If the positive integer x has 8 positive factors, what is the value of

The question states that positive integer X has 8 positive factors. Thus x>0, and all factors are greater than 0. These factors can be either 1*2^8 or 1*2^3 * 3^4 or anything else. We do not know at this stage. We need to find the value of X. Let us analyze the statements we have:

Statement 1:
The product of any two positive factors of X is even.

This statement means that if we multiply any of the 8 factors of X by another factor, we will get an even number. As an even number is the product of two even numbers, or Even*Odd number, it means that it is impossible that two odd numbers can be multiplied by each other. Thus, there is only 1 odd factor of the number X. And other factors are even. Speaking about even factor, all even numbers apart from 2 can be divided further, e.g. 4=2*2 or 12=2*2*3.
Thus, X = 2^7 * odd factor. If there is only one odd factor, it is 1. As a result, X=1*2^7.

Statement 1 is sufficient.

Statement 2:
X has 7 even positive factors.

We were able to identify this information in the analysis of previous statement, thus the information here is also sufficient to identify the number.

Statement 2 is sufficient.

There is no need to combine both statements together, as each one separately was enough to find X.

Answer: D

P.S. Here, again, instead of looking for all possible factors, I neglected 1 initially and for this reason, answered the question wrong. However, the question was not telling anything about prime factors and for this reason, 1 should have been also counted as a factor.

(1) The product of ANY two positive factors of x is even
--> All factors are even except 1
--> x has to be 2^7

Sufficient

(2) x has 7 even positive factors
--> 7 even factors & factor '1'
--> x has to be 2^7

Sufficient

IMO Option D

Pls Hit Kudos if you like the solution

Let x=a^p, where 'a' is prime. hence 'x' has (p+1) nos of factors.
Here , p+1=8 or p=7

question stem:- x=?

st1:- The product of ANY two positive factors of x is even

factor1*factor2=EVEN implies that anyone of the factor must be even.
Hence, 'a' must be the smallest positive even Prime.
So, x=2^7.
Sufficient.

St2:- x has 7 even positive factors
Implies that there are 7 +ve even factors & '1' as the 8th factor.
So 'a' has to be smallest even prime with p=7.
Hence x=2^7
Sufficient

Ans. (D)

If the positive integer x has 8 positive factors, what is the value of x?
Given:
x >= 1 & integer
x has 8 factor, 8 = 1*8=2*4
case-1: x= p^7 or case-2: x = q^1*r^3, where p, q & r are prime factors of x

(1) The product of ANY two positive factors of x is even --> correct:
case-1: any two positive factors will be even, so x = 2^7, then x's factors are, 1,128,2,64, 4,32, 8,16. so product of any two positive factors is even
case-2: x = 3*2^3 = 24, then x's factors are, 1,24, 2,12, 3,8, 4,6, but 1*3 = odd, so x has to be only 2^7
(2) x has 7 even positive factors --> correct
case-1: x = 2^7 =128, then x's factors are, 1,128,2,64, 4,32, 8,16, so 7 are even positive factors & 1 odd positive factors. it's true ,
case-2, x = 3*2^3=24, then x's factors are, 1,24, 2,12, 3,8, 4,6, then x has 6 even positive & 2 odd positive factor. so x has to be only 2^7

Answer: D

Two scenarios that x has 8 positive factors are possible here:
--> Scenario S1: x has 2 or 3 different prime factors (n,m,k) such that x=n*m^3 or x=n*m*k
--> Scenario S2: x has only one prime factor (n) such that x=n^7

(1) The product of ANY two positive factors of x is even
--> This can only be true IF all seven factors of x are even, beside odd factor of 1
--> This rule does NOT apply to Scenario S1 (x=n*m^3 or x=n*m*k)
e.g. x = 24 = 3*2^3 has factors of 1,2,3,4,6,8,12,24 --> the product of 1 x 3 = 3 is odd.
e.g. x = 30 = 2*3*5 has factors of 1,2,3,5,6,10,15,30 --> the product of 3 x 5 = 15 is odd.
--> Only scenario S2 (x=n^7) can accommodate this rule, provided that n is the only even prime factor, which is 2
Thus, x = 128 = 2^7 has factors of 1,2,4,8,16,32,64,128 --> the product of any two positive factors of 128 is always even
(1) is SUFFICIENT

