Maths 2 Mark Question

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.