Is a*b*c divisible by 32?

Hi..

you are correct with your method and answer but two errors..
[1) a*b*c needs to be\(\geq{32]\)..
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

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

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

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.

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