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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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Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   17 Jul 2023, 08:00
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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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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   17 Jul 2023, 08:28
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n = (p^3)*(q^6)*(r^7)

for factors which are perfect square, they can be expressed in EVEN powers of p,q,r respectively.

p has 2 possible even powers - 0,2.
q,r have 4 possible even powers each - 0,2,4,6

Therefore - 2*4*4 = 32 possibilities. but note than (p^0)*(q^0)*(r^0) = 1. We need to count squares greater than 1. therefore answer is 31.

PS - other method is brute force, where one can write down the number of possiblities when we take 1 prime number, 2 prime number and all 3 AT A TIME and add up. but the above one is the shorter and smarter approach.
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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   17 Jul 2023, 08:13
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Asked: 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 ?

Let us consider the expression: -

\((1+p^2)(1+q^2+q^4+q^6)(1+r^2+r^4+r^6)\)
All terms except 1 are valid squares of an integer

Number of factors of n which are squares of an integer greater than 1 = 2*4*4 - 1 = 32 - 1 = 31

IMO 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   17 Jul 2023, 08:17
n=p^3∗q^6∗r^7

For n to be perfect square, n needs to have even powers:

Therefore, the possible values of:
p = 0,2 = 1 value
q = 0,2,4,6 = 3 values
r = 0,2,4,6 = 3 values

(Exclude 0 as the integer must be greater than 1)

So, the number of possibility = 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   17 Jul 2023, 08:33
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Since n = p^3 x q^6 x r^7, the factors which are squares of an integer other than n have to be of the form (p^2) x (q^2) x (r^2)
Taking P = p^2, Q = (q^2), R = (r^2), max possible value of N which is a square is = P x Q^3 X R^3

So number of possible factors = (1+1) x (3+1) x (3+1) - 1 (because question mentions squares greater 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   17 Jul 2023, 08:39
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let x be a factor of n which is a perfect square. x = p^(0/2) * q^(0/2/4/6) * r^(0/2/4/6). So there are 2*4*4=32 combinations which result in x being a perfect square. One of these combinations is p^0*q^0^r^0 which is 1. x must be >1 so the number of factors greater than 1 is 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   17 Jul 2023, 08:40
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
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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   17 Jul 2023, 08:40
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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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Powers that p can take = 0, 2

Powers that q can take = 0, 2, 4, 6

Powers that r can take = 0, 2, 4, 6

Total possibilities = 2 * 4 * 4 = 32

However this also includes = p^0 * q^0 * r^ 0 = 1

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   Updated on: 18 Jul 2023, 08:28
For squares, we can have pairs of p, q, or r. We have 2 usable p's, 6 usable q's, and 6 usable r's.

for p -> we have 2 options, either take the pair or not
for q -> we have 4 options, either take 0,1,2,3 pairs
for r -> we have 4 options, either take 0,1,2,3 pairs

total available squares (as the p,q,r are distinct primes) 2 x 4 x 4 = 32

and we don't want the case where nothing is selected so -1

So the ans - > D

Originally posted by JohnSmith07 on 17 Jul 2023, 08:45.
Last edited by JohnSmith07 on 18 Jul 2023, 08:28, edited 1 time in total.
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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   17 Jul 2023, 08:49
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Inorder to make a number square of any other number that it has to has even exponent. This is the underlying concept here.
Since n is \(p^3*q^6*r^7\)
We can have only 0,2 as exponent of p. Similarly 0,2,4,6 as exponent of q and 0,2,4,6 as exponent of r.
We know total combinations selecting A and B = Number of ways to select A * Number of ways to select B
Using this we have total combinations of p,q,r is \(2*4*4=32\).
But it also involve case when exponent of p,q,r will all have 0. But we need squares of number greater than 1. Thus answer would be 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   17 Jul 2023, 08:49
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To have a perfect square we need only even powers:
So,
Possible values of powers of P= 0, 2 (2 values )
Similarly Possible values of powers of q= 0,2,4,6 (4 values)
Similarly Possible values of powers of r= 0,2,4,6 (4 values)

Total number combination= 2*4*4 - 1( since we have to exclude the case when power of p, q and r is zero)= 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   17 Jul 2023, 08:52
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factors of n are square of some integer.
So sqrt(n) = \(p^1*q^3*r^3\) after ignoring odd factors.
total no. of factors of sqrt(n) = (1+1)(3+1)(3+1) = 32.
But in these 32 factors, 1 is also counted.
Hence factors greater than 1 = 32-1 = 31.
IMO D.
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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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink] New post   17 Jul 2023, 08:57
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squares are even powers. thus in
p: 0, 2 => 2 numbers
q: 0, 2, 4, 6 => 4 numbers
r: 0, 2, 4, 6 => 4 numbers

thus 2*4*4 = 32.

Now 32 includes '1' as well. thus 32 - 1 = 31.

Hence ans 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   17 Jul 2023, 08:57
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n=p^3∗q^6∗r^7n and p, q, and r are different prime numbers

To find factors of n which are squares of an integer greater than 1 we need to split out p, q and r raised to maximum possible even powers -

p^2 ... q^2 ... q^4 ... q^6 ... r^2 ... r^4 and r^6. ATTENTION :death: q^6 and r^6 have to be eliminated from the list as q^2 x q^4 and r^2 x r^4 will give me q^6 and r^6 when I will apply the combination formula.

So we have p^2 ... q^2 ... q^4 ... r^2 ... and r^4 in our list (i.e. 5 elements)

Therefore, total number of factors of n which are squares of an integer greater than 1 can be obtained by combination method...

5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 5 + 10 + 10 + 5 + 1 = 31 (Answer 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   17 Jul 2023, 08:57
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The problem stem tells us that - p,q,r are different prime numbers. We know that prime numbers have only two factors, 1 and the number itself. Therefore the factors of n = (2*3) + (2*6) + (2*7) = 32. And the factors of n are there which are squares of an integer greater than 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   17 Jul 2023, 08:59
In p^3*q^6*r^7, perfect square will be from (p*q^3*r^3)^2

possible powers of p --> 0, 1 (count 2)
possible powers of q,r --> 0, 1,2,3 (count 4)

values of n = 2*4*4 =32
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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   17 Jul 2023, 09:04
Answer happens to be 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   17 Jul 2023, 09:04
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Factors of n are there which are squares of an integer greater than 1: cased of p (power 0 or 2)* cases of q (power 0,2,4,6)* cases of r (power 0,2,4,6)
=2*4*4-1 (1 is reduced to account for a case when the resultant multiple is 1)
=31. Hence 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   17 Jul 2023, 09:05
I think B is the answer

(3+1),(6+1),(7+1)

Since the question asks for square - 2+3+4=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   17 Jul 2023, 09:06
The number of factors of n is determined by adding 1 to each of the exponents in the prime factorization of n and then multiplying these results together.
For a factor to be a square, each of the exponents in its prime factorization must be even.
So, for p^3, we have two possibilities: p^0 and p^2.
For q^6, we have four possibilities: q^0, q^2, q^4, and q^6.
And for r^7, we have four possibilities: r^0, r^2, r^4, and r^6.
Multiplying these together, we get 2 * 4 * 4 = 32.
So, the answer is E. 32.
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Re: Around the World in 80 Questions (Day 1): If n = p^3*q^6*r^7, where p [#permalink]
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