of the following is a composite number ?

1. 23

2. 29

3. 87

4. 97

Answer

A composite number has more than two factors.

23 $\longrightarrow$ Factors are 1 and 23. It has exactly two factors.

29 $\longrightarrow$ Factors are 1 and 29. It has exactly two factors.

87 $\longrightarrow$ Factors are 1, 3, 29 and 87. It has more than two factors.

97 $\longrightarrow$ Factors are 1 and 97. It has exactly two factors.

Since 87 has more than two factors, it is a composite number.

Hence, Option 3 is the correct option.

Which of the following is a pair of co-primes ?

1. 13, 91

2. 23, 46

3. 19, 20

4. 63, 84

Answer

Two numbers are co-primes if their H.C.F. is 1.

Using the prime factorisation method, we need to find the H.C.F. of each pair of numbers:

H.C.F. of 13 and 91:

$\begin{array}{r|r} 13 & 13 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 7 & 91 \\ \hline 13 & 13 \\ \hline & 1 \end{array}$

So,

13 = 1 × 13;

91 = 7 × 13

∴ H.C.F. of 13 and 91 is 13.

H.C.F. of 23 and 46 :

$\begin{array}{r|r} 23 & 23 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 46 \\ \hline 23 & 23 \\ \hline & 1 \end{array}$

So,

23 = 1 × 23;

46 = 2 × 23

∴ H.C.F. of 23 and 46 is 23.

H.C.F. of 19 and 20 :

$\begin{array}{r|r} 19 & 19 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

So,

19 = 1 × 19 ;

20 = 2 × 2 × 5

∴ H.C.F. of 19 and 20 is 1.

H.C.F. of 63 and 84 :

$\begin{array}{r|r} 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So,

63 = 3 × 3 × 7 ;

84 = 2 × 2 × 3 × 7

∴ H.C.F. of 63 and 84 = 3 × 7 = 21.

As the H.C.F. of 19 and 20 is 1, 19 and 20 are co-prime.

Hence, Option 3 is the correct option.

The H.C.F. of 72, 90 and 96 is

1. 18

2. 12

3. 6

4. 9

Answer

By prime factorisation method, we get:

$\begin{array}{r|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|r} 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

So,

72 = 2 × 2 × 2 × 3 × 3;

90 = 2 × 3 × 3 × 5 ;

96 = 2 × 2 × 2 × 2 × 2 × 3

The common factors are 2 and 3.

∴ H.C.F. = 2 × 3= 6

Hence, Option 3 is the correct option.

The H.C.F. of 147, 210 and 294 is

1. 7

2. 14

3. 28

4. 21

Answer

Using the long division method, first, we will find the H.C.F. of 147 and 210.

$\begin{array}{l} 147\overline{\smash{\big)}210 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{147} \\ \phantom{1444)}\overline{63\smash{\big)}147 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{126} \\ \phantom{1444 5678)}\overline{21\smash{\big)}63 \smash{\big(}3} \\ \phantom{space 144 5678 }\underline{63\phantom{1}} \\ \phantom{space 1444 567)}0\ \end{array}$

∴ H.C.F. of 147 and 210 is 21.

Next, we find the H.C.F. of 21 and 294.

$\begin{array}{l} 21\overline{\smash{\big)}294 \smash{\big(}}14 \\ \phantom{14}\phantom{)}\underline{21} \\ \phantom{14}\phantom{1)}{84} \\ \phantom{14}\phantom{2)}\underline{84} \\ \phantom{ 1141)}0\ \end{array}$

The H.C.F. of 21 and 294 is 21.

Hence, Option 4 is the correct option.

$\dfrac{289}{391}$ when reduced to the lowest terms is

1. $\dfrac{11}{23}$

2. $\dfrac{13}{31}$

3. $\dfrac{17}{31}$

4. $\dfrac{17}{23}$

Answer

To reduce a fraction to its lowest terms, we divide both numerator and denominator by their H.C.F.

