M31-22

Question

For prime numbers  x  and  y ,  x^3*y^5=z^4 . How many positive factors does positive integer  z  have?

A. One
B. Two
C. Three
D. Four
E. Five

Bunuel · Accepted Answer

Official Solution:

For prime numbers  x  and  y ,  x^3*y^5=z^4 . How many positive factors does positive integer  z  have?

A. One
B. Two
C. Three
D. Four
E. Five

Note that  z^4  is a perfect square and, therefore, must have an odd number of factors. Since  z^4  equals  x^3y^5 , then  x^3*y^5  must also have an odd number of factors.
 
Now, if  x  and  y  are  different  primes, then the number of factors of  x^3y^5  would be  (3+1)(5+1)=24 , which is not odd. Thus,  x  and  y  must be the same prime, and we can write  x^3y^5=x^8=z^4 .
 
In this case  x^3*y^5 = x^3*x^5 = x^8 = z^4 . Taking the fourth root of both sides of  x^8 = z^4  gives  x^2=z , since  x  is prime,  z  has  (2+1)=3  factors. Therefore, we conclude that  z  has three positive factors.

Answer: C

HunterJ · Answer

"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

Bunuel · Answer

1. Any integer in even power is a perfect square.
2.  z^4  is a perfect square. A perfect square has odd number of factors. Since  z^4  equals to  x^3*y^5 , then  x^3*y^5  (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer :

First make prime factorization of an integer  n=a^p*b^q*c^r , where  a ,  b , and  c  are prime factors of  n  and  p ,  q , and  r  are their powers.
 
The number of factors of  n  will be expressed by the formula  (p+1)(q+1)(r+1) .  NOTE:  this will include 1 and n itself.

Example:  Finding the number of all factors of 450:  450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is  (1+1)*(2+1)*(2+1)=2*3*3=18  factors.

BACK TO YOUR QUESTION: 
In case of a perfect square, say for  6^2 = 2^2*3^2 , the powers of primes will be always even, so when counting the number of factors we'd have  (even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd .

Tips about the perfect square :
1. The  number of distinct factors  of a perfect square is ALWAYS ODD.  The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square ;

2. The  sum of distinct factors  of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an  ODD number of Odd-factors , and  EVEN number of Even-factors .  The reverse is also true: if a number has an  ODD number of Odd-factors , and EVEN number of Even-factors then it's a perfect square . For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has  even powers of its prime factors .  The reverse is also true: if a number has even powers of its prime factors then it's a perfect square . For example:  36=2^2*3^2 , powers of prime factors 2 and 3 are even.

Hope it helps.

sveniga4 · Answer

z cannot be a prefect square unless x=y. x and y are prime numbers, we cant get a prefect square with unequal powers of prime numbers. hence, x=y.

Vikram_Katti · Answer

I think this is a  high-quality  question and I agree with explanation.

mbsingh · Answer

SO  Bunuel  just to reiterate last line. Does that mean the factors are 1,x,z ?

Bunuel · Answer

Absolutely. Say if x = 2, then y = 4 and the factors of 4 are 1, 2, and 4.

culsivaji · Answer

I think this is a  high-quality  question and I agree with explanation.

kishor1 · Answer

This is a high quality question.

T1101 · Answer

Another approach would be to pick numbers, which worked for me...

x=3  and  y=3 , the question does not state that the prime number have to be distinct... this makes it easier to work with the same primes.

3^3*3^5=3^8  so we know that  3^8=z^4 , therefore  9^4=z^4  -->  z=9

Factors of  z=9  --> 1,3,9

Akshit03 · Answer

I solved this Q like this:

x^3 * y^5 = (x*y)^3*y^2

Now y^2 has to be equal to x*y to make it a power of 4.

y^2= x*y  leads to x=y
Therefore same prime numbers.

So  z= (prime^2)^2

Therefore number of factors = 2+1.

SHUBHAM GAUTAM · Answer

Bunuel  : IF z is a positive integer, does 'z^4' not have positive factors 1, z^2, z4, x and y?

Bunuel · Answer

x and y must be equal (check the solution above). So, the factors of z^4 are 1, x, x^2 = z, x^3, x^4 = z^2, x^5, x^6 = z^4, x^7, and x^8 = z^4.

kaladin123 · Answer

I think this is a  high-quality  question.

This was my approach as well.

sam12rawat · Answer

I think this is a  high-quality  question and I agree with explanation.

Abhishekgmat87 · Answer

I have a very easy and helpful approach in case if one stuck in such questions.

x and y are prime and it's nowhere given that they are distinct. We can assume both as 2. 
z^4 = 2^2*2^5 = 2^8 = 4^4
z = 4
No. of factors of z = 3 (1,2 and 4)

Option C

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

hbhura · Answer

I like the solution - it’s helpful.

chattyyee · Answer

This is a great question that’s helpful for learning and I like the solution - it’s helpful. I am confused if the question is hard, why has it been labelled as a Sub 505 question? The tag is confusing.

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