Lesson: The GCD-LCM Formula

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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
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

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