(n)(n+1) = 15 to 200

For Least put N as least Positive Integer to make multiplication greater than 15

(4)(4+1) ==> So, "4"

Now, we see that C & D are the possible cases.

Let's try the greater one, to be sure for elimination

(14)(14+1) = 14^2 + 14 = 200+

So, [spoiler]{C}[/spoiler]

VGSoni89 wrote:In a certain deck of cards, each card has a positive integer. In a multiplication game, a person draws a card and multiplies that integer by next larger integer.

If the product is between 15 and 200, the least and greatest integer on the cards could be?

A) 3 & 15
B) 3 & 20
C) 4 & 13
D) 4 &14
E ) 5 & 14


I picked B and it was C.

I tried to wrap my around what the reasoning behind the solution could be and I still can not understand why that is the answer.

If someone could help me that would be awesome!

This particular Problem can be solved very quickly -

As we know product here means nothing but N *( N + 1 )

So try exploring the options -

a. 3 & 15

Just try with the smaller value , 3 in this case

3 * ( 3 + 1 ) = 12

This is not possible since product must be greater than = 15

b. Same as a

The minimum criteria is well fulfilled by the options C , D and E

But explore the maximum criteria now -

c. 4 & 13

With 13 given in hand you can square it ( Just to approximate a bit to save time ) = 169

Possible condender

d. 4 &14

14^2 = 196 , its remotely possible that product of 14 *15 will be less than 200 , so it's out..

e. Same as d



Now just go a bit further -

C. Find if the maximum criteria stated holds true or not , 13*14 = 182

Find if the minimum criteria is fulfilled or not , 4*5 = 20


Hence C is definitely what we are looking for..

Testing the answers given would be the fastest option for this question, but here's a way of solving it for ANY values, not just integers, positive, negative or nought.

Here are the parameters:
15 < L * (L + 1) where L is the lowest card value.
and H * (H + 1) < 200 where H is the highest card value.

These 2 quadratic inequalities can be solved using the quadratic formula thus:

L^2 + L - 15 > 0.
L = [-1 +/- SQRT(1 - -60)]/2 = 3.4, 4.4

and

H^2 + H - 200 < 200
H = [-1 +/- SQRT(1 - - 800)]/2 = 13.7, 14.7


As 3 would be lower than the minimum of 3.4, it is too low. 4 is the next integer.
As 14 would have to be the higher value (of either the card, or card + 1), then the highest card is 13.

So the answer is 4 and 13.