Maths 2 Mark Question

18, 17
Prime factorization of 18 = 2 × 3 × 3
Prime factorization of 17 = 17
Therefore, Required LCM = 2 × 3 × 3 × 17 = 306

180, 384, 144
Prime factorization of 180 = 2 × 2 × 3 × 3 × 5
Prime factorization of 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
Therefore, Required LCM = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 5,760

504 and 980
Prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7
Prime factorization of 980 = 2 × 2 × 5 × 7 × 7
Therefore, HCF = 2 × 2 × 7 = 28

21063
Sum of the digits of the given number = 2 + 1 + 0 + 6 + 3 = 12 which is divisible by 3.
Hence, 21,063 is divisible by 3.
Again, a number is divisible by 7 if the difference between twice the one’s digit and the number formed by the other digits is either 0 or a multiple of 7. 2,106 - (2 × 3) = 2,100 which is a multiple of 7. Thus, 21,063 is divisible by 21.

468
We have:

2	468
2	234
3	117
3	39
13	13
	1

Therefore, Prime factorization of 468 = 2 × 2 × 3 × 3 × 13

Those numbers with only two factors, i.e., 1 and the number itself, are known as prime numbers.
Examples: 2, 3, 5, 7. 11 and 13
The prime numbers between 1 and 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

13915
We have:

5	13915
11	2783
11	253
23	23
	1

Therefore, Prime factorization of 13915 = 5 × 11 × 11 × 23

A number which has more than two factors is called a composite number.
For example, the numbers 4, 6, 8, 9 10 and 15 are composite numbers.
Yes, a composite number can be an odd number. The smallest odd number is 9.

35 and 50
35 = 1 × 35
35 = 5 × 7 i.e., the factors of 35 are 1, 5, 7 and 35.
Again, 50 = 1 × 50
50 = 2 × 25
50 = 5 × 10
i.e., the factors of 50 are 1, 2, 5, 10, 25 and 50.
Therefore, the common factors of the two numbers are 1 and 5.

28, 36, 45, 60
Prime factorization of 28 = 2 × 2 × 7
Prime factorization of 36 = 2 × 2 × 3 × 3
Prime factorization of 45 = 3 × 3 × 5
Prime factorization of 60 = 2 × 2 × 3 × 5
Therefore, Required LCM = 2 × 2 × 3 × 3 × 5 × 7 = 1,260

225 and 450
Prime factorization of 225 = 3 × 3 × 5 × 5
Prime factorization of 198 = 2 × 3 × 3 × 5 × 5
Therefore, HCF = 3 × 3 × 5 × 5 = 225

5, 15 and 25
Factors of 5 are 1 and 5
Factors of 15 are 1, 3, 5 and 15
Factors of 25 are 1, 5 and 25
Therefore, the common factors of 5, 15, and 25 are 1 and 5.

110011
The sum of the digits at the odd places = 1 + 0 + 1 = 2
The sum of the digits at the even places = 1 + 0 + 1 = 2
The difference of the two sums = 2 - 2 = 0
Therefore, 1, 10,011 is divisible by 11 because the difference of the sums is zero.

399 and 437
We have dividend = 399 and divisor= 437

Clearly, the last divisor is 19.
Hence, HCF of the given number is 19

420
We have:

2	420
2	210
3	105
5	35
7	7
	1

Therefore, Prime factorization of 420 = 2 × 2 × 3 × 5 × 7

15 and 25
15 = 1 × 15
15 = 3 × 5
i.e., the factors of 15 are 1, 3, 5 and 15.
Again, 25 = 1 × 25
25 = 5 × 5 i.e., the factors of 25 are 1, 5 and 25.
Therefore, the common factors of the two numbers are 1 and 5.

Rule: A number is divisible by 6 if it is divisible by 2 as well as 3.
7020
Here, the units digit = 0
Thus, the given number is divisible by 2.
Also, the sum of the digits = 7 + 0 + 2 + 0 = 9 which is divisible by 3. So, the given number is
divisible by 3. Hence, 7,020 is divisible by 6.

66784 *
Sum of the given digits = 6 + 6 + 7 + 8 + 4 = 31
The multiple of 9 which is greater than 31 is 36.
Therefore, the smallest required number = 36 - 31 = 5

The number 2 is the smallest prime number.
It is an even prime number. Except 2, all other even numbers are composite numbers.

