Lesson: The GCD-LCM Formula

Comment on The GCD-LCM Formula

Hi Brent,

Can you please answer this question?

In a certain deck of cards, each card has a positive integer written on it. In a multiplication game, a child draws a card and multiplies the integer on the card by the next larger integer. If each possible product is between 15 and 200, then the least and greatest integers on the cards could be

(A) 3 and 15

(B) 3 and 20

(C) 4 and 13

(D) 4 and 14

(E) 5 and 14
gmat-admin's picture

Key: The question says "...each possible product is BETWEEN 15 and 200"
So, this means the products do no necessarily EQUAL 15 and 200

Let x be the number on the selected card.
So, x+1 = the next larger integer

Let's first test the LOWER LIMITS
We need (x)(x + 1) ≥ 15
Let's test some answer choices...

(A) 3 and 15
If x = 3, then (x + 1) = 4
Here, we get (x)(x + 1) = (3)(4) = 12
Since it is NOT the case that 12 ≥ 15, we know that the smallest number cannot be 3
ELIMINATE A and B

What about C?

(C) 4 and 13
If x = 4, then (x + 1) = 5
Here, we get (x)(x + 1) = (4)(5) = 20
Since it IS the case that 20 ≥ 15, we know that the smallest number must be 4
ELIMINATE E

Now let's test the UPPER LIMITS
We need (x)(x + 1) ≤ 20
(D) 4 and 14
If x = 14, then (x + 1) = 15
Here, we get (x)(x + 1) = (14)(15) = 210
Since it is NOT the case that 210 ≥ 200, we know that the biggest number cannot be 14
ELIMINATE D

By the process of elimination, the correct answer is C

Cheers,
Brent

Hey Brent,

I have a DS question and I do not understand the thinking behind this one.

If -5x = –x – 2 + 58x, and x does not equal zero, does x = -9?

(1) x does not equal 7
(2) x is less than zero

The answer is E, but I can not find an explanation.

Kind regards,
Niek
gmat-admin's picture

Hi Niek,

Sorry, but the equation in your question was spread out over about 10 lines.
I tried to reconstruct it (above), but it doesn't look right.
If you can tell be the equation, I'm sure I can help.

Cheers,
Brent

Hi Brent,

In your video lessons, you teach how to find LCM and GCF for given numbers, however there are a lot of problems that require to do the opposite, to find pairs of possible numbers for given LCM. Do you have a video explaining how to do that? You mention that we should master it, however, I find it hard to master something without a method. I see lots of people on GMAT forum easily suggest pairs for a given LCM, however I cannot understand how to come up with such pairs fast.
gmat-admin's picture

To find numbers that meet certain criteria (LCM or GCD), we need to keep in mind what each term means.

For example, if N is the GCD of integers J and K, we know that:
- N is a divisor of J and K
- In other words, N is "hiding" in the prime factorization of J and K
- N is the BIGGEST divisor of both J and K

So, for example, let's find two values (J and K) with a GCD of 15
- So 15 is "hiding" in the prime factorization of J and K
- Since 15 = 3 x 5, we know that a 3 and a 5 are "hiding" in the prime factorization of J and K

We can write:
J = (3)(5)(?)(?)(?)
K = (3)(5)(?)(?)(?)

At this point, we can see there are many possibilities....
For example, we could have:
J = (3)(5) = 15
K = (3)(5) = 15
J and K could both be 15

Or we could have:
J = (3)(5)(2) = 30
K = (3)(5) = 15

Or we could have:
J = (3)(5)(2)(7) = 210
K = (3)(5) = 15

Or we could have:
J = (3)(5)(2) = 30
K = (3)(5)(11) = 165

Or we could have:
J = (3)(5) = 15
K = (3)(5)(2)(2)(2) = 120

Or we could have:
J = (3)(5)(5) = 75
K = (3)(5)(2)(2)(2) = 120

In GENERAL, here are some quick and pairs of values that satisfy a given condition:
If N is the GCD of integers J and K, then some easy pairs of values include:
J = N and K = N
J = N and K = some multiple of N

-------------------------------

Now let's examine LCMs

For example, if N is the LCM of integers J and K, we know that:
- J and K are both divisors N
- J and K are both "hiding" in the prime factorization of N
- N is the SMALLEST multiple of both J and K

So, for example, let's find two values (J and K) with an LCM of 120
120 = (2)(2)(2)(3)(5)
So, J and K are both "hiding" in the product (2)(2)(2)(3)(5)

At this point, we can see there are many possibilities....
For example, we could have:
J = (2)(2)(2) = 8
K = (3)(5) = 15

Or we could have:
J = (2)(2)(2)(3)(5) = 120
K = (2)(2)(2)(3)(5) = 120

Or we could have:
J = (2)(2)(5) = 20
K = (2)(2)(2)(3) = 24

etc

In GENERAL, here are some quick and pairs of values that satisfy a given condition:
If N is the LCM of integers J and K, then some easy pairs of values include:
J = N and K = N
J = N and K = some divisor of N
----------------------------

Does that help?

Cheers,
Brent

https://gmatclub.com/forum/x-and-y-are-positive-integers-if-the-greatest-common-divisor-of-x-and-233373.html

For this question, why is it wrong to solve it this way . Given 3x and 9y LCM is 81, x =27 and y=9. I assume this is because we don't know for certain x =27 and y=9?
gmat-admin's picture

Question link: https://gmatclub.com/forum/x-and-y-are-positive-integers-if-the-greatest...

Your approach of choosing values is perfectly fine.
It's true that x = 27 and y = 9 satisfy the condition that the LCM of 3x and 9y is 81.
HOWEVER, the other condition is that the greatest common divisor (GCD) of x and 3y is 9
If x = 27 and y = 9, then the GCD of x and 3y is 27 (not 9)

Cheers,
Brent

Add a comment

Office Hours

Have questions about your preparation or an upcoming test? Need help modifying the Study Plan to meet your unique needs? No problem. Just book a Skype meeting with Brent to discuss these and any other questions you may have. 

Change Playback Speed

You have the option of watching the videos at various speeds (25% faster, 50% faster, etc). To change the playback speed, click the settings icon on the right side of the video status bar.

Have a question about this video?

Post your question in the Comment section below, and I’ll answer it as fast as humanly possible.

Free “Question of the Day” emails!