(2) x has 7 even positive factors
--> This means that x has 7 even positive factors and an odd factor of 1
--> This rule does NOT apply to Scenario S1 (x=n*m^3 or x=n*m*k).
e.g. x=24=3*2^3 has factors of 1,2,3,4,6,8,12,24 --> only 6 even positive factors.
e.g. x = 30 = 2*3*5 has factors of 1,2,3,5,6,10,15,30 --> only 4 even positive factors.
--> Only scenario S2 (x=n^7) can accommodate this rule, provided that n is the only even prime factor, which is 2
Thus, x=128= 2^7 has factors of 1,2,4,8,16,32,64,128 --> exactly 7 even positive factors.
(2) is SUFFICIENT


CORRECT ANSWER IS (D)

First, how do w find number of factors of a number?

Consider number, K = p1^a * p2^b * p3^c * .... (where p1, p2, p3 are prime numbers)

Thus, the total number of factors or K = (a + 1) * (b + 1) * (c + 1)

Now given that, positive integer x has 8 positive factors.

There, x can be of either of the following 3 forms --

1. x = p1^1 * p2^1 * p3^1.
=> This will result in 2 * 2 * 2 = 8 factors of x (from formula above)

2. x = p1^3 * p2^1
=> This will result in 4 * 2 = 8 factors of x (from formula above)

3. x = p1^7
=> This will result in 8 factors of x (from formula above)

Note that out of the 8 fctors of x, 1 is one of the factors as 1 divides all numbers. Now let us consider the options ->

Option 1: The product of ANY two positive factors of x is even

We know that 1 is an ODD number and is 1 of the factors. Since as per this option any 2 positive factors will multiply and give us an even number, the remaining factors should all be evern. Now let us consider the 3 cases listed above -

Case 1: x = p1^1 * p2^1 * p3^1.
=> Here we know that p1 can be 2 and p2, p3 will both be odd prime numbers thus not following the rule given in this option that the product of any 2 positive factors will be even (Ex - 2^1 * 3^1 * 5^1 will have multiple odd factors and hence their product will not be even)

Case 2: x = p1^3 * p2^1
=> Here again we know that p1 can be 2 and p2 will both be odd prime number thus not following the rule given in this option that the product of any 2 positive factors will be even as remember that the 8th factor of x is 1 which is odd (Ex - 2^3 * 3^1 will have some odd factors (3^1 * 1) and hence their product will not be even).

Case 3: p1^7
=> Here the rule of option 1 will always hold true for p1 = 2 as any form of 2 multiplied by the only odd factor of x which is 1 will always be even.

Hence, option 1 is sufficient.

Option 2: x has 7 even positive factors

This straight away gives us from our 3 cases that the only case which holds true in this option is p1^7 when p1 = 2.

Hence, this option 2 sufficient.

Answer: D

x has 8 positive integers.
x can be in either \(a^7\) or \(a*b^3\), where a and b are distinct prime numbers

Statement 1
The product of ANY two positive factors of x is even, implying that all the positive factors of x are even except 1. Hence x must be \(2^7\)
Sufficient

Statement 2
x has 7 even positive factors and eighth one is 1. It's only possible when x=\(2^k\)=\(2^8\), where k is positive integer.
Sufficient

IMO D

For any number \(n=p_1^a.p_2^b.p_3^c\), where \(p_1,p_2,p_3\) are prime numbers, the number of factors is given by (a+1)(b+1)(c+1)

x has 8 positive factors. So x can be of the following 2 forms only:
a. \(x_1^7\)
b. \(x_1^1.x_2^3\)
This is because 8 can be factorized as (0+1)(7+1) or (1+1)(3+1) only (expressed in terms of the above formula).