The smallest odd prime number is 3.
No, every odd number is not a prime number. For example, 9 is an odd number but it is not a prime number because its three factors are 1, 3 and 9.

Rule: A number is divisible by 6 if it is divisible by 2 as well as 3.
56423
Here, the units digit = 3 Thus, the given number is not divisible by 2.
Also, the sum of the digits = 5 + 6 + 4 + 2 + 3 = 20 which is not divisible by 3.
So, the given number is not divisible by 3. Since 3,56,423 is neither divisible by 2 nor by 3, it is
not divisible by 6.

20 and 28
20 = 1 × 20
20 = 2 × 10
20 = 4 × 5
i.e., the factors of 20 are 1, 2, 4, 5, 10 and 20.
Again, 28 = 1 × 28
28 = 2 × 14
28 = 7 × 4
i.e., the factors of 28 are 1, 2, 4, 7, 14 and 28.
Therefore, the common factors of the two numbers are 1, 2 and 4.

106, 159 and 265
Prime factorization of 106 = 2 × 53
Prime factorization of 159 = 2 × 53
Prime factorization of 265 = 5 × 53
Therefore, HCF = 53

Twin primes: Two prime numbers are said to be twin primes if there is only one composite number between them.
For example, (3, 5) and (5, 7) are twin primes.
Twin primes between 50 and 100 are (59, 61) and (71, 73).

84, 120 and 138
Prime factorization of 84 = 2 × 2 × 3 × 7
Prime factorization of 120 = 2 × 2 × 2 × 3 × 5
Prime factorization of 138 = 2 × 3 × 23
Therefore, HCF = 2 × 3 = 6

144 and 198
Prime factorization of 144 = 2 × 2 × 2 × 3 × 3
Prime factorization of 198 = 2 × 3 × 3 ×11
Therefore, HCF = 2 × 2 × 3 = 18

48, 60
Prime factorization of 48 = 2 × 2 × 2 × 2 × 3
Prime factorization of 60 = 2 × 2 × 3 × 5
Therefore, Required LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240

Since the number is divisible by 24, it will be divisible by all the factors of 24.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Hence, the number is also divisible by 1, 2, 3, 4, 6, 8 and 12.

Primes: Natural numbers which have exactly two distinct factors, i.e., 1 and the number itself are called prime numbers.
Hence, (59, 61) and (71, 73) are pairs of prime numbers.

67 *19
Sum of the given digits = 6 + 7 + 1 + 9 = 23
The multiple of 9 which is greater than 23 is 27.
Therefore, the smallest required number = 27 - 23 = 4

The given number is 1,00,00,001.
The sum of the digit at the odd places = 1 + 0 + 0 + 0 = 1
The sum of the digits at the even places = 0 + 0 + 0 + 1 = 1
Their difference = 1 - 1 = 0
Therefore, 1,00,00,001 is divisible by 11.

25
The first five multiples of 25 are as follows:
25 × 1 = 25
25 × 2 = 50
25 × 3 = 75
25 × 4 = 100
25 × 5 = 125

1 and the number itself are not included in the prime factorization of a composite number.
Example: 4 is a composite number.
Prime factorization of 4 = 2 × 2.

504 and 980
Prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7
Prime factorization of 980 = 2 × 2 × 5 × 7 × 7
Therefore, HCF = 2 × 2 × 7 = 28

The smallest 4-digit number is 1000.
1000 = 2 × 500
=2 × 2 × 250
=2 × 2 × 2 × 125
=2 × 2 × 2 × 5 × 25
=2 × 2 × 2 × 5 × 5 × 5
Therefore, 1000=2 × 2 × 2 × 5 × 5 × 5

145 and 232
Prime factorization of 145 = 5 × 29
Prime factorization of 232 = 2 × 2 × 2 × 29
Therefore, Required HCF of 145 and 232 = 289
Therefore, Required LCM of 145 and 232 = 2 × 2 × 2 × 5 × 29 = 1160

One prime and one composite number
One prime and one composite number are not always co-prime
Example: 3 and 21 are not co-primes to each other.

The given number is 5,335.
The sum of the digit at the odd places = 5 + 3 = 8
The sum of the digits at the even places = 3 + 5 = 8
Their difference = 8 - 8 = 0
Therefore, 5,335 is divisible by 11.

125
125 = 1 × 125
125 = 5 × 25
Therefore, the factors of 125 are 1, 5, 25 and 125.

