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firas92
How many odd factors does the integer n have?

(1) 16 is the highest power of 2 that divides n

(2) n has a total of 68 factors and 3 prime factors.

1) 16 is the highest power of 2 that divides n
In other words...
\(2^{16} \text{ is the largest power of 2 which divides n}\)

Possibilities of n:
\(n = 2^{16}\), 0 prime factors
\(n = 2^{16} * 3\), 1 prime factor

Insufficient

(2) n has a total of 68 factors and 3 prime factors.
However, we do not know if these primes are all odd, or contain an even prime (2).
Insufficient

Combine (1) and (2)
Recall the formula for finding total number of factors: https://gmatclub.com/forum/math-number-theory-88376.html#p666609

\(n=a^p*b^q*c^r \text{, where a, b, and c are prime factors of n and p, q, and r are their powers.}\)
\(\text{Total Number of Factors} = (p+1)(q+1)(r+1)\)

So therefore:
\(68 = (p+1)(q+1)(r+1)\), results in
\(68 = (16+1)(1+1)(1+1) = 17*2*2\)

We know that the prime factorization of 2 results in a power of 16. Thus, the other 2 numbers must be odd primes to the first power.
\(n=2^{16}*odd^1*odd^1\)

Sufficient
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firas92
How many odd factors does the integer n have?

(1) 16 is the highest power of 2 that divides n

(2) n has a total of 68 factors and 3 prime factors.

This question has a bit of ambiguity. "16 is the highest power of 2 that divides n" could mean two different things:

2^16 divides evenly into n, but 2^17 doesn't;
The number 16 (which equals 2^4) divides evenly into n, but 32 (which equals 2^5) doesn't.

The GMAT would also most likely use the phrasing "unique prime factors" in the second statement. I think it's reasonably clear from context that the statement is referring to the unique prime factors, but the GMAT is usually as clear as it possibly can be in order to avoid challenges to official problems!
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firas92
How many odd factors does the integer n have?

(1) 16 is the highest power of 2 that divides n

(2) n has a total of 68 factors and 3 prime factors.

1) 16 is the highest power of 2 that divides n
In other words...
\(2^{16} \text{ is the largest power of 2 which divides n}\)

Possibilities of n:
\(n = 2^{16}\), 0 prime factors
\(n = 2^{16} * 3\), 1 prime factor


Insufficient

(2) n has a total of 68 factors and 3 prime factors.
However, we do not know if these primes are all odd, or contain an even prime (2).
Insufficient

Combine (1) and (2)
Recall the formula for finding total number of factors: https://gmatclub.com/forum/math-number-theory-88376.html#p666609

\(n=a^p*b^q*c^r \text{, where a, b, and c are prime factors of n and p, q, and r are their powers.}\)
\(\text{Total Number of Factors} = (p+1)(q+1)(r+1)\)

So therefore:
\(68 = (p+1)(q+1)(r+1)\), results in
\(68 = (16+1)(1+1)(1+1) = 17*2*2\)

We know that the prime factorization of 2 results in a power of 16. Thus, the other 2 numbers must be odd primes to the first power.
\(n=2^{16}*odd^1*odd^1\)

Sufficient

\(n = 2^{16}\), 0 1 prime factors
\(n = 2^{16} * 3\), 1 2 prime factor. 2 is a prime number.
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LeoN88
firas92
How many odd factors does the integer n have?

(1) 16 is the highest power of 2 that divides n

(2) n has a total of 68 factors and 3 prime factors.

1. n can be k. 2^16, where
if k=3, 1 odd factor.
if k=15, 2 odd factors. Insufficient.

2. 68= (a+1)(b+1)(c+1) since it's told that there are 3 prime factors. a*b*c not = 0
68= 2*2*17

So n can be= 7 * 5 * 3^16
or n= 2 * 5 * 3^16. Insufficient.


1+2 tells us power of 2 is 16

n= (odd factor) * (another odd factor) * (2^16), as 2 is the only even prime factor.
Sufficient, answer is 2 odd factors. C
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From statement 1
N= 2^16× k
Where k is some odd number we don't know that is why given
statement is insufficient.
From statement 2
Whether N is a multiple of 2 or not we don't know that is given statement is insufficient.
Combining both the statements
N=2^16×a^1×b^1
Where a and b are odd prime numbers
Total factors =17×2×2=68
Odd factors =1×2×2=4
Sufficient
Answer is C

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