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a, b, c are consecutive integers, so that a < b < c. Is the

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a, b, c are consecutive integers, so that a < b < c. Is the [#permalink]

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a, b, c are consecutive integers, so that a < b < c. Is the product a*b*c divisible by 8?

(1) a*b*c is divisible by 12
(2) b is a prime number
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Re: divisibility [#permalink]

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eladshush wrote:
a,b,c are consecutive integers , so that a < b < c

Is the product a*b*c divisible by 8?

1. a*b*c is divisible by 12
2. b is a prime number


Note that if \(b=odd\) then \(a\) and \(c\) become two consecutive even numbers and the product of two consecutive even numbers is always divisible by 8 (as one of these even numbers would be multiple of 4 too). So if we could determine that \(b=odd\) it would be sufficient to answer that \(abc\) is divisible by 8.

(1) a*b*c is divisible by 12 --> if \(a=0\), \(b=1\), and \(c=2\) then answer would be YES but if \(a=3\), \(b=4\), and \(c=5\) then answer would be NO. Not sufficient.

(2) b is a prime number --> if \(b=2=even \ prime\) then \(abc=6\) and the answer is NO but if \(b=3\) then \(abc=24\) and the answer is YES. Not sufficient.

(1)+(2) If \(b=2=even \ prime\) then \(abc=6\) and 6 is not divisible by 12, so \(b\) is odd prime --> \(abc\) is divisible by 8. Sufficient.

Answer: C.
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Re: divisibility [#permalink]

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New post 04 Oct 2010, 09:27
1) a*b*c/12
a*b*c= 3*4*5/12 = 60/12 =5 , but 60 is not divisible by 8.
a*b*c= 4*5*6/12 = 120/ 12=10, but 120 is divisible by 8. (not sufficient)

2) b is prime number(not sufficient)

(1)+ (2) a*b*c= 4*5*6=120,120/8=15. (b is prime)
a*b*c= 12*13*14=2184/8= 273(b is prime)

answer is c.

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Re: a,b,c are consecutive integers , so that a < b < c Is [#permalink]

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New post 22 Jan 2014, 07:19
eladshush wrote:
a,b,c are consecutive integers , so that a < b < c

Is the product a*b*c divisible by 8?

1. a*b*c is divisible by 12
2. b is a prime number


Since these three numbers are consecutive integers, we need to know if either 'a' AND 'c' are even or if b is a multiple of 7

Statement 1

abc is multiple of 12, as these are three consecutive integers they will always be divisible by 3. So we know that abc is also divisible by 4. This means two things, either abc has two even factors and thus our answer would be YES. (Eg. (2)(3)(4)) or b is a multiple of 4 (Eg. ((3)(4)(5)=60) with an answer of NO

Hence insuff

Statement 2

B is a prime number, if B is any odd prime number then the answer is yes because other two factors will be even and thus divisible by 2^3. If B is 2 then (1)(2)(3)=6 and not divisble by 8

Statements 1 and 2

We know that if B is a prime number and if the product of the three numbers must be a multiple of 12 then b has to be an odd prime number and thus this two statements together are sufficient

C

Hope it helps
For any doubts please reach out and don't forget to leave your Kudos if helpful

Cheers!
J :)

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Re: a,b,c are consecutive integers , so that a < b < c Is [#permalink]

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New post 22 Jan 2014, 17:24
eladshush wrote:
a,b,c are consecutive integers , so that a < b < c

Is the product a*b*c divisible by 8?

1. a*b*c is divisible by 12
2. b is a prime number


The answer is C
1. a *b*C is divisible by 12 - Insufficient
if a,b,c =3,4,5 ---------- not divisivle by 8
if a,b,c =7,8.9 ----------- divisible by 8
2. b is a prime number --- Insufficient
let us consider primes --- 2,3,5,7,11,.............
if we consider even prime i.e. 2
a,b,c =1,2,3
abc is not divisible by 8
if b=3
a,b,c =2,3,4
and abc=24 which is divisible by 8
if we consider 1 & 2 together abc is divisible by 12 and b is prime it will eliminate considering b as 2 because when b is 2 abc =6 which is eliminated.
so now abc is always multiple of 8

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Re: a, b, c are consecutive integers, so that a < b < c. Is the [#permalink]

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Re: a, b, c are consecutive integers, so that a < b < c. Is the   [#permalink] 26 Sep 2017, 04:16
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