Problems on H.C.F and L.C.M

Correct Answer: 2

Explanation:

Let the numbers 13 a and 13 b

Then, 13a x 13b = 2028

⟹ ab = 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 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Correct Answer: 4

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

  = H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Correct Answer: 1

Explanation:

Go through options.
Put m = 1 in two given expression,we have
x³ - 10x² + 31x - 30 and x² - 8x + 15
x² - 8x + 15 = (x - 5) (x - 3)
x³ - 10x² + 31x - 30 = (x - 2) (x - 5) (x - 3)
? (x - 5) (x - 3) = x² - 8x + 15 is the HCF of both expressions.
? m = 1