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GMAT Club Olympics 2025 (Day 3): If a positive integer n has 24

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Bunuel

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If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Bunuel

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Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

GMAT Club Official Explanation:

If the prime factorization of a positive integer is a^x * b^y * c^z *... where a, b, and c are prime numbers and x, y, and z are their powers, then the number of positive factors of the integer is given by (x + 1)(y + 1)(z + 1)... So, to find the number of positive factors, we add 1 to the powers of the distinct primes in the prime factorization and multiply.

We are given that n has 24 factors. To find how many prime factors n can have, we consider how 24 can be written as a product of such terms.

The least number of primes n can have is 1, if n = prime^23, which gives the number of factors as 23 + 1 = 24. For example, n could be 2^23.

For the maximum number of primes, break 24 into the maximum number of integers greater than 1: 24 = 2 * 2 * 2 * 3. In this case, n would have 4 primes: n = (prime_1) * (prime_2) * (prime_3) * (prime_4)^2, which gives the number of factors as (1 + 1)(1 + 1)(1 + 1)(2 + 1) = 2 * 2 * 2 * 3 = 24. For example, n could be 2 * 3 * 5 * 7^2.

So, the range is 4 - 1 = 3.

Answer: C.

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Number of positive factors = n! where n is the number of distinct prime factors

24 = 4!
hence there can be maximum 4 distinct prime factors

Hence D is correct

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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for the GMAT Club Olympics Competition

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total positive factors 24
possible when power of numbers is 2*2*2*3
i.e. number has to be prime to have 2 has power ; so total prime numbers will be 3
OPTION C, 3

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Information given:
- A positive integer n has 24 positive factors

Question:
- What is the maximum possible range in the number of distinct prime factors that n can have

Solution:
- The number of factors formula for n = p1^a * p2^b * p3^c ... is factors = (a + 1)(b + 1)(c + 1)
- To maximize the number of distinct primes, split 24 into as many integer factors greater than or equal to 2 as possible
- 24 = 2 x 2 x 2 x 3 = 4 factors (and 4 primes)
- Therefore, the maximum possible distinct prime factors is 4
- The minimum number of distinct prime factors occurs when you use only one prime raised to a power (e.g. 2^23), which would give 1 distinct prime factor
- The maximum possible range therefore is 4 - 1 = 3

Answer: C, 3

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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for the GMAT Club Olympics Competition

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Ans: C (3)

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

number of positive factors = prime^a * prime^b * prime^c = (a+1)* (b+1) * (c+1) * (d+1)

n has 24 positive factors = 24 = 3*2*2*2 = (a+1)* (b+1) * (c+1) * (d+1)
minimum (some value)^23 will give us (23+1 = 24 factors) so one prime value

max when all 4 ( (a+1)* (b+1) * (c+1) * (d+1) ) are on different prime factors
in this case 4 distinct prime factors

range = max - min
4-1 = 3

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The number of factors of an integer is found by taking the exponents in its prime factorization, adding 1 to each, and multiplying the results. We are given that this product is 24. The number of distinct prime factors is the number of terms in this product.

To find the minimum number of distinct prime factors, we use the fewest possible terms to make 24. We can use one term: 24.

This corresponds to a number like p^23, which has only 1 distinct prime factor.

To find the maximum number of distinct prime factors, we use the most possible terms. We break 24 into its smallest factors (greater than 1): 24 = 2 * 2 * 2 * 3.

This product has four terms, corresponding to a number like p1^1 * p2^1 * p3^1 * p4^2, which has 4 distinct prime factors.

The number of distinct prime factors can range from a minimum of 1 to a maximum of 4.

The range is the difference: 4 - 1 = 3.

The correct answer is (C) 3.

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If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

For minimum number of prime factors, n will be of the form p^23
For maximum number of prime factor, n will be of the form p1*p2*p3*p4^2

Maximum possible range of the number of distinct prime factors that n can have = 4 - 1 = 3

IMO C

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First, I'll recall how to determine the number of factors of a number. If a number n has prime factorization: n=paxpox pcx... *pk
Then the total number of factors of n is:
(a+1) * (b+1) × (c+1)x ... * (k+1)
Since we know the number has exactly 24 factors, I need to find the prime factorization combinations that give us 24 factors.
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
So the possible combinations for (a+1) × (b+1) x ... that equal 24 are:
* 24 = 24 × 1 → This means one prime raised to power 23
* 24 = 12 × 2 → This means one prime raised to power 11 and another to power 1
* 24 = x x 3→ This means one prime raised to power 7 and another to power 2
* 24=6 x 4→ This means one prime raised to power 5 and another to power 3
* 24=6×2 x 2→ This means one prime raised to power 5 and two others each to power 1
* 24=4x3 × 2 → This means one prime raised to power 3, another to power 2, and a third to power 1
* 24=3x2x2x2→ This means one prime raised to power 2 and three others each to power 1
* 24=2x2x2x3*1→ This means four primes each raised to power 1 and one prime raised to power 2
The last case I listed isn't valid because it would require 5 distinct primes but the product of exponents would be 2x2x2x3x1 = 24, not 24 factors.
From these combinations:
* Minimum number of distinct prime factors: 1 (from the first case)
* Maximum number of distinct prime factors: 4 (from the second-to-last case)
Therefore, the range in the number of distinct prime factors is 4 - 1 = 3.
The answer is C. 3.

