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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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04 Jun 2015, 05:18

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Stmt 1:

acd is even. No info about b. Not suff (Eg: if a =2, c=2 and d=2 acd =8; If b =6 then abcd is divisible by 6. If b =7 then abcd not divisible by 6) abd is odd. No info about c . Not suff (Eg: if a =3, b=3 and d=3 abd =27; If c =6 then abcd is divisible by 6. If b =7 then abcd not divisible by 6)

Combining both:

abd is odd which means a, b and d are odd. Plus Statement 1 says acd is even, so c must be even. So abcd will be even. Not sufficient to say if the product is divisible by 6.

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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Kindly press Kudos if the explanation is clear. Thank you Ambarish

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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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05 Jun 2015, 09:59

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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.

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

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05 Jun 2015, 20:04

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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

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.

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

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13 Mar 2017, 03:25

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