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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod

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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?

(1) The product ac is even.

(2) The product bc is a multiple of 4.

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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 27 Jul 2017, 07:31
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carcass wrote:
If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?

(1) The product ac is even.

(2) The product bc is a multiple of 4.



hi,

If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8
TWO cases..
A) a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24.. ans will be YES
B) a and c are ODD and b is even.. if b is multiple of 8.. YES otherwise NO


lets see the statements
(1) The product ac is even.
this MEANS a and c are even..
falls under category A above..
always YES
sufficient

(2) The product bc is a multiple of 4

either b is multiple of 4, ans can be YES or NO case B
OR c is multiple of 4, ans will be YES.. case A
different answers possible
insuff

A
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 27 Jul 2017, 07:34
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I Think answer is A

Statement 1:
ac is even -> a and c are even and b is odd, minimum possible values of abc is 2*3*4.
Sufficient

Statement 2:
bc multiple of 4 --> b can be 4.. c can be 5.... so a can be 3, which gives abc=60
not sufficient

Answer A
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 01 Dec 2017, 14:58
I don't get why a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24, could you explain me please
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 01 Dec 2017, 19:38
analuisa wrote:
I don't get why a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24, could you explain me please


Hi analuisa

as the numbers are consecutive, so \(a\), \(b\), \(c\) can take two options -

Case 1: Even, Odd, Even OR

Case 2: Odd, Even, Odd

Statement 1: as \(ac\) is even, so it has to be Case 1

Now let \(a=2k\) (even number) so \(b=2k+1\) and \(c=2k+2=2(k+1)\), where \(k\) is any number greater than \(0\) (because \(0<a<b<c\))

Hence \(ac=2k*2(k+1)=4k(k+1)\)

so if \(k\) is odd then \(k+1=even\), hence \(4k(k+1)\) will be a multiple of \(8\)

if \(k\) is even then \(4k\) will be a multiple of \(8\). Therefore \(abc\) is a multiple of \(8\) because \(ac\) is a multiple of \(8\). Sufficient

Statement 2: \(bc\) is a multiple of \(4\). so it could be multiple of \(8\) such as \(16,24,32\) etc. or it could not be a multiple of \(8\) such as \(20, 28\) etc.. Insufficient

Option A
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 01 Dec 2017, 20:29
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carcass wrote:
If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?

(1) The product ac is even.

(2) The product bc is a multiple of 4.


The Best would be to plug in by numbers.

Let b = n, a = n-1, c = n+1

Stmnt 1: ac = even, \(n^2\) - 1 =even, Here n can be only a odd number starting from 3,5,7,9,11 etc.. So abc will be Divisible by 8.
Stmnt 2: n(n+1) = 4m.. Here, n can be 3,4,7,8,9,11,13,14.. etc. when n = 3, abc Divisible by 8 but n= 4, abc not divisible by 8.

Hope this Helps.
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 01 Dec 2017, 20:48
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analuisa wrote:
I don't get why a and c are even AND b is ODD.... abc will always be multiple of 8 or rather 24, could you explain me please


Hi...
Three consecutive integers and two even means even, odd, even...
Now since three are consecutive, one of them would surely be a MULTIPLE of 3.
Two of them even means one will be MULTIPLE of 2 and other will be MULTIPLE of atleast 4 so product will be MULTIPLE of atleast 3*2*4=24
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 23 Aug 2018, 22:36
Hi,

Another way to solve this question is trying out numbers.

Always try out few numbers and understand the pattern of the answers.

Here in the question, it’s given, a, b and c are consecutive integers and also positive (0 < a < b < c),

Question: Is product abc a multiple of 8 ?

To be a multiple of 8 , it has to have minimum of three 2’s because of 8 can be written in prime factorization as 2^3

Now, lets try out some numbers,

a, b and c minimum has to be 1,2 and 3. Here the product abc is 6. Not a multiple of 8.

If a, b and c is 2,3 and 4. Then the product abc is 24. Multiple of 8.

Okay, then lets check the next set.

If a,b and c is 3,4 and 5. Then the product abc is 60. Not a multiple of 8.

If a, b and c is 4,5 and 6. Then the product abc is 120. Multiple of 8.

Note 1:

We can note from the above pattern, everytime if a and c is even, then the product abc is a multiple of 8(Infact 24).

Note 2:

Also, if a and c are odd, then the product abc need not be a multiple of 8 but it is an even number.

Statement I is sufficient.

Given, the product ac is even.


Since, a, b and c are consecutive integers. If ac is even, then both a and c has to be even.

So, from the note 1, we can see that product is a multiple of 8.

So sufficient.

Statement II is insufficient.

Given, the product bc is a multiple of 4.


bc is a multiple of 4 means, either b is a multiple of 4 and c is odd or b is odd and c is a multiple of 4.

For example,

If a,b and c is 3,4 and 5. Then the product abc is 60. Not a multiple of 8. But here 4*5 = 20 is a multiple of 4.

If a, b and c is 2,3 and 4. Then the product abc is 24. Multiple of 8. Here 3*4= 12 also a multiple of 4.

So it is insufficient.

So the answer is A.

I alone sufficient.
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Re: If a, b, and c are consecutive integers and 0 < a < b < c, is the prod  [#permalink]

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New post 25 Aug 2018, 20:08
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carcass wrote:
If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ?

(1) The product ac is even.

(2) The product bc is a multiple of 4.


a, b, and c are consecutive integers and 0 < a < b < c --> b=a+1; c=a+2
Question: Is \(a(a+1)(a+2)\) a multiple of 8?

(1) The product \(ac\) is even. -> \(a(a+2)\) is even. Since \(a\) and \(a+2\) are either both odd or both even, we can conclude that \(a\) and \(a+2\) are both even in this case. The product of two consecutive even numbers is a multiple of 8, so \(a(a+2)\) is a multiple of 8. --> \(a(a+1)(a+2)\) is a multiple of 8. --> Sufficient.

(2) The product \(bc\) is a multiple of 4 -> (a+1)(a+2) is multiple of 4.
    If a=2 then a(a+1)(a+2)=2*3*4 -> YES
    If a=3 then a(a+1)(a+2)=3*4*5 -> NO
-> Statement 2 alone is not sufficient.

Answer A
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