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Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Taking n as 24 ,we know there are 8 prime numbers in 24 therefore from the choices the larger number is 5 which is less than 8, hence we take 5 as the maximum possible prime

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Ok for min prime factor we know we can have p^23 so 1
for max prime factors lets break down 24 = 3 * 2* 2* 2 => here p1^2 * p2 * p3 * p4 so 4
Range = 4-1 = 3
hence Ans C

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Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Wow great onw, Almost got me wrong.

Always clearly define what we are looking for : Range = (Max #Prime factors - Min #Prime factors).

So there is a number that has 24 factors we need to find max,min prime factors for it to get the solution.

We know that #factors of N = $p1^(x1) + p2^(x2)+p3^(x3)................$ is (x1+1)(x2+1)......

So here to get min prime factors the number can be$ p^23 $ . So factors are 24. (1,p, p^2,p^3.....p^23)

To get the max prime factors we need have as more x1,x2,..
So 24 can be written as 2*2*2*3 => So $ (p1) * (p2) * (p3) * (p4)^2$ can be the number, Max prime factors is 4.

Hence, the range is 4-1 =3, IMO C

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number of factors = (p+1)*(q+1)*...

where p, q are power of prime numbers in factorization

24 = 2*2*2*3
24 = (p+1)(q+1)(r+1)(s+1) , where p=1, q=1, r=1, s=2

therefore max 4 distinct prime factors are possible

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Here the integer has 24 factors.
We can write the integer as distinct prime factors
Like i=p1^a1*p2^a2*p3^a3........(here p1,p2 are prime numbers like 2,3 etc)
But to get the total factors from this equation we have to do (a1+1)(a2+1)(a3+1).....=24(given from question)

Next we have to figure out a1,a2

They have asked range.Lets think of maximum.If a1=23,it satisfies the equation. That means i=p1^23
What about minimum ,we can break 24 as 2*2*2*3,So that means a1=1,a2=1,a3=1,a4=2,that means i=p1^1*p2^1,p3^1,p4^2

Finally, we can now see that we have 1 prime number in the first case and 4 prime numbers in the second case.
Range is 4-1=3
Answer is C.

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Integer: a whole number (not a fraction or decimal) that can be positive, negative, or zero.
Factor: number that divides another number evenly, resulting in a whole number with no remainder
Distinct prime factors: the unique prime numbers that divide that number.

When it comes to prime factors we can do 3*2*2*2
Lowest range is 1. Highest is 4.
So C - 3

Answer C

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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if we want to calculate the the number of factors

n=2^(a+1) 3(b+1).....

if we put all 24 factor in one prime
a+1=24 ,the prime number is only 1
if we separate 24 factors into different prime
each one just 1
just like
2*3*5*7....
(a+1)(b+1)(c+1)...

2x2x2x3=24
that means we can have 4 prime numbers
range=4-1=3

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For any number who has prime no in factors their total factors are +1 to their exponent and then all powers are multiplied together.

We know total factors are 24
so
24 = (23+1) ---> 1 factor only
24= 2 * 12---> (1+1)(11+1) --> 2 factor
24= 2*6*2-->(1+1)(5+1)(1+1)-->3 factor
24=2*2*3*2--->4 factor

range = 3

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Total Number of Factors = Product of (Powers of distinct prime factors + 1)

24= 2*2*2*3
This means, maximum no of Distinct Prime factors of n is 4
(n= a^1* b^1* c^1* d^2 where a,b,c,d are distinct prime factors)

Minimum no of distinct Prime factors of n is 1
(n= e^23)

Range= Max-Min= 4-1= 3

Ans- 3

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Number of factors: (a+1)(b+1)(c+1)...=24. Maximum possible range in the number of distinct prime factors when break 24 into many 2 as possible: 24=2^3x3=2x2x2x3, so 4 different prime factors is maximum. Minimum is 1 prime factor when n = a prime ^ 23, n will has 24 factors: (23+1)=24=> Range: max-min: 4-1=3 => C

Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

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Bunuel

If a positive integer n has 24 positive factors, what is the maximum possible range in the number of distinct prime factors that n can have?

A. 1
B. 2
C. 3
D. 4
E. 5

This question was provided by GMAT Club
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If a number X can be expressed as, X= p1^a1 * p2^a2 ... pn^an

where p1, p2, ... pn are distinct prime factors

Then total factors = (a1+1)*(a2+1)*....(an+1)

24 can be expressed as

24 = (23+1), n could be p^23, distinct prime factor = 1

It can also be expressed as 24 = 2 * 2 * 2 * 3 = (1+1) * (1+1) * (1+1) * (2+1) , n could be p1^1 *p2^1 * p3^1 *p4^2, distinct prime factors =4

Maximum Range = 4-1 = 3

Answer choice B is the correct answer

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