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Child's play (GMAT Prep)

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Child's play (GMAT Prep)

by alex.gellatly » Sat Jun 30, 2012 3:16 am
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

3 and 15
3 and 20
4 and 13
4 and 14
5 and 14
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by Anurag@Gurome » Sat Jun 30, 2012 3:37 am
alex.gellatly wrote: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
Tricky Approach:
Look at the options.
If 3 cannot be the least integer as 3*4 = 12 < 15
But 4 can as 4*5 > 15

Hence, the correct answer is either C or D.
Now, 14 cannot be the greatest integer as 14*15 = 210 > 200
But 13 can as 13*14 = 182 < 200

The correct answer is C.
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by Anurag@Gurome » Sat Jun 30, 2012 3:50 am
Algebraic Approach:
Just note that the product is a product of two consecutive integers.
Hence, it is of the form n(n + 1) = (n² + n)
Therefore, 15 ≤ (n² + n) ≤ 200

Now, least integer which is square of an integer and greater than 15 is 16 = 4²
Hence, least integer = 4

And, greatest integer which is square of an integer and less than 200 is 196 = 14²
But, (14² + 14) > 200
Hence, greatest integer = (14 - 1) = 13

The correct answer is C.
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by sandeep_thaparianz » Sat Jun 30, 2012 5:30 am
+1 for c
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by alex.gellatly » Sun Jul 01, 2012 5:56 pm
sandeep_thaparianz wrote:+1 for c
Instead of always just saying +1 for the correct answer that the expert already explained, maybe it would be useful to explain how you solved the question.
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