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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
Yes it is E. Nice question. Thank you Bunuel.

Bunuel wrote:
If a, b, c, and d are each integers greater than 1, is the product abcd divisible by 6?

(1) acd is even.
(2) abd is odd.
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
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If a, b, c, and d are each integers greater than 1, is the product abcd divisible by 6?

is abcd by 2*3 ?

(1) acd is even.
a=2
c=7
d=3
yes, divisible

a=2
c=7
d=1
no, not divisible
Insufficient

(2) abd is odd.
a=11
b=11
d=11 and c=6
yes divisible

a=11
b=11
d=11 and c = 2
not divisible
Insufficient

Combined, We still are uncertain if its divisible or not, a,b,d=11 c=6 or 2
Answer: E
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
1
Kudos
stmt 1 - we get acd is even so one of them/two of them or all of them can be even and no info about b so not sufficient.
stmt 2 - abd is odd - so all of them are odd but no info about c - not sufficient
combining - we get a d b are odd and c is even, for the product to be divisible by 6 it should be divisible by 2 and 3 with c being even it is divisible by 2 but we are not sure about the other odd numbers, for they can be 13,17,19 or 3 9 15.

hence ans E
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If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
1
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Q If a, b, c, and d are each integers greater than 1, is the product abcd divisible by 6?
in other words does abcd have factors of 2, 3

stmt (1) acd is even.
    at-least one integer is 2 or factor of 2 ; no info about 3 as a factor. ----------- Not Sufficient

stmt (2) abd is odd
    one integer out of (abd) can be factor of 3 or none (abd) factor of 3; no info about 2 as a factor. ----------- Not Sufficient
    (a b d) ~ 5 * 7 * 11
    (a b d) ~ 3 * 5 * 7

together stmt (1) & (2) , same as above :
    2* 5 * 7 * 11 = no
    2* 3 * 5 * 7 = yes ------------ Not Sufficient

Ans : E
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
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Bunuel wrote:
If a, b, c, and d are each integers greater than 1, is the product abcd divisible by 6?

(1) acd is even.
(2) abd is odd.


MANHATTAN GMAT OFFICIAL SOLUTION:

Another way of saying that an integer is divisible by 6 is to say that it is a multiple of 6. Multiples of 6 must have all of the prime factors of 6 (6 = 2 × 3) and could have additional prime factors. Thus, our rephrased question is “Does the product abcd have at least one 2 and one 3 as prime factors?”

(1) INSUFFICIENT: This tells us that at least one of the integers a, c, and d must be even. Thus we have at least one 2 as a prime factor. However, we do not know anything about the remaining factors, and cannot determine whether there is one 3 among the prime factors of a, b, c, and d.

(2) INSUFFICIENT: This tells us that a, b, and d are all odd, which means there is no factor of 2 among their prime factors. Without information about c, we are uncertain about whether abcd has a factor of 2. Additionally, we have no information about the number of 3s among the prime factors of a, b, and d. It is possible that abd is 105, for example, and we would have the 3 required for divisibility by 6. On the other hand, abd could be 125 and we would have no 3s as factors.

(1) AND (2) INSUFFICIENT: If acd is even and if abd is odd, it must be true that c is even and that abcd has at least one factor of 2. Neither of the statements gives us a conclusive answer about the number of 3s among the prime factors of a, b, c, and d, however, and combining the statements does nothing to resolve that uncertainty.

The correct answer is E.
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
Hello

I think official solution is flawed since it consider 125 in explaining statement 2.

It is mentioned as a b c d so we presume that these integers are distinct. Please correct me if I am wrong.
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
Expert Reply
vikasbansal227 wrote:
Hello

I think official solution is flawed since it consider 125 in explaining statement 2.

It is mentioned as a b c d so we presume that these integers are distinct. Please correct me if I am wrong.


Yes, you are wrong.

Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
Bunuel wrote:
If a, b, c, and d are each integers greater than 1, is the product abcd divisible by 6?

(1) acd is even.
(2) abd is odd.


Statement 1

Too many possibilities

insuff

Statement 2

This just tells us abd are all odd integers...no info about what integers they are

insuff

Statement 1 & 2

5 * 7 * 11 * some even integers
3* 5 * 11 * some even integer

If there were a restriction that one of the integers had to be 3 then both would be suff but otherwise

insuff
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
Bunuel wrote:
If a, b, c, and d are each integers greater than 1, is the product abcd divisible by 6?

(1) acd is even.
(2) abd is odd.

We need at least one \(3\) and one \(2\) for a number to be divisible by \(6\)

S1: acd is even [Insufficient]
1. \(a*c*d = 2*5*7\) and \(b = 3\), then \(abcd\) is divisible by \(6\)
2. \(a*c*d = 2*5*7\) and \(b = 11\), then \(abcd\) is not divisible by \(6\)

S2: abd is odd [Insufficient]
1. \(a*b*d = 3*5*7\) and \(b = 2\), then \(abcd\) is divisible by \(6\)
2. \(a*b*d = 11*5*7\) and \(b = 2\), then \(abcd\) is not divisible by \(6\)

S1 + S2 [Insufficient]
1. \(a*b*c*d = 3*5*2*7\), then \(abcd\) is divisible by \(6\)
2. \(a*b*c*d = 11*5*2*7\), then \(abcd\) is not divisible by \(6\)

Ans. E
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Re: If a, b, c, and d are each integers greater than 1, is the product abc [#permalink]
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