Hi all –
I don’t understand the solution to this question. Couldn’t the answer technically be anything? For example:
24 x 32 x 40 = 30, 720.
This product is a multiple of 8. You could go on like this with even bigger multiples of 8. So why is 3072, specifically, the answer?
The solution cites the reason to be because “the answer can be written in terms of 8 x 16 x 24”, but I’m not really sure what that means.
This is an interesting question that I prefer to solve using algebra. Let’s start with the phrase “consecutive multiples” of 8, which can be represented as 8y, 8y + 8, and 8y + 16. This ensures that any value substituted for y yields consecutive multiples of 8.
Now, we need to find the largest number that the product (8y)(8y + 8)(8y + 16) is a multiple of. We can express this as:
Here assume m to be the divisor that we have to find.
(8y)(8(y + 1))(8(y + 2)) / m
Combining the 8s gives us (512y(y + 1)(y + 2))/m. Since we need at least three 8s for it to be an integer, m must contain at least three 8s. Now next thing is, finding the number which is always a divisor of y(y + 1)(y + 2)
We know that, the product of three consecutive integers y, y + 1, and y + 2 will always be a multiple of both 2 and 3. Thus, it is also a multiple of 6. The product 512 * 6 equals 3072. We multiplied by 6 and not 3 or 2 because we are looking for the largest number.Therefore, m must have at least 512 and 6 as factors.
While higher numbers may yield larger multiples, to satisfy the requirement across all values of y, 3072 is the largest guaranteed divisor. We cannot assume that consecutive integers will always be divisible by numbers beyond 6. For instance, while pairs of consecutive integers with one odd in the middle will always include two even integers and thus be a multiple of 8, 512 * 8 (4096) but this does not ensure that it the largest number divisor or for that matter divisor at first place for all values of y for eg what if we get 3 consecutive multiples of 8 where the middle one isn’t odd? Hence, we conclude that in “must” scenario, 3072 must be our largest divisor. If we go beyond this, we can’t ensure it works for every value of y.
Does this make sense?
Hello! Thank you for taking the time to reply. I hadn’t considered taking an algebraic approach as I haven’t reached that portion of my studies yet, but I do have previous knowledge, so this made some sense. I suppose I was just unsure of how to tackle this problem using the knowledge from the videos. Just for some clarification:
3072 is the only number that satisfies the conditions of (1) being divisible by 512 and (2) being divisible by 6?
And furthermore, would you suggest tackling problems of this same nature in the exact same way? I mean, using algebra and doing all of the steps you did.
In a “must” scenario, it is the only guaranteed largest divisor. If you have just started the plan, you will learn more about “must” scenarios in the strategy videos section on Prepswift. Greg has explained it beautifully, so there’s no need to worry.
There isn’t a fixed template to follow. Choose an approach that you are comfortable with and that suits the question best. Be open to using different methods to solve problems. You’ll gain a better understanding when you watch the strategy videos. Happy learning!
I referred to this video for explaining the consecutive integer thingy - Three Consecutive Integers I - GregMat
“Must” scenarios- MUST problems
Why can’t the answer be whatever the product is :-
product = (8y)(8y + 8)(8y + 16).
answer = (8y)(8y + 8)(8y + 16) ?
because (8y)(8y + 8)(8y + 16) is always an multiple of (8y)(8y + 8)(8y + 16).
And the highest value of this will go to infinity.
Or is it that in GRE multiple cannot be the number itself?
Let z = (8y)(8y + 8)(8y+16) = 8^3 \underbrace{(y)(y+1)(y+2)}_{6n}, where n \in \mathbb{Z}^+. Thus, we know that z is always a multiple of 8^3 \cdot 6 = 3072.
You know there exists integers such that their multiples could always coincide with the multiples of 3072, but those integers can’t be greater than 3072 itself. For example, all multiples of 3072 are always multiples of 8, by default. However, if u pick a number like 6144 then all multiples of 3072 are not multiples of 6144 even though “half of them” work.
Dk what the logic behind this is, but what i’ve written above specifically addresses why this isn’t the case.
Thanks for your reply! I must have watched the Three Consecutive Integers videos three times over and still did not understand how to solve this question. I understand your explanation well enough, so I thank you for taking the time out to reply.
Good to know you understood it!
thanks
