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## 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

## Key: The question says "..

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

## Hi Niek,

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.

## To find numbers that meet

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

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?

## Question link: https:/

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

## I think Brent is the best

## Ha! Thanks Dan!

Ha! Thanks Dan!

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