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06 Feb 2019, 18:36
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
59% (01:23) correct
41%
(01:24)
wrong
based on 240
sessions
History
Date
Time
Result
Not Attempted Yet
If a and b are positive integers, is a a multiple of b?
(1) Every distinct prime factor of b is also a distinct prime factor of a.
(2) Every factor of b is also a factor of a.
(1) Every distinct prime factor of b is also a distinct prime factor of a.
(2) Every factor of b is also a factor of a.
Question: is a = bk+0/a? In other words, can we divide a by b with 0 remainder?
1) Take 2,3,5 as the prime factorization.
2*3*5/2*3*5 = 1 so yes, a/b ONLY if they have the same amount of each prime, but we don't know this.
If we change the power of the primes in b, (e.g. 2*3*5/2^2*3*5) then it's no since it's a non-int.
Not sufficient.
2) If every factor of b is in a, then a is a multiple of b.
Take 2,3,5 again as the prime factorization. Factors are 1*30, 2*15, 3*10, 5*6. Any way we choose we get a=30, b=30, 30/30 = 1, so Yes.
Sufficient.
Official explanation:
To solve this yes-or-no problem, your best bet is to plug in values that meet the requirements given in the statements to determine whether Plugging In always yields the same answer to the question.
For Statement (1), for example, we could Plug In 4 for both a and b—that’s one easy way to make sure that every prime factor of b is also a prime factor of a—and since every number is a multiple of itself, the answer to the question is “yes.” However, if we leave b = 4 but make a = 2, we can still satisfy the requirement of Statement (1)—4 has only one prime factor, 2, which is also a prime factor of 2—but now our answer is “no.” Since Statement (1) yields different answers, it’s insufficient, and the possible
answers are BCE.
Similar attempts in Statement (2), however, will always yield the answer “yes.” We could, of course, again Plug In 4 for both variables, and we have our first “yes.” If we leave b = 4, though, we can’t make a = 2, since 4 isn’t a factor of 2; a is a number such as 4 (which we’ve already used), 12, 16, or some other number that has 1, 2, and 4 as factors. Whichever one we pick, though, our answer is “yes”; Statement (2) is therefore sufficient, and the answer to the problem is (B).
1) Take 2,3,5 as the prime factorization.
2*3*5/2*3*5 = 1 so yes, a/b ONLY if they have the same amount of each prime, but we don't know this.
If we change the power of the primes in b, (e.g. 2*3*5/2^2*3*5) then it's no since it's a non-int.
Not sufficient.
2) If every factor of b is in a, then a is a multiple of b.
Take 2,3,5 again as the prime factorization. Factors are 1*30, 2*15, 3*10, 5*6. Any way we choose we get a=30, b=30, 30/30 = 1, so Yes.
Sufficient.
Official explanation:
To solve this yes-or-no problem, your best bet is to plug in values that meet the requirements given in the statements to determine whether Plugging In always yields the same answer to the question.
For Statement (1), for example, we could Plug In 4 for both a and b—that’s one easy way to make sure that every prime factor of b is also a prime factor of a—and since every number is a multiple of itself, the answer to the question is “yes.” However, if we leave b = 4 but make a = 2, we can still satisfy the requirement of Statement (1)—4 has only one prime factor, 2, which is also a prime factor of 2—but now our answer is “no.” Since Statement (1) yields different answers, it’s insufficient, and the possible
answers are BCE.
Similar attempts in Statement (2), however, will always yield the answer “yes.” We could, of course, again Plug In 4 for both variables, and we have our first “yes.” If we leave b = 4, though, we can’t make a = 2, since 4 isn’t a factor of 2; a is a number such as 4 (which we’ve already used), 12, 16, or some other number that has 1, 2, and 4 as factors. Whichever one we pick, though, our answer is “yes”; Statement (2) is therefore sufficient, and the answer to the problem is (B).
06 Feb 2019, 19:53
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If a and b are positive integers, is a a multiple of b?
(1) Every distinct prime factor of b is also a distinct prime factor of a.
This tells us that if \(b=x^p*y^q\), then \(a=x^r*y^s*z^t...\)
Now distinct prime factors x and y are also factors of a, but we cannot say about their power, p, q, r and s.
If \(p\leq{r}\) and \(q\leq{s}\), answer is YES... \(b=x^p*y^q=2^4*3^1=48\), then \(a=x^r*y^s*z^t...=2^5*3^1*5^2=2400\) and 48*50 is 2400
If \(p>{r}\) or \(q>{s}\) or both, answer is NO... \(b=x^p*y^q=2^4*3^1=48\), then \(a=x^r*y^s*z^t...=2^3*3^1*5^2=600\) and 600/48=12.5
Hence, insufficient.
(2) Every factor of b is also a factor of a.
This tells us that if \(b=x^p*y^q\), then \(a=x^r*y^s*z^t...\) and \(p\leq{r}\) and \(q\leq{s}\).
Answer is always YES
Sufficient
B
(1) Every distinct prime factor of b is also a distinct prime factor of a.
This tells us that if \(b=x^p*y^q\), then \(a=x^r*y^s*z^t...\)
Now distinct prime factors x and y are also factors of a, but we cannot say about their power, p, q, r and s.
If \(p\leq{r}\) and \(q\leq{s}\), answer is YES... \(b=x^p*y^q=2^4*3^1=48\), then \(a=x^r*y^s*z^t...=2^5*3^1*5^2=2400\) and 48*50 is 2400
If \(p>{r}\) or \(q>{s}\) or both, answer is NO... \(b=x^p*y^q=2^4*3^1=48\), then \(a=x^r*y^s*z^t...=2^3*3^1*5^2=600\) and 600/48=12.5
Hence, insufficient.
(2) Every factor of b is also a factor of a.
This tells us that if \(b=x^p*y^q\), then \(a=x^r*y^s*z^t...\) and \(p\leq{r}\) and \(q\leq{s}\).
Answer is always YES
Sufficient
B
18 Jun 2022, 08:41
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energetics
Took me 3:30 minutes to solve it. Is there any quicker method? Bunuel
I always get stuck in such kind of questions. What should be the structured and organised way to tackle such questions? when we talk about multiple and factors?
