If x is the product of integers a, b, c, and d, is x divisible by 128?

Statement 1: A= 2^3*8. Therefore, we need another four 2s from the other three integers. Not sufficient.
Statement 2: A, B, C, D are consecutive even integers. Lowest possible option will be 2, 4, 6, and 8 (or their negative counterparts), which has 2^7 in it, or even 0, 2, 4, 6 (0 is even) in which case X is 0 and 0/128=0, so answer to the question is yes. Sufficient.

B.

VERITAS PREP OFFICIAL SOLUTION:

B. While 128 may seem an arbitrary number, look a bit closer and recognize that it's a power of 2. 27=128. If you didn't see this right away, there are clues in the answer choices; 24 from statement 1 is 3x8, and 8 is 23, and in statement 2 you're given several even numbers, so your mind might see those as clues that 2 is an important number here.

Statement 1 is insufficient alone, as without any knowledge of the other three numbers you can only account for the three 2s contained within 24.

Statement 2, however is sufficient, and you might find that by looking at patterns in numbers. Consider a string of consecutive even numbers:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24....

if you're thinking in terms of "2 to a big exponent" like 128 is, you should be conscious of where you get more than one factor of 2. If you list those numbers out in terms of factors of 2, you have:

2, (2 * 2), (2 * 3), (2 * 2 * 2), (2 * 5), (2 * 2 * 3), (2 * 7), (2 * 2 * 2 * 2), (2 * 9), (2 * 2 * 5), (2 * 11), (2 * 2 * 2 * 3)

And you should see that within any stretch of four of those, you get two that contain one 2, but one that contains at least two 2s (the multiples of 4) AND one that contains at least three 2s (the multiples of 8). So at a minimum, any string of four even numbers will give you at least seven factors of 2, meaning that just knowing you have four consecutive even numbers will be sufficient.

And a note - since 0 is divisible by 128 (and every other integer other than 0), even crossing 0 with your consecutive integers will still work, as if one of those evens is 0 the product will still be divisible by 128.

Great Question.
Here is what i did in this one =>

We are told that a,b,c,d are all integers.
We need to see if x=a*b*c*d is divisible by 128 or not.

Statement 1->
Lets use test cases.
Case 1=>
a=24
b=1
c=1
d=1
x=24 => Clearly it is not divisible by 128.
Case 2=>
a=24
b=1
c=1
d=128
x=24*128 => divisible by 128.
Hence not sufficient.

Statement 2->
Consecutive evens =>
a=2n
b=2n+2
c=2n+4
d=2n+6


Taking the Product => x=2n(2n+2)(2n+4)(2n+6)=> 8*n(n+1)(n+2)(n+3)

Property -> Product of n consecutive integers is always divisible by n!

Hence x=8*24k For some integer k=> 128k'
Hence x must be divisible by 128.

Hence sufficient.

Hence B.

@Bunnel, what if we consider 0,2,4,6 for statement 2? Why are we not considering this string of integers? In this case, 128 wont be divisible.

What am I missing here?

In that case the product will be 0 and 0 is divisible by every integer (except 0 itself).

\(128=2^7\)

(1) Insufficient. If x=24*1*1*1, the answer is no. If x=24*4*4*4, the answer is yes

(2) Sufficient. Let a=2n, then b=2n+2, c=2n+4, d=2n+6. So, x=2n(2n+2)(2n+4)(2n+6)=2^4n(n+1)(n+2)(n+3). n, n+1, n+2, n+3 are consecutive 4 integers. Therefore, there are exactly two even numbers. Moreover, among them there is exactly one number that is divisible by 4, that gives us divisibility by extra 2*4=2^3.

The answer is B.

Solution:

Statement One Alone:

a = 24

Knowing a = 24 does not allow us to determine whether x is divisible by 128. For example, if b = c = d = 1, then x = 24 is not divisible by 128. However, if b = c = 1 and d = 128, then x = 24 * 128 is divisible by 128.

Statement one alone is not sufficient.

Statement Two Alone:

a, b, c, and d are consecutive even integers

If a, b, c, and d are consecutive even integers, then one of them must have (at least) 3 factors of 2, another one of them must have (at least) 2 factors of 2, and the remaining two must have (at least) 1 factor of 2. Therefore, if x is a product of a, b, c, and d, then x must have (at least) 3 + 2 + 1 + 1 = 7 factors of 2. Since 128 = 2^7, then x is divisible by 128.

Note that even if one of the even integers is equal to 0, the product a * b * c * d will equal 0, which is divisible by 128, since 0/128 = 0.

Statement two alone is sufficient.

Answer: B

Question Stem:

x = a * b * c * d

Prime factorisation: 128 = \(2^{7}\)

The question boils down to, "Is (a * b * c * d) divisible by \(2^{7}\)?"
We need to answer the question with a definite YES or NO.

If the number x has a minimum of seven 2s(2^7) in it, then x is divisible by 128.
If the number x has less than seven 2s(2^7) in it, then x is NOT divisible by 128.


Evaluate Statement 1 ALONE: a = 24

Prime Factorisation: 24 = \(2^{3}\) * 3

x = 24 * b * c * d = 24 * bcd
x = \(2^{3}\)* 3 * bcd

There is no information available regarding the number of 2s in b, c, and d.
Without knowing them it is not possible to calculate the number of 2s in the number x.

We are not able to answer the question with a definite YES or NO.
Statement 1 alone is NOT sufficient.


Evaluate Statement 2 ALONE: a, b, c, and d are consecutive even integers

Let us approach this question with a counter-example.

Example: a = 2; b = 4; c = 6; d = 8
x = 2 * \(2^{2}\) * (2 * 3) * \(2^{3}\)
x = \(2^{7}\) * 3
The number x has seven 2s\(2^{7}\) in it. (a * b * c * d) is divisble by \(2^{7}\).

Counter Example: a = -18; b = -16; c = -14; d = -12 (The product of four negative integers results in a positive integer)
x = (2 * \(3^{2}\)) * (\(2^{4}\)) * (2 * 7) * (\(2^{2}\) * 3)
x = \(2^{8}\) * \(3^{3}\) * 7
The number x has eight 2s(\(2^{8}\)) in it. i.e., the number has more than seven 2s(\(2^{7}\)) in it. (a * b * c * d) is divisble by (\(2^{7}\)).

We can answer the question with a definite YES.
Statement 2 alone is sufficient.

Note: Even if anyone of the integers is zero, then the value of x will be zero. Therefore, x is divisible by 128.

Choice B is the correct answer.


Key Takeaway:

The product of every four consecutive even integers (excluding zero) will have a \(2^{7}\) in it.

In a set of every four consecutive even integers, there will be:

(i) at least one multiple of 8 - It contains \(2^{3}\)(three 2s) in it.
(ii) at least one multiple of 4 - It contains \(2^{2}\)(two 2s) in it.
(iii) Other two integers will have at least \(2^{1}\) in each of them.

Pick any four consecutive even integers(excluding zero) and go for it!