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Question

Let N=24×35×52×74×11 be resolved as a product of two factors (order immaterial), then

A
number of cases in which the two factors are coprime is 16
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B
number of cases in which their H.C.F. is 3, is 32
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C
number of cases in which their H.C.F. 5, is 16
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D
number of cases in which their H.C.F is odd is 180
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Solution

The correct options are
A number of cases in which the two factors are coprime is 16
D number of cases in which their H.C.F is odd is 180
Number of cases in which two factors are coprime:
For this case, we need to divide the numbers such that two factors do not have any common factor. Let's say the factors are a and b and a×b=24×35×52×74×11
Here, we need to understand that if a is take 2something as its factor, then that 'something' must be 4. Because if a takes 22 as factor then the remaining 22 has to be factor of b, in this case a and b are not coprime.

Here,
242 choices, either go to a or b352 choices, either go to a or b522 choices, either go to a or b742 choices, either go to a or b112 choices, either go to a or b
No. of cases 252=24=16 we will divide by 2 because order doesn't matter,a=24×35,b=52×74×11 is same as b=24×35,a=52×74×11

Number of cases in which their H.C.F. is 3:
Let a and b be two parts then a=3k1 & b=3k2 where k1 and k2 are coprime
Remaining ways 252=16

Number of cases in which their H.C.F. is 5:
no. of cases 242=8

Number of cases in which their H.C.F. is odd:
no. of cases =2×6×3×5×22=180

as 24 has 2 ways35 has 6 ways52 has 3 ways74 has 5 ways111 has 2 ways⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪of dividing into 2 factors

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