HCF and LCM

Find the least number which when divided by 12, 27 and 35 leaves 6 as a remainder:
Let find the answer:
Number = LCM (12, 17, 35) + 6
= 3780 + 6
= 3786

Let the two numbers be 13a and 13b respectively.
Then, 13a×13b = 2028
⇒ab = 2028 / (13×13) = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13×1, 13×12) and (13×3, 13×4). Clearly, there are 2 such pairs.

Given: HCF of two numbers = 12
Let numbers be 12a and 12 b where a and b are co-prime numbers.
12a×12b = 2160
ab = 2160/(12×12)
ab = 15
Possible pairs of a, b are (1, 15), (3, 5).
∴ numbers are (12, 180), (36, 60).
From the options, (36, 60) is the correct answer.

Multiples of 42
= 252, 294, 336, 378
So, * = 7

First number × second number = LCM × HCF
52×Second number = 520 × 4
Second number = (520×4)/52
= 40
Then the other number is 40.

Find the least number which when divided by 2, 3, 4 and 5 leaves a remainder 3. But when divided by 9 leaves no remainder:
Explanation :
LCM of 2, 3, 4 and 5 is 30,
let number be 30k + 3 put k = 2,
we get 63 which is divisible by 9

Let the numbers be 2x and 3x. Then, their L.C.M. = 6x. So, 6x = 48 or×= 8. The numbers are 16 and 24. Hence, required sum = (16 + 24) = 40.

Let the numbers be 'a' and 'b'.
Product of two numbers = LCM × HCF
Product of two numbers = 48
ab = 48
Difference of two numbers = 8
a - b = 8
Factors of ab are (1,48) , (2,24), (3,16),(4,12),(6,8).
Here, the only pair which satisfies the given condition is (4, 12).

Let the numbers be 3x and 4x.
LCM of 3x and 4x = 12x
Given, LCM = 240
12x = 240
x = 240/12 = 20
Smaller number = 3x = 3 × 20 = 60

99 = 1×3×3×11
101 = 1×101
176 = 1×2×2×2×2×11
182 = 1×2×7×13
So, divisors of 99 are 1, 3, 9, 11, 33, .
99 Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.