No, it is not correct.
We know that HCF of two co-prime number is 1.
4 and 15 are co-prime numbers because the only factor common to them is 1.
Thus, HCF of 4 and 15 is 1.

The given number is 7,01,69,803.
The sum of the digit at the odd places = 7 + 1 + 9 + 0 = 17
The sum of the digits at the even places = 0 + 6 + 8 + 3 = 17
Their difference = 17 - 17 = 0
Therefore, 7,01,69,803 is divisible by 11.

81 and 117
Prime factorization of 81 = 3 × 3 × 3 × 3
Prime factorization of 117 = 3 × 3 × 13
Therefore, HCF = 3 × 3 = 9

234 and 572.
Prime factorization of 234 = 2 × 3 × 3 × 13
Prime factorization of 572 = 2 × 2 × 11 × 13
Therefore, Required HCF of 234 and 572 = 226
Therefore, Required LCM of 117 and 221 = 2 × 2 × 3 × 3 × 11 × 13 = 5148

538 * 8
Sum of the given digits = 5 + 3 + 8 + 8 = 24
The multiple of 9 which is greater than 24 is 27.
Therefore, the smallest required number = 27 - 24 = 3

60
60 = 1 × 60
60 = 2 × 30
60 = 3 × 20
60 = 4 × 15
60 = 5 × 12
60 = 6 × 10
The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

Since 60 = 30 x 2.
30 = 10 × 3 and 10 = 5 × 2 we have:

170 and 238
Prime factorization of 170 = 2 × 5 × 17
Prime factorization of 238 = 2 × 7 × 17
Therefore, HCF = 2 × 17 = 34

1100011
the sum of the digits at the odd places = 1 + 0 + 0 + 1 = 2
The sum of the digits at the even places = 1 + 0 + 1 = 2
The difference of the two sums = 2 - 2 = 0
Therefore, 11, 00,011 is divisible by 11 because the difference of the sums is zero.

216
We have:

2	216
2	108
2	54
3	27
3	9
3	3
	1

Therefore, Prime factorization of 216 = 2 × 2 × 2 × 3 × 3

Product of two numbers = HCF of two numbers × LCM of two numbers
⇒ 55545 = 23 × LCM of two numbers
⇒ LCM of two numbers = 5554523 = 2415

45
The first five multiples of 45 are as follows:
45 × 1 = 45
45 × 2 = 90
45 × 3 = 135
45 × 4 = 180
45 × 5 = 225

A number which is exactly divisible by 2 is called an even number. Therefore, 42 and 144 are even numbers.
A number which is not exactly divisible by 2 is called an odd number. Therefore, 89 and 321 are odd numbers.

729
729 = 1 × 729
729 = 3 × 243
729 = 9 × 81
729 = 27 × 27
Therefore, the factors of 729 are 1, 3, 9, 27, 81, 243 and 729.

No. We know that HCF of the given two numbers must exactly divide their LCM.
But 16 does not divide 380 exactly.
Hence, there can be no two numbers with 16 as their HCF and 380 as their LCM.

7325
We have:

5	7325
5	1465
293	293
	1

Therefore, Prime factorization of 732 5= 5 × 5 × 293

35
The first five multiples of 35 are as follows:
35 × 1 = 35
35 × 2 = 70
35 × 3 = 105
35 × 4 = 140
35 × 5 = 175

Two prime numbers.
Two prime numbers are always co-primes to each other.
Example: 7 and 11 are co-primes to each other.

42, 63
Prime factorization of 42 = 2 × 3 × 7
Prime factorization of 63 = 3 × 3 × 7
Therefore, Required LCM = 2 × 3 × 3 × 7 = 126

The largest 4-digit number is 9999.
We have:

3	9999
3	3333
11	1111
101	101
	1

Hence, the largest 4-digit number 9999 can be expressed in the form of its prime factors as:
3 × 3 × 11 × 101.

84 and 98
Prime factorization of 84 = 2 × 2 × 3 × 7
Prime factorization of 98 = 2 × 7 × 7
Therefore, HCF = 2 × 7 = 14

Two numbers are said to be co-primes if they do not have any common factors other than 1.
For example, (2, 3), (3, 4), (4, 5), (5, 7) and (13, 17) are co-primes.
Two co-primes numbers need not be both prime numbers.
e.g., (3, 4), (6, 7) and (4, 13).

