MATHS 3 Marks Question

LCM(3, 4, 5,7) = 3 × 4 × 5 × 7 = 420
The number must be a multiple of 420, so the number can be written as 420k, where k is a whole number.
N = 420 × 1 = 420
When divided by 11 leaves a remainder of 2
11 × 38 + 2 = 420
But we require 10 as a remainder
∴ 2k = 10
⇒ k = 5
Hence number = 420 × 5 = 2100

Longest number that perfectly divides (306, 36) = HCF (306, 36)

HCF (306, 36) = 2 × 3 × 3 = 18.
∴ (c) is the correct option.

No. of cows is divisible by 3, 5, 7.
This means the number of cows is a multiple of the LCM(3, 5, 7).
The total number of cows is also less than 200.
LCM (3, 5, 7) = 3 × 5 × 7 = 105
No. of cows must be a multiple of 105.
Multiples of 105 are 105, 210, 315.
The problem states that the cowherd had fewer than 200 cows.
Only multiples of 105 that are less than 200 are 105.
Hence cowherd had 105 cows.

For two numbers a and b, the product of the numbers is equal to the product of their LCM and HCF.
∴ a × b = HCF(a, b) × LCM(a, b)
Here HCF = 1, LCM = 66
⇒ a × b = 1 × 66 = 66
∴ Pairs of factors of 66 are (1, 66), (2, 33), (3, 22), (6, 11)
Any of the following pairs of numbers will satisfy the conditions: (1, 66), (2, 33), (3, 22), (6, 11).

The positions of the black star in the first row are 4 and 10.
The difference in position = 10 – 4 = 6.
So, the next positions of the black star in the first row are: 4, 10, 16, 22, 28, …
The positions of the black star in the second row are 4 and 8.
The difference in position = 8 – 4 = 4.
So, the next positions of the black star in the second row are: 4, 8, 12, 16, 20, …
The 16th position is common in both rows.
Hence, the black stars meet again at the 16th position.

Two Co-prime Numbers
Examples:
(i) (4, 9)
LCM (4, 9) = 36
(ii) (5, 8)
LCM (5, 8) = 40
(iii) (7, 10)
LCM (7, 10) = 70
Observation: The LCM of two co-prime numbers is equal to their product.
Reason: Co-prime numbers do not share any common factors except 1, so the smallest number that contains both is simply their product.
General Statement: The LCM of two co-prime numbers is equal to their product.
Note: Co-prime numbers are any two natural numbers that have no common factor other than 1.

Two Consecutive Numbers
Examples:
(i) (7, 8)
LCM (7, 8) = 56
(ii) (9, 10)
LCM (9, 10) = 90
(iii) (10, 11)
LCM (10, 11) = 110
Observation: The LCM of two consecutive numbers is equal to their product.
Reason: Consecutive numbers have no common factors other than 1, so their product is the smallest number divisible by both.
General Statement: The LCM of two consecutive numbers is their product.

Two Multiples of 3
Examples:
(i) (6, 9)
LCM (6, 9) = 18
(ii) (9, 12)
LCM (9, 12) = 36
(iii) (12, 18)
LCM (12, 18) = 36
Observation: The LCM of two multiples of 3 is also a multiple of 3.
Reason: Since both numbers are divisible by 3, their common multiples will also be divisible by 3.
Hence, the LCM must include 3 as a factor.
General Statement: The LCM of two multiples of 3 is always a multiple of 3.

General Statement: For any two positive integers, let’s call them a and b, the Least Common Multiple (LCM) will be equal to one of the numbers (specifically, the larger number) if and only if the smaller number is a factor (or a divisor) of the larger number.
In the original example, the LCM of 3 and 24 is 24 because 3 is a factor of 24 (24 ÷ 3 = 8)
Algebraic Description: Let the two positive integers be a and b, where a The LCM of a and b will be b if and only if b is a multiple of a.
This can be expressed using algebra as: LCM(a, b) = b if and only if b = k × a, where k is a positive integer.

No, one cannot say that 72 and 144 have no common factor other than 1 because their factorisations have composite numbers.
Here 72 = 6 × 12
144 = 8 × 18
These are not prime factorisations.
Both 6 and 12 are composite numbers, as are 8 and 18.
Prime factorisation of 72 = 2 × 2 × 2 × 3 × 3
Prime factorisation of 144 = 2 × 2 × 2 × 2 × 3 × 3
∴ HCF(72, 144) = 2 × 2 × 2 × 3 × 3 = 8 × 9 = 72
Since the HCF is 72, which is greater than 1, the numbers have common factors other than 1.

Here 81 and 243

∴ 81 = 3 × 3 × 3 × 3 and 243 = 3 × 3 × 3 × 3 × 3
∴ Common factors = 3 × 3 × 3 × 3 and HCF (81, 243) = 3 × 3 × 3 × 3 = 81.

Here, 370 and 592

∴ 370 = 2 × 5 × 37 and 592 = 2 × 2 × 2 × 2 × 37
∴ Common factors = 1, 2, 37, and HCF (370, 592) = 2 × 37 = 74

Here, 77 and 725

∴ 77 = 7 × 11 and 725 = 5 × 5 × 29
∴ Common factor = 1 and HCF (77, 725) = 1 (as there are no common prime factors)

Here 140 and 275

∴ 140 = 2 × 2 × 5 × 7 and 275 = 5 × 5 × 11
∴ Common factor of 140 and 275 = 5, and HCF of 140 and 275 = 5.

Here, 50

∴ Common factors of 50 and 60 are 2, 5, and HCF (50, 60) = 2 × 5 = 10.