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

Official Solution:


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

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

The above implies that \(x\) must not have two odd factors (every integer has one odd factor, which is 1) because if it does then the product of these two factors would be odd, not even. So, \(x\) must be of the form \(2^n\): 2, 4, 8, 16, ... For each of those numbers, the product of ANY two positive factors will be even. Now, \(x=2^n\) having 8 factors means that \(n+1=8\) and thus \(n=7\). So, \(x=2^n=2^7\). Sufficient.

(2) \(x\) has 7 even positive factors.

The toal number of factors of \(x\) is given to be 8, so \(x\) has 7 even positive factors and one odd factor, 1. \(x=2^7\). Sufficient.


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

(1) The product of ANY two positive factors of x is even
(2) x has 7 even positive factors

Option 1: if product of any two positive factors is even then atleast 7 factors of X are even. Now even factors like 6 will not be there. because if 2 and 6 are factor, 3 will be a factor. Hence even factors will only constitute powers of 2. So the highest factor will be 128. hence the number will be 128. - Sufficient

Option 2: X has 7 even positive factors. One of the factors will be 1. Hence rest 7 are even. It follows same explanation as above. - Sufficient

Answer - D

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.

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)
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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

(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

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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