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How many sets of three distinct factors of the number N = 2^{6} x 3^{4} x 5^{2} can be made such that the factors in each set have a highest common factor of 1 with respect to every other factor in that set?

- a)236
- b)360
- c)104
- d)380

Correct answer is option 'A'. Can you explain this answer?

N = 2^{6} × 3^{4} × 5^{2}

Case: [A]

when none of the elements is 1 ⇒ the three factors must be some powers of 2, 3, 5 respectively.

⇒ total number of such sets = 6 × 4 × 2 = 48

Case: [B]

when one of the elements is 1

⇒ the other two factors could be of the form (2^{a}), (3^{b}) − number of sets = 6 × 4 = 24

(2^{6}), (5^{2}) − number of sets = 6 × 2 = 12

(3^{4}),(5^{2}) − number of sets = 4 × 2 = 8

(2^{6} × 3^{4} × 5^{2}) − number of sets = 6 × 4 × 2 = 48

(2^{6} × 5^{2} × 3^{4}) − number of sets = 6 × 2 × 4 = 48

(3^{4} × 5^{2} × 2^{6}) − number of sets = 4 × 2 × 6 = 48

∴ total number of such sets = 188

combining both the cases, total sets possible

= 188 + 48 = 236

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How many sets of three distinct factors of the number N = 26 x 34 x 52 can be made such that the factors in each set have a highest common factor of 1 with respect to every other factor in that set?a)236b)360c)104d)380Correct answer is option 'A'. Can you explain this answer? for CAT 2023 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about How many sets of three distinct factors of the number N = 26 x 34 x 52 can be made such that the factors in each set have a highest common factor of 1 with respect to every other factor in that set?a)236b)360c)104d)380Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for CAT 2023 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How many sets of three distinct factors of the number N = 26 x 34 x 52 can be made such that the factors in each set have a highest common factor of 1 with respect to every other factor in that set?a)236b)360c)104d)380Correct answer is option 'A'. Can you explain this answer?.

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N = 26 × 34 × 52Case: [A]when none of the elements is 1 ⇒ the three factors must be some powers of 2, 3, 5 respectively.⇒ total number of such sets = 6 × 4 × 2 = 48Case: [B]when one of the elements is 1⇒ the other two factors could be of the form (2a), (3b) − number of sets = 6 × 4 = 24(26), (52) − number of sets = 6 × 2 = 12(34),(52) − number of sets = 4 × 2 = 8(26 × 34 × 52) − number of sets = 6 × 4 × 2 = 48(26 × 52 × 34) − number of sets = 6 × 2 × 4 = 48(34 × 52 × 26) − number of sets = 4 × 2 × 6 = 48∴ total number of such sets = 188combining both the cases, total sets possible= 188 + 48 = 236