If m and p are positive integers and m^2 + p^2 < 100, what is the

My solution
m^2 + p^2 = (m - p)^2+2mp<100, so if we assume that (m - p)^2 = 0 in order to maximize the value of 2mp, then 2mp < 100, mp < 50
answer 49

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

Does this mean that in GMAT questions we can always assume x may be equal to y, unless specifically mentioned that they are 'distinct numbers'?

Unless it is explicitly stated otherwise, different variables CAN represent the same number.

AM >= GM
\((m^2+p^2 )/ 2 >= \sqrt{m^2*p^2}\)
\(mp =< (m^2+p^2 )/ 2 <100/2\)
mp < 50

Greatest possible value = 49

Answer D.

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

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:
Cheers,
Brent

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

Please give Kudos if you like this explanation

Answer: Option D

Video solution by GMATinsight

Video solution from Quant Reasoning:
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­Utilising the propert of positive integers.
Between two numbers a & b, the Arithmetic Mean (AM) is always greater than or equal to the Geometric Mean (GM).

So, 
\(\frac{[m](X+5)}{(X+4)\)}[/m]

Similarly
 ­

­The question can also be solved with the following method:

\( AM \geq{GM}\)
\(\frac{(m+p)}{2} \geq{\sqrt{mp}}\)

Squaring both sides,
\( (m+p)^2 \geq{4mp}\)
\(m^2+p^2+2mp \geq{4mp} \)
\(m^2+p^2 \geq{2mp} \)
\(2mp \leq{m^2+p^2} \)

As per the question,
\(2mp \leq{m^2+p^2} < 100 \)
\(2mp < 100 \)
\(mp < 50 \)
Hence, mp can be max 49

Correct Option: D­

For whatever reason, I solve this problem thinking that the two variables had to be unique and this is why I picked a. Good lesson this is high-quality question.

Posted from my mobile device

any reason as to why we are assuming (m-p)^2 = 0?

Given that (m - p)^2 + 2mp < 100, our goal is to maximize mp. To do that, we need to minimize (m - p)^2. Since it’s the square of a number, its smallest possible value is 0, which occurs when m = p.

Lets say m^2 + p^2 = 100, then m^2 - 100 + p^2 = 0

and we know (m-p)^2 = 0 => m^2 - 2mp + p^2 = 0, hence 2mp = 100 --> mp = 50, mp has to be less than 50.