Power Prep Number Properties Woes :(

[shilotilo]

Hi,
I have a problem with each of the following problems that came up in the Power Prep:

1) If If r lies between 0 and 1 and s lies between 1 and 2, which of the following must be less than 1?
I r/s
II rs
III s-r
My answer is I and II, while the answer given is "only I". How come?

2)Is there a fast way to do this one ( a POE kind of thing?). It took me a while trying to figure out the units digits of which 3 squares can add upto a unit digit of 5, etc, etc ----
The number 75 can be written as the sum of the squares of 3 different integers. What is the sum of these 3 integers?
A) 17, B)16, C)15, D) 14, E)13

And I have no clue how to go about this one ----
3) n is a multiple of 5 and n=(p^2) q. p and q are prime numbers. Which must be a multiple of 25?
A) p^2 B) q^2 C) pq D) (p^2)(q^2) E) (p^3)q


Can someone please help? Thanks.

Shilo

[g_beatthegmat]

1) If If r lies between 0 and 1 and s lies between 1 and 2, which of the following must be less than 1?
I r/s
II rs
III s-r
My answer is I and II, while the answer given is "only I". How come?

Plug in number for this. Question says r is in between 0 and 1 and s is in between 1 and 2. Say r = 0.9 and s is 1.9

I) r/s = 0.9/1.9 CORRECT, less than 1
II) rs = 0.9 * 1.9 > 1 Not less than 1
III) s-r = 1.9 - 0.9 = 1 Not less than 1

Thus, I only.

2)Is there a fast way to do this one ( a POE kind of thing?). It took me a while trying to figure out the units digits of which 3 squares can add upto a unit digit of 5, etc, etc ----
The number 75 can be written as the sum of the squares of 3 different integers. What is the sum of these 3 integers?
A) 17, B)16, C)15, D) 14, E)13

See https://www.beatthegmat.com/gmat-prep-t9319.html

And I have no clue how to go about this one ----
3) n is a multiple of 5 and n=(p^2) q. p and q are prime numbers. Which must be a multiple of 25?
A) p^2 B) q^2 C) pq D) (p^2)(q^2) E) (p^3)q

See https://www.beatthegmat.com/gmat-prep-multiples-and-primes-t10432.html

[shilotilo]

Hi g_beatthegmat,
Thank you for your prompt reply!

However, I still have a question about the first problem. I thought a "must be true" question is the one in which the the answer always has to meet the given criterion.
I understand that 1.9*0.9 is greater than 1. But if I try 1.5*0.5, I will get an answer less than 1. In fact, this "less than 1" thingy seems to appear for a whole lot of values when I multiply a proper fraction (fractions between 0 and 1) with any other number.
Is there a paradox here somewhere or am I missing something?
Help!

Shilo

[shilotilo]

Anybody?????

[AleksandrM]

I found it useful to pick numbers for this one.

0 < r < 1 and 1 < s < 2

I. r/s try 1/2 divided by 3/2 = 2/6, which is <1

try 8/9 divided by 11/10 = 80/99, which is <1

You notice that because the number in the denominator is more than 1, when flipped, it always gives you a number less than 1.

II that's easy: 3/2 x 1/2 = 3/4, but 3/2 x 8/9 = 24/18.

III again 3/2 - 1/2 = 1.

So, you are left with just I.

I'll be honest, took me a little under 2 minutes, whereas it could have probably taken someone else under a minute to solve this using logic.

But, we each do what we can.

[Ian Stewart]

Hi g_beatthegmat,
Thank you for your prompt reply!

However, I still have a question about the first problem. I thought a "must be true" question is the one in which the the answer always has to meet the given criterion.
I understand that 1.9*0.9 is greater than 1. But if I try 1.5*0.5, I will get an answer less than 1.

Shilo

All you're demonstrating above, by choosing r = 0.5 and s = 1.5 is that rs might be less than 1. That doesn't mean it must be less than 1, which is what the question asks. If you can find a single example where 0<r<1, 1<s<2, and rs is greater than 1, certainly it is not true that rs must be less than 1- you've just proven that it sometimes is not. It might be less than 1, but it might not be. I don't think I understand what you're finding paradoxical here?

[shilotilo]

Hi Ian,
Thank you for your reply. Now that you've explained it so clearly, there is no paradox.

Shilo

[ksh]

hi shilotilo,

For your sl.no. 2 & 3 we can get as below:
2. 2^2+5^2+6^2=75
so, sum of nos. would be 2+5+6=13

3. if n is divisible by 5 and n=p^2q then for n to be divisible by 25 it should be p^3q

Hope u get it.

[shilotilo]

Hi krsh,
Thank you for your reply.

I solved Q.2. My query was whether there was a quick way to know which squares we were looking for in the first place.

About Q3, The answer is p^2q^2. The solution is in the link in one of the posts above.

Shilo

[beeparoo]

@ksh: You made an error in your calculations.

2. 2^2+5^2+6^2=75
so, sum of nos. would be 2+5+6=13

2^2+5^2+6^2 does not equal 75.
However, you coincidentally got the right answer (i.e. 13) by accident...

As for Question 3 and why you are also incorrect with that one, someone posted a good link above to the correct solution.

[Thouraya]

75= 5^2 x 3
So Im assuming the sum of the integers is 5+5+3, but I am not sure how these are "different"? theyre only 2 different not 3..