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OG IF m and p are positive integers

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by Brent@GMATPrepNow » Tue Aug 22, 2017 7:17 am
AbeNeedsAnswers wrote:If m and p are positive integers and m^2 + p^2 < 100, what is the greatest possible value of mp ?

A. 36
B. 42
C. 48
D. 49
E. 51
D
Let's test some pairs of values.
Since m and p are POSITIVE INTEGERS, we won't have a ton of options

Try m = 9 and p = 4 (aside: if m = 9, then 4 is the biggest possible value of p)
In this case, mp = (9)(4) = 36

Try m = 8 and p = 5 (aside: if m = 8, then 5 is the biggest possible value of p)
In this case, mp = (8)(5) = 40

Try m = 7 and p = 7 (aside: if m = 7, then 7 is the biggest possible value of p)
In this case, mp = (7)(7) = 49

Try m = 6 and p = 7 (aside: if m = 6, then 7 is the biggest possible value of p)
In this case, mp = (6)(7) = 42

Try m = 5 and p = 8
At this point, we can see that, if we continue, we'll be duplicating the work we did earlier.
That is, this case (m = 5 and p = 8) is the SAME as the 2nd case we examined.
If we continue, the next case we test will be m = 4 and p = 9. which is the SAME as the 1st case we examined, etc.

Since we've now tested all possible (and relevant) cases, we can see that the maximum value of mp is 49

Answer: D

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Brent
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by Scott@TargetTestPrep » Fri Aug 25, 2017 9:53 am
AbeNeedsAnswers wrote:If m and p are positive integers and m^2 + p^2 < 100, what is the greatest possible value of mp ?

A. 36
B. 42
C. 48
D. 49
E. 51

D
Among all the pairs of integers that satisfy m^2 + p^2 < 100, the product mp is greatest if m and p are as close to each other as possible, and optimally, if they are equal. Thus, we can let p = m and our inequality becomes m^2 + m^2 < 100. Let's solve it:

2m^2 < 100

m^2 < 50

m < √50

Since m is a positive integer and the largest positive integer less than √50 is 7, m = 7. In that case, p is also 7. Thus, the greatest possible value of mp is 7 x 7 = 49.

Alternate Solution:

Let's test each answer choice, starting from the greatest, which is 51.

Notice that 51 = 3 x 17, so our only choices for m and p are 3 and 17 or 1 and 51. Neither of these choices satisfies m^2 + p^2 < 100, and therefore mp cannot equal 51.

Next, let's test 49. Since the choice m = 49 and p = 1 does not satisfy m^2 + p^2 < 100, let's take a look at m = 7 and p = 7. Since m^2 + p^2 = 49 + 49 = 98 < 100, mp can equal 49. Since we are looking for the greatest possible value of mp, 49 is correct.

Answer: D

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by [email protected] » Sat Jun 16, 2018 3:32 pm
Hi All,

We're told that If M and P are POSITIVE INTEGERS and M^2 + P^2 < 100. We're asked for the GREATEST possible value of (M)(P). This question can be solved in a couple of different ways, including by TESTing VALUES. Since the question asks for the greatest possible PRODUCT of M and P, we can use the Answer choices 'against' the prompt and TEST THE ANSWERS.

To start, you should recognize that two of the Answers are 'perfect squares': 36 and 49.

IF... M =6 and P = 6.... 6^2 + 6^2 = 72 which IS less than 100. Is it possible that one of the other answers would also 'fit' the given information AND lead to a bigger product? We have to check.

IF... M =7 and P = 7.... 7^2 + 7^2 = 98 which IS less than 100 and is really, really close to it.

With a product of 51, we would either have (1)(51) or (3)(17), but neither of those options fits the 'restriction' that M^2 + P^2 < 100, so Answer E can't be correct.

Final Answer: D

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Rich
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