Using the long division method to find the H.C.F. :

$\begin{array}{l} 289\overline{\smash{\big)}391 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{289} \\ \phantom{1444}\overline{102\smash{\big)}289 \smash{\big(}2} \\ \phantom{144 4444}\phantom{)}{204} \\ \phantom{1444 5678)}\overline{85\smash{\big)}102 \smash{\big(}1} \\ \phantom{144 444478911}\phantom{)}{85} \\ \phantom{1444 56787891)}\overline{17\smash{\big)}85 \smash{\big(}5} \\ \phantom{space 144 56785677 }\underline{85\phantom{1}} \\ \phantom{space 1444 567 5678)}0\ \end{array}$

The H.C.F. of 289 and 391 is 17.

Dividing both numerator and denominator of $\dfrac{289}{391}$ by 17,

$\dfrac{289}{391} = \dfrac{289 \div 17}{391 \div 17} = \dfrac{17}{23}$

Hence, Option 4 is the correct option.

The L.C.M. of 12, 15, 20, 27 is

1. 270

2. 360

3. 480

4. 540

Answer

By common division method, we get:

$\begin{array}{r|rrrr} 2 & 12, & 15, & 20, & 27 \\ \hline 2 & 6, & 15, & 10, & 27 \\ \hline 3 & 3, & 15, & 5, & 27 \\ \hline 5 & 1, & 5, & 5, & 9 \\ \hline & 1, & 1, & 1, & 9 \\ \end{array}$

The L.C.M. of 12, 15, 20, 27 = 2 × 2 × 3 × 5 × 9 = 540

Hence, Option 4 is the correct option.

If a and b are co-primes, then their L.C.M. is

1. a + b

2. ab

3. $\dfrac{a}{b}$

4. 1

Answer

If $a$ and $b$ are co-primes, their H.C.F. is 1.

For any two numbers $a$ and $b$,

L.C.M. = $\dfrac{a \times b}{\text{H.C.F.}} = \dfrac{a \times b}{1} = ab$

Hence, Option 2 is the correct option.

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

1. 69

2. 46

3. 92

4. none of these

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.

Hence, Option 1 is the correct option.

The product of two numbers is 2160 and their H.C.F. is 16. Then, the L.C.M. of these numbers is

1. 270

2. 540

3. 135

4. 405

Answer

It is given that,

H.C.F. = 16

Product of two numbers = 2160

We know that,

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

= $\dfrac{2160}{16}$ = 135

Hence, the L.C.M. is 135.

Hence, Option 3 is the correct option.

The smallest 4-digit number divisible by each one of 4, 5 and 6 is

1. 1060

2. 1040

3. 1020

4. 1080

Answer

First, we find the L.C.M. of 4, 5, and 6:

$\begin{array}{r|rrr} 2 & 4, & 5, & 6 \\ \hline 2 & 2, & 5, & 3 \\ \hline & 1, & 5, & 3 \\ \end{array}$

The L.C.M. of 4, 5, and 6 = 2 × 2 × 5 × 3 = 60

The smallest 4-digit number is 1000. Dividing 1000 by 60 gives a remainder of 40.

So the Required number = 1000 + (60 - 40) = 1020.

Hence, Option 3 is the correct option.

The least number divisible by each of the numbers 15, 20, 24, 32 and 36 is

1. 1660

2. 1440

3. 2490

4. 2160

Answer

The least number divisible by the given numbers is their L.C.M.

By common division method:

$\begin{array}{r|rrrrr} 2 & 15, & 20, & 24, & 32, & 36 \\ \hline 2 & 15, & 10, & 12, & 16, & 18 \\ \hline 2 & 15, & 5, & 6, & 8, & 9 \\ \hline 3 & 15, & 5, & 3, & 4, & 9 \\ \hline 5 & 5, & 5, & 1, & 4, & 3 \\ \hline & 1, & 1, & 1, & 4, & 3 \\ \end{array}$

The L.C.M. of 15, 20, 24, 32, 36 = 2 × 2 × 2 × 3 × 5 × 4 × 3 = 1440

Hence, Option 2 is the correct option.

