Is a*b*c divisible by 32?

Question

(1) a, b and c are consecutive even integers.

(2) a*c < 0

Bunuel · Accepted Answer

If  a ,  b  and  c  are integers, is  a*b*c  divisible by 32? 
 
First of all, notice that  32=2^5 .
 
(1)  a ,  b , and  c  are consecutive even integers.
 
In this case,  a*b*c=(2k-2)*(2k)*(2k+2)=2^3*(k-1)*k*(k+1) , for some integer  k . Now,  (k-1)k(k+1)  is the product of three consecutive integers, so it must be divisible by 6 ( the product of  n  consecutive integers is ALWAYS divisible by  n! ; therefore, the product of three consecutive integers must be divisible by  3!=6  ). Thus,  a*b*c  MUST be divisible by  2^4  ( a*b*c=2^3*(k-1)*k*(k+1)=2^3*6*something=2^4*3*something ) but not necessarily by  2^5 . Hence, this statement is not sufficient.
 
For example:
 
If  a = 2 ,  b = 4 , and  c = 6 , then  abc = 2^4*3  and in this case,  a*b*c  is NOT divisible by 32.
 
If  a = 4 ,  b = 6 , and  c = 8 , then  abc = 2^6*3  and in this case,  a*b*c  IS divisible by 32.
 
(2)  ac \lt 0 . 
 
This merely tells us that  a  and  c  have different signs, which is clearly insufficient to answer the question.
 
(1)+(2) From (2), we deduced that either  a  or  c  is negative. And since (1) states that  a ,  b , and  c  are consecutive even integers, then these three integers must be -2, 0, and 2 (this is the only set of THREE consecutive EVEN integers that includes both positive and negative numbers). Thus,  abc = 0 , which is divisible by 32. Sufficient.

Answer: C

gmatiscoming · Answer

Statement I: insufficient but helpful

Statement two says that a*c <0 so one of the numbers must be negative, all could be negative as well:

-8 * -6 * -4 = -192 which is divisible by 32 as per GMAT definition y=xq+r when q and r are unique integers and 0<=r<x ( -192=32*-6 + 0)

-6*-4*-2 =-48 which is not divisible by 32.

insufficient.

Taken together we have that they are consecutive, and that one is negative and the other is not. Because of the restriction of being consecutive and even we have only one possible set of numbers: -2,0,2.

abc =0. 0 is divisable by any integer except zero. so the answer was C.

tricky problem and I would have no doubt gotten this wrong on the test if i didn't have 10 minutes to sit and think about it

satsurfs · Answer

a*b*c needs to be \geq{32]  and divisble by 32.

1) States that a, b, c are consecutive even integer - it can be -2,2,4 or 2,4,6 (not divisible by 32) or 4,6,8 (divisible by 32) NOT sufficient 
2) A*C<0 that means A is negative and C is positive or vice versa NOT sufficient

Combine both 
A,C, B = -2,2,4 ( A*B*C is not divisible by 32) SUFFICIENT! C answer

chetan2u · Answer

Hi..

you are correct with your method and answer but two errors..
 \geq{32] ..
 Not necessary as 0 can also be one value of a*b*c 
2) a,b,c are consecutive even integers
 so -2,2,4 is wrong.. it will be -2,0,2

and hence product a*b*c= -2*0*2=0
and o is div by all numbers because 0*any integer = 0

Ramyanag · Answer

The question is to find whether a*b*c is divisible by 32.

Statement 1: a, b and c are consecutive even integers.
We know 32=2^5. That is, to be divisible by 32 the numerator has to contain at least 2^5.

Case 1: Let's take values for a,b and c as 4,6 and 8 respectively.
4= 2^2
6= 2^1 *3
8=2^3
When we add all the powers of 2 and3 we get: (2^6 * 3^2)/2^5
This is divisible by 32. --> YES

Case 2: Let's take values for a,b and c as 2,4 and 6 respectively.
2=2^1
4=2^2
6=2^1 * 3^1
Which gives, (2^4 * 3^1)/2^5
This is not divisible by 32. --> NO
As we get both YES and NO from statement 1, this is not sufficient.

Statement 2: a*c < 0
We are just given a*c is negative and nothing else is given about these numbers.
Case 1: a,b and c can be -8,2 and 10 => (-8*2*10) is divisible by 32 --> YES
Case 2: a,b and c can be -3,1,5 => (-3*1*5) is not divisible by 32 --> NO
Thus statement 2 is insufficient.

Combining both, we have a,b and c are consecutive even integers and a*c<0.
This means, a<0, b is equal to 0 (0 is even and these are consecutive even numbers) and c>0. We don't have to pick numbers and check as anything multiplied by 0 is equal to 0. Also, 0 is a multiple of every number. Thus, 0 is a multiple of 32. Therefore, a*b*c is divisible by 32.

Thus answer is C.

Check this to know more about this property of 0: https://gmatclub.com/forum/is-0-zero-to ... 04179.html

vv65 · Answer

Check statement 1:

If a, b, c are -2, 0, 2: Yes, a*b*c is divisible by 32
If a, b, c are 2, 4, 6: No, a*b*c is not divisible by 32

Statement 1 is insufficient

Check Statement 2:
If a, b, c are -2, 0, 2: Yes, a*b*c is divisible by 32
If a, b, c are -2, 4, 6: No, a*b*c is not divisible by 32

Statement 2 is insufficient

Check Statements 1 and 2 together:
For a*c<0, one of a and c has to be negative and one has to be positive.
Since a, b, c are consecutive even integers, b=0
That means a*b*c=0
That means a*b*c is divisible by 32

So the two statements together are sufficient.
Answer is C

Posted from my mobile device

shubham1206 · Answer

To find whether a*b*c is divisible by 32?

Consider statement 1: a, b and c are consecutive even integers.

Let the numbers a, b and c be 2k, 2(k+1), and 2(k+2), where k is any integer.

a*b*c = 2k*2(k+1)*2(k+2)

Case 1, k = 0
a*b*c = 0 (Divisible by 32)

Case 2, k = 1
a*b*c = 2*4*6 = 48 (Not divisible by 32)

Hence, statement 1 is not sufficient.

Consider statement 2: a*c < 0,

Now, a and c are numbers with opposite signs. They can be irrational numbers, rational numbers and what not. Product of these numbers
might be divisible by 32, might not be.

Hence, statement 2 is also not sufficient.

Consider statement 1 and 2 together.

a,b and c are consecutive even numbers and a*c <0

=> 2k*2(k+2)<0, where k is any integer
k*(k+2)<0 => -2<k<0.
-1 is the only integer between -2 and 0, therefore putting k= -1 into assumed values of a, b, c

a=-2 , b=0, c=2.

a*b*c = -2 *0 * 2 = 0 (Divisible by 32)

Thus, statement 1 and 2 together are sufficient. Option C is the correct answer.

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Is abc divisible by 32?

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