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In the fractions a/b and c/d, where a, b, c and d are positive integer 01 Jun 2020, 04:16
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In the fractions a/b and c/d, where a, b, c and d are positive integers, both b and d have two prime factors each. What is the number of prime factors in the product of b and d?

(1) The least common denominator of a/b and c/d is half the product of b and d
(2) The highest integer that divides both b and d completely is 2


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In the fractions a/b and c/d, where a, b, c and d are positive integer 01 Jun 2020, 07:39
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Bunuel
In the fractions a/b and c/d, where a, b, c and d are positive integers, both b and d have two prime factors each. What is the number of prime factors in the product of b and d?

(1) The least common denominator of a/b and c/d is half the product of b and d
(2) The highest integer that divides both b and d completely is 2
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Using Statement 2, if 2 is the largest divisor of b and d, then clearly 2 is a prime divisor of each. Since b and d share no other divisors, the second prime divisor of b must be different from the second prime divisor of d. So the prime divisors of b are 2 and p, and the prime divisors of d are 2 and q, where p and q are different primes. The product bd will therefore have three prime divisors, 2, p and q, and Statement 2 is sufficient.

Using Statement 1, when fractions are completely reduced, their least common denominator will be the LCM of the two denominators. Since, for two numbers, LCM*GCD = product, then if the LCM of the denominators is half their product, the GCD of the two denominators would be 2. Then we have the situation of Statement 2, discussed above, and we have three prime divisors of bd in total. The problem though is that our fractions might not be reduced. We know a '2' cancels somewhere when we find the least common denominator of the two fractions. It might cancel, as above, when the denominators share a prime divisor of 2 -- the fractions might be 1/6 and 1/10, say. Then the least common denominator is 30, half of the product of 6 and 10, and bd = 60 has three prime divisors. But the 2 might instead cancel because one of the fractions can be reduced. The fractions might instead be 2/14 and 1/15, say. Then the least common denominator is the least common denominator of 1/7 and 1/15, which is 105, and that is (correctly) half the product of 14 and 15. But in this case bd has four prime divisors, 2, 3, 5 and 7, not three. So Statement 1 is not sufficient, and the answer is B.
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In the fractions a/b and c/d, where a, b, c and d are positive integer 01 Jun 2020, 08:52
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Bunuel
In the fractions a/b and c/d, where a, b, c and d are positive integers, both b and d have two prime factors each. What is the number of prime factors in the product of b and d?

(1) The least common denominator of a/b and c/d is half the product of b and d
(2) The highest integer that divides both b and d completely is 2

Using Statement 2, if 2 is the largest divisor of b and d, then clearly 2 is a prime divisor of each. Since b and d share no other divisors, the second prime divisor of b must be different from the second prime divisor of d. So the prime divisors of b are 2 and p, and the prime divisors of d are 2 and q, where p and q are different primes. The product bd will therefore have three prime divisors, 2, p and q, and Statement 2 is sufficient.

Using Statement 1, when fractions are completely reduced, their least common denominator will be the LCM of the two denominators. Since, for two numbers, LCM*GCD = product, then if the LCM of the denominators is half their product, the GCD of the two denominators would be 2. Then we have the situation of Statement 2, discussed above, and we have three prime divisors of bd in total. The problem though is that our fractions might not be reduced. We know a '2' cancels somewhere when we find the least common denominator of the two fractions. It might cancel, as above, when the denominators share a prime divisor of 2 -- the fractions might be 1/6 and 1/10, say. Then the least common denominator is 30, half of the product of 6 and 10, and bd = 60 has three prime divisors. But the 2 might instead cancel because one of the fractions can be reduced. The fractions might instead be 2/14 and 1/15, say. Then the least common denominator is the least common denominator of 1/7 and 1/15, which is 105, and that is (correctly) half the product of 14 and 15. But in this case bd has four prime divisors, 2, 3, 5 and 7, not three. So Statement 1 is not sufficient, and the answer is B.
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Can't we just take it like since B and D consist of two prime each and according to statement 2, B and D both have two in common. So B = 2X ans D = 2Y
Where X and Y are prime.
So product BD have 3 prime factors. Namely (2*X*Y)
Kindly revert.

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In the fractions a/b and c/d, where a, b, c and d are positive integer 01 Jun 2020, 12:24
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yashikaaggarwal

Can't we just take it like since B and D consist of two prime each and according to statement 2, B and D both have two in common. So B = 2X ans D = 2Y
Where X and Y are prime.
So product BD have 3 prime factors. Namely (2*X*Y)
Kindly revert.
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That's the right idea, and is roughly what I said, but I needed to use a few more words because it's not quite right. All we know is that 2 and some other prime are the only prime divisors of b, so b could, say, be equal to 2^9 * p^5; it doesn't need to be true that b = 2p. Besides that, conceptually your solution is right (and is essentially what I said above).
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