The L.C.M. of 87 and 145 is

1. 435

2. 870

3. 1305

4. 1740

Answer

By prime factorisation, we have:

$\begin{array}{r|r} 3 & 87 \\ \hline 29 & 29 \\ \hline & 1 \end{array} \quad \text{and} \quad \begin{array}{r|l} 5 & 145 \\ \hline 29 & 29 \\ \hline & 1 \end{array}$

87 = 3 × 29;

145 = 5 × 29

So, L.C.M. of 87 and 145 = 3 × 5 × 29 = 1305

Hence, Option 3 is the correct option.

$\dfrac{561}{748}$ when reduced to the lowest terms is

1. $\dfrac{3}{4}$

2. $\dfrac{11}{14}$

3. $\dfrac{13}{14}$

4. $\dfrac{23}{24}$

Answer

To reduce a fraction to its lowest terms, we divide both numerator and denominator by their H.C.F.

Using the long division method to find the H.C.F. :

$\begin{array}{l} 561\overline{\smash{\big)}748 \smash{\big(}}1 \\ \phantom{144}\phantom{)}{561} \\ \phantom{1444}\overline{187\smash{\big)}561 \smash{\big(}3} \\ \phantom{space 1442 }\underline{561\phantom{1}} \\ \phantom{space 1444 )}0\ \end{array}$

The H.C.F. of 561 and 748 is 187.

Dividing both numerator and denominator of $\dfrac{561}{748}$ by 187,

$\dfrac{561}{748} = \dfrac{561 \div 187}{748 \div 187} = \dfrac{3}{4}$

Hence, Option 1 is the correct option.

The H.C.F. of $\dfrac{3}{16}, \dfrac{5}{12}$ and $\dfrac{7}{8}$ is

1. $\dfrac{1}{4}$

2. $\dfrac{1}{48}$

3. $\dfrac{105}{4}$

4. $\dfrac{105}{48}$

Answer

As per the hint provided,

H.C.F. of $\dfrac{3}{16}, \dfrac{5}{12}$ and $\dfrac{7}{8}$ = $\dfrac{\text {H.C.F. of } 3, 5, 7 }{ \text{L.C.M. of } 16, 12, 8}$

Since 3, 5, and 7 are co-prime, their H.C.F. is 1.

To find the L.C.M. of 16, 12, 8:

By common division method:

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

The L.C.M. of 8, 12, 16 = 2 × 2 × 2 × 3 × 2 = 48

H.C.F. of $\dfrac{3}{16}, \dfrac{5}{12}$ and $\dfrac{7}{8}$ = $\dfrac{\text {H.C.F. of } 3, 5, 7 }{ \text{L.C.M. of } 16, 12, 8}$ = $\dfrac{1}{48}$

Hence, Option 2 is the correct option.

The L.C.M. of $\dfrac{1}{3} , \dfrac{5}{6}, \dfrac{5}{9},$ and $\dfrac{10}{27}$ is

1. $\dfrac{5}{24}$

2. $\dfrac{5}{27}$

3. $\dfrac{5}{3}$

4. $\dfrac{10}{3}$

Answer

As per the hint provided,

L.C.M. of $\dfrac{1}{3} , \dfrac{5}{6}, \dfrac{5}{9},$ and $\dfrac{10}{27}$ = $\dfrac{\text {L.C.M. of } 1, 5, 5, 10 }{ \text{H.C.F. of } 3, 6, 9, 27}$

L.C.M. of 1, 5, 5, 10:

$\begin{array}{r|rrrr} 5 & 1, & 5, & 5, & 10 \\ \hline & 1, & 1, & 1, & 2 \\ \end{array}$

The L.C.M. of 1, 5, 5, 10 = 5 × 2 = 10

H.C.F. of 3, 6, 9, 27 :

The prime factorizations are :

3 = 1 × 3

6 = 2 × 3

9 = 3 × 3

27 = 3 × 3 × 3

So the H.C.F. of 3, 6, 9, 27 is 3.

L.C.M. of $\dfrac{1}{3} , \dfrac{5}{6}, \dfrac{5}{9},$ and $\dfrac{10}{27}$ = $\dfrac{\text {L.C.M. of } 1, 5, 5, 10 }{ \text{H.C.F. of } 3, 6, 9, 27}$ = $\dfrac{10}{3}$

Hence, Option 4 is the correct option.