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Re: If m and p are positive integers and m^2 + p^2 < 100, what is the [#permalink]

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17 Jun 2017, 10:49

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Given m and p are postive integers and we need to find out the greatest value of mp?

Since \(m^2 + p^2\) < 100, evaluating the answer options.. 36 = 6*6, 42 = 6*7, 48 = 6*8, 49 = 7*7 and 51 = 3*17

Of the available only 51(3*17) has a value of \(m^2 + p^2\) greater than 100 Since we have to find out the greatest value of mp, 49, where m=p=7 (Option D) will have value of \(m^2 + p^2\) =\(7^2 + 7^2\) = 98 which is less than 100
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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

The product of two positive integers is greatest if they are as close as possible, that is, 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 satisfy 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, it is 49.

Answer: D
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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

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

Re: If m and p are positive integers and m^2 + p^2 < 100, what is the [#permalink]

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05 Sep 2017, 04:38

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AbdurRakib 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

I learned one of these concepts from Bunnel

(\((M-P)^2>= 0\) Therefore M^2 +P^2 >= 2MP . Greatest Value of 2MP = M^2+P^2 . Great Value of M^2+P^2 = 99 or 98 There MP = 99/2 = Integer (not possible - Integer X Integer = Integer) or 98/2 =49

Re: If m and p are positive integers and m^2 + p^2 < 100, what is the [#permalink]

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05 Nov 2017, 22:54

I think the answer should be 42. The question says m and p are integers. 'Are integers' should mean two distinct integers, not a single integer. So, we cant take 7 as the value of both m and p. Rather we can take 6 and 7 where the sum of their square is 85 that is lower than 100. And 7 times 6 equals to 42. So, the answer is 42.

I think the answer should be 42. The question says m and p are integers. 'Are integers' should mean two distinct integers, not a single integer. So, we cant take 7 as the value of both m and p. Rather we can take 6 and 7 where the sum of their square is 85 that is lower than 100. And 7 times 6 equals to 42. So, the answer is 42.

You are wrong. Unless it is explicitly stated otherwise, different variables CAN represent the same number. Please re-read the discussion above.
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