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Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p

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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 04:43
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There are 2 ways to select p: p^2, or no p
There are 4 ways to select q: q^2, q^4, q^6, or no q
There are 4 ways to select r: r^2, r^4, r^6, or no r
There are 2*4*4 = 32 ways
There is 1 way to select none of the three so there are 32-1=31 ways

Ans D
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 06:49
Since we are looking for squares, so to do that, we need the even powers of primes.
p can have 0 or 2 as values so total 2 values.
q can have 0 or 2 or 4 or 6 as values so total 4 values.
r can have 0 or 2 or 4 or 6 as values so total 4 values.
Total values = 4*4*2 = 32.
So n can have a total of 32 factors.
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 07:14
Bunuel wrote:
If \(n = p^3*q^6*r^7\), where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

A. 8
B. 9
C. 16
D. 31
E. 32


 


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To get number of factors for a number - add 1 to each exponents of the prime factors of the number and multiply them
If the number is a perfect square, then each of the exponents must be even.

When p^3, the possible exponents for a perfect square factor are 0 and 2.
When q^6, the possible exponents are 0, 2, 4, and 6.
When r^7, the possible exponents are 0, 2, 4, and 6.
So, the total = (2)(4)(4) = 32.

So, the answer is E. 32.

( Mods: My account seems deactivated and hence the old post is null and void, hence reposting please do not report this post.)
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 07:26
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[quote="Bunuel"]If \(n = p^3*q^6*r^7\), where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

A. 8
B. 9
C. 16
D. 31
E. 32

We have \(n = p^3*q^6*r^7\)
Basis this we get \(n = p*r[p^2*q^6*r^6]\)

The squares from the root equation are \([p*q^3*r^63]^2\)
The number of factors for this would be 2*4*4 = 32 but this would include 1 hence it would be 31

Option D
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 07:44
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p can be taken 0 times or 2 times. so 2 cases.
q can be taken 0 times, 2 times, 4 times or 6 times.. total 4 cases.
Similarly r can be taken 0 times, 2 times, 4 times or 6 times. total 4 cases.
So total cases 2*4*4=32. Now one case is p, q and r each taken 0 times. SO we need to subtract that. Hence total possible cases 9Greater than 1)= 32-1=31.
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 07:45
Bunuel wrote:
If \(n = p^3*q^6*r^7\), where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

A. 8
B. 9
C. 16
D. 31
E. 32


 


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p can take values= 0,1,2,3; we need squares therefore we will need one value i.e. 2 from here
q can take values= 0,1,2,3,4,5,6; we need squares we will need three values from here i.e. 2,4,6
Similarily, values from r we need= 3 i.e. 2,4,6

Total squares= 1*3*3= 9
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 07:56
The number of factors of n is (3+1)(6+1)(7+1) = 24.
A perfect square factor of n must have an even exponent for each prime factor of n.
The exponents of p, q, and r in n are 3, 6, and 7, respectively.
The number of perfect square factors of n is (3/2)(6/2)(7/2) = 9.
So the answer is B.
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 08:00
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Bunuel wrote:
If \(n = p^3*q^6*r^7\), where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

A. 8
B. 9
C. 16
D. 31
E. 32


 


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We have to find factors of n which are square of an integer greater than 1.
We can solve this ussing P&C.
for the prime number "P" we have 2 choices i.e (either p^1 or p^0) similarly for prime number "Q" we have 4 choices and For prime number R we have 4 choices

Therefore total cases =2*4*4=32
we need to subtract the case in which the number is 1.
So the final answer is 32-1=31.
Option D
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   18 Jul 2023, 08:02
p can take 2 values 0,2
q can take 4 values 0,2,4,6
r can take 4 values 0,2,4,6

Total number of square factors = 2*4*4 = 32
However, one combination will have 0 for all three, resulting in 1
So, no. of factors = 32-1 = 31

(D)
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   20 Jul 2023, 22:16
Hello Guys

Simply speaking the question wants us to find out the number of squares GREATER THAN 1

therefore we will find out no. of all possible squares excluding 1
No. of all possible squares= P(0,2) Q(0,2,4,6) R(0,2,4,6)
= PQR(2*4*4) = 32

We have to subtract 1 case where in P^0 Q^0 R^0 will give 1

Hence answer will be 32-1 =31
option D
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   24 Jul 2023, 18:27
We should find how many factors of n there are which are squares of an integer greater than 1. This means to reduce the degree of the prime numbers to the square and to find the all positive divisors of the reduced number and then exempt 1.

Extract p^2 (a square of p) from p^3;
q^6 is itself a square of q^3;
Extract r^6 (a square of r^3) from r^7.

Then, k=p*q^3*r^3 is created. (1+1)(3+1)(3+1)=32. 1 is in this. So, 32-1=31. D
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   25 Jul 2023, 09:38
Expert Reply
Green2k1 wrote:
If n=p3∗q6∗r7, where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

A. 8
B. 9
C. 16
D. 31
E. 32

since all p, q and r are different prime number. it can be written as p*p2*q2*3*r*r2*3*
no of factor = 2*4*4 = 32

Correct answer option E



It should be 1 less than that - so 31 (since we need numbers greater than 1)
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   25 Jul 2023, 09:43
Expert Reply
Bunuel wrote:
If \(n = p^3*q^6*r^7\), where p, q, and r are different prime numbers, how many factors of n are there which are squares of an integer greater than 1 ?

A. 8
B. 9
C. 16
D. 31
E. 32


 


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for the Around the World in 80 Questions

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Since we need squares, we need to combine
a) the factors of p^3 that have even powers, i.e. p^0 and p^2
b) the factors of q^6 that have even powers, i.e. q^0, q^2, q^4 and q^6
c) the factors of r^7 that have even powers, i.e. r^0, r^2, r^4 and r^6

Thus, the total number of combinations = 2 x 4 x 4 = 32
However, out of these, 1 of them is the number 1 itself formed by combining p^0 x q^0 x r^0
Thus the number of factors greater than 1 is 32 - 1 = 31

Answer D
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