If x and y are odd positive integers, and x and y both have an odd

Question

If x and y are positive odd integers, and both numbers have an odd number of positive divisors, which of the following could be the value of x-y?

A) 4818
B) 5174
C) 5320
D) 5482
E) 5566

BrentGMATPrepNow · Accepted Answer

Key concept:    Most integers have an EVEN number of positive divisors. 
For example, 12 has 6 divisors, 35 has 4 divisors, and 13 has 2 divisors. 
However, SQUARES of integers always have an ODD number of divisors. 
For example, 25 has 3 divisors, 36 has 9 divisors, and 81 has 5 divisors.

So, if and x and y both have an odd number of positive divisors, then x and y squares of integers. 
So, let’s let x = j² and let y = k² for integers j and k

NOTE: If x is odd, then  j must be odd , and if y is odd, then  k must be odd.

Our goal is to find a possible value of x – y
We can write: x – y = j² - k² 
Factor to get: (j + k)(j – k)
Since j and k are both ODD, we know that (j + k)(j – k) = (odd + odd) (odd - odd) = (even)(even) = ( 2  times some integer)( 2  times some integer) = (  4  )(some integer)

ASIDE: If an integer is even, we can rewrite that number as  2  times some number. For example, we can take 10 and rewrite it as ( 2 )(5)

So, it must be the case that the correct answer must be divisible by   4

Key concept: If integer N is divisible by 4, then the number created by the last 2 digits of N must be divisible by 4.

Check the answer choices.
A) 48 18 . 18 is NOT divisible by 4. Eliminate A. 
B) 51 74 . 74 is NOT divisible by 4. Eliminate B.
C) 53 20 . 20 IS divisible by 4. Keep.
D) 54 82 . 82 is NOT divisible by 4. Eliminate D.
E) 55 66 . 66 is NOT divisible by 4. Eliminate E.

Answer: C

Cheers,
Brent

nick1816 · Answer

x and y are both odd and both have an odd number of integers; hence, x and y are both perfect perfect square of odd integers.

x=(2a+1)^2 , where a is an integer
 y=(2b+1)^2 , where b is an integer

x-y= 4a^2+4a+1-4b^2-4b-1  = 4(a^2+a-b^2-b) 
hence, value of x-y must be a multiple of 4

Only option C is a multiple of 4

IMO C

Kinshook · Answer

Given: 
1. x and y are positive odd integers
2. Both numbers have an odd number of positive divisors.

Asked: Which of the following could be the value of x-y?

x, y are positive odd integers of the form a^2, b^2 respectively

x - y = a^2 - b^2 = (a+b)(a-b) 
Since a & b are odd ; a+b & a-b are even; (a+b)(a-b) is a multiple of 4

70^2 = 4900 
71^2 = 5041
72^2 = 5184
73^2 = 5329
74^2 = 5476
75^2 = 5625

A) 4818; NOT a multiple of 4
B) 5174; NOT a multiple of 4
C) 5320; MULTIPLE of 4; 5320 = 5329 - 9 = 73^2 - 3^2 = 76*70 
D) 5482; NOT a multiple of 4
E) 5566; NOT a multiple of 4

IMO C

Archit3110 · Answer

from the given info we know that both x& y are odd integers and are perfect squares as well.
we can test for few values of given condition
3^2-1^2=8
5^2-3^2= 16
5^2-1^2= 24
13^2-11^2 = 48
21^2-11^2= 320
we observe that all the integers are divisible by 8 
seeing the answer options 
A) 4818 ; has only 1 factor of 2
B) 5174; has only 1 factor of 2
C) 5320; has 3 factors of 2 hence divisible by 8 sufficient
D) 5482; has only 1 factor of 2
E) 5566' has only 1 factor of 2

IMO C

Shishou · Answer

This is quite simple,really.When you subtract two odd perfect squares ,you get an even number. To be more specific let's take a look at this :
25-9=(5+3)(5-3)=8 × 2.
Another: 81-25=(9-5)(9+5)=4×14.
What you will realise is that 8×2 and 4×14 must be divisible by at least 4 or 8 since the difference of two squares of two odd numbers is the product of 2 even integers.

