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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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### HideShow timer Statistics 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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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
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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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

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

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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.

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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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.

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