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

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 because both are even and one of a or c will surely be a multiple of 4....ans will be YES
numbers can be 2a, 2a+1, 2a+2....product = 2a*(2a+1)*2(a+1)=4*a*(a+1)*(2a+1)...one of a or a+1 will be even.
B) a and c are ODD and b is even.. if b is multiple of 8.. YES otherwise NO
2a-1,2a,2a+1.....if a is a multiple of 4, 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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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

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

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

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.

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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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.

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

given a, b, c are consecutive integers ;

#1 product ac is even ; so they will be in 2^3 least value eg 2<3<2^2 ; 2^2<5<2*3 ; so always yes


#2 product bc is multiple of 4 ; 2,3,4 ; yes ; 3,4,5 ; no insufficient ;

option A is sufficient

St 1 will include all sets where a and c both are even like
(2,3,4) (4,5,6) and not like (7,8,9)

So it helps

St 2

For st 2 one set can be (4,5,6) this fulfils 1 but breaks st2 hence insufficient

Posted from my mobile device

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

Three consecutive integers can be even * odd * even or odd * even * odd. If a = even, then product abc will be a multiple of 8.

(1) The product ac is even.

This statement tells us even * odd * even. SUFFICIENT.

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

Lets test numbers:
a = 2, b = 3, c = 4; multiple of 8
a = 3, b = 4, c = 5; not multiple of 8

INSUFFICIENT.

Answer is A.
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Similar question: https://gmatclub.com/forum/a-b-c-are-co ... fl=similar

Solution:

We need to determine whether the product abc is a multiple of 8, given that a, b, and c are consecutive integers.
If a is even, then abc is a multiple of 8 since a is a multiple of 4 and c is a multiple of 2, or vice versa.
If a is odd, then b must be a multiple of 8 in order for abc to be a multiple of 8.

Statement One Alone:

Since ac is even, either a or c is even. However, since a, b and c are consecutive integers, then ac is even means both a and c are even. From our stem analysis, we see that ac itself is a multiple of 8 and hence abc is a multiple of 8. Statement one alone is sufficient.

Statement Two Alone:

This is not sufficient. For example, if bc = 12 with c = 4, then abc = 2 x 3 x 4 = 24 is a multiple of 8. However, if bc = 20 with b = 4, then abc = 3 x 4 x 5 = 60 is not a multiple of 8.

Answer: A
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