Only C is divisible by 4 because its last 2 digits form a number divisible by 4. ANSWER IS C

cpn · Answer

The clue lies in that the odd integers having odd factors...take 3 for example...it has 2 factors...but 3^2=9 has 3 factors 1,3&9...true for all odd squares...so x and y are perfect squares and have to be odd. Now using the formula a^2-b^2=(a+b)(a-b) and the property odd plus odd is even and odd minus odd is even, we get x-y as the product of 2 even numbers.. so a multiple of 4. ...C is the only number who's last 2 digits are divisibke by 4. Hence C

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ScottTargetTestPrep · Answer

Solution:

Since both x and y have an odd number of positive divisors, they are both perfect squares. Since the answer choices are in the high 4000s to 5000s, we see that x must be the square of a number in at least the 70s. Since x and y are both odd, let’s write down the perfect squares of odd numbers beginning from 71^2:

71^2 = 5041, 73^2 = 5329, ...

At this point we see that 5329 is 9 more than 5320 (choice C); therefore, 5320 can be written as x - y where x = 5329 = 73^2 and y = 9 = 3^2.

Answer: C

Sabby · Answer

Hi,

As X and Y have odd number of factors, they are perfect squares
As X and Y could be any number, there must be a pattern; we can try out with small numbers and extrapolate the result to one of the answer choices

Case 1: x= 25 and y=9 
x-y = 25-9=16

Case 2: x=169 (or 13^2) and y= 121 (or 11^2)
x-y= 169-121=48

As we can see, the pattern can't be about unit digits because both are different here. 
So what could be the pattern? well, 48 is divisible by 4 and so is 16.

Rule : For a number to be divisible by 4, the two last digits must be divisible by 4

Let's check out our answer choices by scanning the two last digits : only one is divisible by 4 and it is 5320.

Answer C)

Ekland · Answer

For me, the highlighted part is actually the hard part of this question.
I first thought that it meant "both" and not "each", which is actually what it means.

The quant language says one thing and means another.
Bcos if you go by both(i.e. the sum of the number of divisors is odd; thus one is odd, the other even and set up  (2a+1)-(2b+1)^2 ) you will only arrive at the conclusion that whatever is the answer is a multiple of 2, which wont help at all. So I went back and interpreted the stem in a grammatically incorrect way which is the correct way and correctly set up the equation  (2a+1)^2-(2b+1)^2 . The wording of the quant questions is actually the hard part not the math.

Dear experts, I have used taking notes and writing "quant grammars" on flashcards. they seem infinite, I'm store them in a big carton, sort them daily. I hope it helps at the end of the day.

I also thought about using 1(being an odd perfect square and having odd number of divisor)  1^2 - 1^2 = 0  and chosen C... but realized It was just luck and very dangerous.

Sabby · Answer

Hi  Ekland ,

I guess you got confused with "both" and "together". 
you can simplify as following "x and y both have an odd number of integers"
 
Sabby and Ekland are Canadian, and both candidates have 2 siblings. 
I have 2 siblings and you have 2 siblings as well. We both have 2 siblings.

x has an odd number of divisors and y has an odd number of divisors as well.
Both x and y have an odd number of divisors

Does this analogy helps?

Foreheadson · Answer

Very hard question. divisibility by 4 is only one characteristic when there is difference of perfect squares. I don't undestand how one must magically come up with this idea on test day in two minutes.

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Posted from my mobile device

Kinshook · Answer

Asked: If x and y are positive odd integers, and both numbers have an odd number of positive divisors, which of the following could be the value of x-y?

Since x and y are odd and have odd number of positive divisors, each is an odd square.

Let x = a^2 and y=b^2
x - y = a^2 - b^2 = (a-b)(a+b)

Since both a and b are odd; a-b & a+b are both even; x - y is divisible by 4.

A) 4818: Not divisible by 4: Incorrect
B) 5174: Not divisible by 4: Incorrect
C) 5320: Divisible by 4: Correct
D) 5482: Not divisible by 4: Incorrect
E) 5566: Not divisible by 4: Incorrect

IMO C

AndiTT · Answer

Same here, how on earth am I supposed to know all of that? That is by far the hardest question I´ve seen...

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