Statement (1) The product of ANY two positive factors of x is even

x cannot take the form \(x_1^1.x_2^3\) as either \(x_1\) or \(x_2\) will be odd (as 2 is the only even prime number) and hence, the product of any 2 factors can be odd (one of the factors of x is 1 which when multiplied by the odd prime number gives odd number).

So x can take the form \(x_1^7\) only. Since product of any 2 factors is even, \(x_1 = 2\). Therefore, \(x=2^7=128\).

Sufficient.

Statement (2) x has 7 even positive factors

x can take the form \(x_1^7\) only with \(x_1\)=2 as so many even factors won't be possible in other cases for a number having 8 factors, 1 being one of the factors.

The factors are \(1,2^1,2^2,...,2^7\).

\(x=2^7=128\).

Sufficient.

Hence, option (D).

If the positive integer x has 8 positive factors, what is the value of x?

x = 8 factors
If N = a^m*b^n where a and b are prime numbers
then,
Number of factors = (m+1)*(n+1)

From this we can say that,
8 factors can come from
1. a* b*c (a, b, c are prime numbers)
2. a^3* b (a, b are prime numbers)
3. a^7 (a is prime number)

(1) The product of ANY two positive factors of x is even
if any product should be even then there should be only even number and we have only one even prime number. So from the above 3 possibilities, we can eliminate 1st and 2nd.
Value of x will be 2^7.
Sufficient.

(2) x has 7 even positive factors
factors also include the number as well as 1. So if all factors are even except 1 then we can eliminate first two possibilities which need at least two prime factors.
Value of x will be 2^7.
Sufficient.

Answer: D

To have 8 factors we will need a number with prime factorization either in the form \(a^3*b^1\) or in \(c^7\) (a,b,c can be any prime number)

Statement (1) The product of ANY two positive factors of x is even
Only even prime available is 2, the other one(if any has to be odd) so the possibility of form \(a^3*b^1\) is gone. Only possibility is \(c^7\). And since the product of any 2 has to be even, the only possibility is \(2^7\). Strike off B, C and E

Statement (2) x has 7 even positive factors
By same logic used in statement 1 , to have 7 even positive factors only possible answer is \(2^7\).

Hence, IMO, D is the answer

positive integer x has 8 positive factors (given)

so X can be =\(2^7\)
total factors= 7+1=8 (1 odd factor and 7 even factors)

X = \(2^3*3\)
total factors = (3+1)(1+1)= 8 (more than one odd factor since there is 3 and will also have even factors)

X = \(2*3*5\)
total factors = (1+1)(1+1)(1+1)= 8 (here more than one 0dd factors since there is 3 and 5)

STATEMENT(1) The product of ANY two positive factors of x is even
since 1 is a factor of every integer so there is already one odd factor
there cant be another odd factor ( suppose if there is another odd factor 3 then product of 1 and 3 will be odd and it will contradict the statement)
so all the other factors will be even anD it is possible only
when X = \(2^7\) ( 8 positive factors and product of any 2 factors are even)
SUFFICIENT

STATEMENT (2) x has 7 even positive factors
in this statement we have x ---7 even positive factors (since x has 8 factors then one odd factor must be 1)
and if X = 8 positive factors out of which 7 are even (suppose x has 3 as a factor minimum odd factor 1,3 then there cant be 7 positive even factors this contradicts the statement)

so this case is possible only in X = \(2^7\)
1 odd factor and 7 positive even factor SUFFICIENT

D is the answer

x has 8 positive factors means that it is not a perfect square

x=p^a * q^b where (a+1)(b+1)=8

So, 8 can be expressed as 8*1 (not possible because then one factor will have power 0) or 4*2

statement1, product of any 2 positive factors is even. We know, one of the factors is 1. For the product of 1 and any other factor to be even that factor must itself be even

Here, nos. like 6 are even but can't be factor as 3 is also one of the factors so product of 1 & 3 won't be even. Hence, only possible prime factor is 2. Therefore x=2^7

HENCE, SUFFICIENT

statement2, directly says x has 7 even positive factors. We have total 8 factors So, x=2^7 HENCE, SUFFICIENT

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