Question

How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 3000?

Answer 1

Answer:

We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.

Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.

So h

3

(ab) = 1080 = (2^3)(3^3)(5).

We need to write 1080 as a product of a perfect cube and another number.

Four cases:

h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080).

h = 2, We need to find number of ways of writing (33) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (33) * (5) , (33) and (5)

number of pairs = 2, number of ordered pairs = 4

h = 3, number of pairs = 2, number of ordered pairs = 4

h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.

The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30). Choice (C)

Correct Answer: 9

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Answer 2

Answer:

The pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 3000 is 16 pairs.

Step-by-step explanation:

Let , x= HCF *a, where a is some random integer.

Similarly let y = HCF*b, where b is some random integer.

3000 = HCF*HCF*a*b*HCF = HCF^3*a*b

Now, the Prime Factorization of 3000 is 2³ × 3¹ × 5³

Factors of 3000- 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 25, 30, 40, 50, 60, 75, 100, 120, 125, 150, 200, 250, 300, 375, 500, 600, 750, 1000, 1500 and 3000

Now we can write 3000 in the form of cube*a*b such that a,b are co-primes

1*3000, 8*375, and 600*5

375 -> prime factorization-> 3*5*5*5 -> 2 primes

any number can be expressed as a product of two co-primes in 2^(n–1) ways where n is the no of primes in its factorization, here n =2 , so 2 ways

similarly, for 600 -> 2 × 2 × 2 × 3 × 5 × 5 -> 3 primes = 3 ways

for 3 -> 1 prime = 1 way

for 3000 = (2^3)*(3^3)*5 = 4 ways

thus , total 2+2+1+5 = 8 ways

if (x,y) is not an ordered pair then total ways = 8*2 = 16 ways.

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