The numbers between 1 and 100 having exactly three factors are 4, 9, 25, and 49.
The factors of 4 are 1, 2 and 4.
The factors of 9 are 1, 3 and 9.
The factors of 25 are 1, 5 and 25.
The factors of 49 are 1, 7 and 49.

108, 135, 162
Prime factorization of 108 = 2 × 2 × 3 × 3 × 3
Prime factorization of 135 = 3 × 3 × 3 × 5
Prime factorization of 162 = 2 × 3 × 3 × 3 × 3
Therefore, Required LCM = 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1,620

For a number (greater than 10) to be a prime number, the possible digit in the unit’s place may be 1, 3, 7 or 9.
Example: 11, 13, 17 and 19 are prime numbers greater than 10.

40
The first five multiples of 40 are as follows:
40 × 1 = 40
40 × 2 = 80
40 × 3 = 120
40 × 4 = 160
40 × 5 = 200

Co-primes: Two natural numbers are said to be co-primes numbers if they have 1 as their only common factor.
Hence, all the given pairs of numbers are co-primes.

We have: Since 6 = 2 × 3 and 10 = 5 × 2.

945
We have:

3	945
3	315
3	105
5	35
7	7
	1

Therefore, Prime factorization of 945 = 3 × 3 × 3 × 5 × 7

Composite numbers: Natural numbers which have more than two factors are called composite numbers.
Hence, (55, 57) and (63, 65) are pairs of composite numbers.

20163
Sum of the digits of the given number = 2 + 0 + 1 + 6 + 3 = 12 which is divisible by 3. Hence, 20,163 is divisible by 3.
Again, a number is divisible by 7 if the difference between twice the one’s digit and the number formed by the other digits is either 0 or multiple of 7. 2016 - (2 × 3) = 2010 which is not a multiple of 7. Thus, 20,163 is not divisible by 21.

15, 30, 90
Prime factorization of 15 = 3 × 5
Prime factorization of 30 = 2 × 3 × 5
Prime factorization of 90 = 2 × 3 × 3 × 5
Therefore, Required LCM = 2 × 3 × 3 × 5 = 90

Two composite numbers
Two composite numbers are not always co-primes to each other.
Example: 4 and 6 are not co-primes to each other.

2, 6 and 8
Factors of 2 are 1 and 2
Factors of 6 are 1, 2, 3 and 6
Factors of 8 are 1, 2, 4 and 8
Therefore, the common factors of 2, 6 and 8 are 1 and 2.

76
76 = 1 × 76
76 = 2 × 38
76 = 4 × 19
Therefore, The factors of 76 are 1, 2, 4, 19, 38 and 76.

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132…
Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198…
Therefore, the first two common multiples of 12 and 18 are 36 and 72.

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84…
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96…
Therefore, the first three common multiples of 6 and 8 are 24, 48 and 72.

1,111
The sum of the digits at the odd places = 1 + 1 = 2
The sum of the digits at the even places = 1 + 1 = 2
The difference of the two sums = 2 - 2 = 0
Therefore, 1,111 is divisible by 11 because the difference of the sums is zero.

The HCF of all natural numbers from 200 to 478 is 1 because there are some prime numbers like 211, 233 and so on which can't have common factor other than 1.

861 and 1353
Prime factorization of 861 = 3 × 7 × 41
Prime factorization of 1353 = 3 × 11 × 41
Therefore, Required HCF of 861 and 1353 = 123
Therefore, Required LCM of 861 and 1353 = 3 × 7 × 11 × 41 = 9471

1045 and 1520
We have dividend = 1045 and divisor = 1520

Clearly, the last divisor is 95.
Hence, HCF of given numbers is 95.

56, 65, 85
Prime factorization of 56 = 2 × 2 × 2 × 7
Prime factorization of 65 = 5 × 13
Prime factorization of 85 = 5 × 17
Therefore, Required LCM = 2 × 2 × 2 × 5 × 7 × 13 × 17 = 61,880

174 and 221
Prime factorization of 117 = 3 × 3 × 13
Prime factorization of 221 = 13 × 17
Therefore, Required HCF of 117 and 221 = 13
Therefore, Required LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1989

11011
The sum of the digits at the odd places = 1 + 0 + 1 = 2
The sum of the digits at the even places = 1 + 1 = 2
The difference of the two sums = 2 – 2 = 0
Therefore, 11,011 is divisible by 11 because the difference of the sums is zero.