Lesson: Counting the Divisors of Large Numbers

Comment on Counting the Divisors of Large Numbers

great tip! thank you so much!!

Should we consider 1 as the divisor of a number ?
gmat-admin's picture

1 is a divisor of all positive integers. However, with respect to the technique described in this video, we wouldn't include 1 in the prime factorization of the number for which we're trying to find the total number of divisors.

For example, to determine the total number of divisors of 14000, we first prime factorize 14000 to get 14000 = (2^4)(5^3)(7^1)

We wouldn't write: 14000 = (1^1)(2^4)(5^3)(7^1)

We wouldn't do this because 1 is NOT a prime number, so it has no place in the PRIME factorization of a number.

Also, including 1 would be problematic, because the exponent could be ANY number.

For example, 12 = (1^1)(2^2)(3^1) or 12 = (1^4)(2^2)(3^1) or 12 = (1^33)(2^2)(3^1) etc.

Does that help?

What is the rationale behind adding 1 to each exponent/ raising each prime factor to 0?
gmat-admin's picture

I cover this a 1:50 in the video.

We add 1, because we must also consider the possibility that the exponent is 0.

So, for example, if 2^5 is in the prime factorization of N, then when it comes to possible divisors of N, we must consider 2^5, 2^4, 2^3, 2^2, 2^1 and 2^0.
So there are 5+1 possible exponents.

I got 30 positive divisors for 1008, but I cant seem to get the other two right. Can you show me how you got 16 divisors for 3375, and 49 for 100 000?

I got 20 for 3375 and 36 for 100 000. Where am I going wrong? Hmm.
gmat-admin's picture

3375 = (3)(3)(3)(5)(5)(5) = (3^3)(5^3)
So, the number of divisors = (3+1)(3+1) = 16

1,000,000 = (2)(2)(2)(2)(2)(2)(5)(5)(5)(5)(5)(5) = (2^6)(5^6)
So, the number of divisors = (6+1)(6+1) = 49

Does that help?

Cheers,
Brent

Hi Brent,

Need a little clarification on this problem:

For a certain positive integer N, N³ has exactly 13 unique factors. How many unique factors does N have?
A. 1
B. 2
C. 3
D. 4
E. 5

Specifically, I don't get this part of the rationale: "So, a number with 13 positive divisors must have a prime factorization that looks like this: prime^12"

What I'm I not seeing?
gmat-admin's picture

Question link: https://gmatclub.com/forum/for-a-certain-positive-integer-n-n-3-has-exac...

RULE: if integer N = (p^a)(q^b)(r^c)(etc), where p, q, r etc are prime numbers, then the number of factors of N = (a + 1)(b + 1)(c + 1)...etc

Notice that the number of factors of N equals the PRODUCT of all of the exponents increased by 1.

So, if we say that the number N^3 has 13 factors, then the PRODUCT of all of the exponents (increased by 1) of N^3 must equal 13.

However, there is only ONE WAY to write 13 as a product. That is: 13 = 13 x 1

So, it must be the case that N^3 = p^12 (where p is some prime number)

Let's confirm this.

When we apply the above RULE, we see that the number of factors of N^3 = (12 + 1) = 13

So, if N^3 = p^12 (where p is some prime number), then we can also conclude that N = p^4.

So, if N = p^4, then we can apply the RULE to conclude that the number of factors of N = (4 + 1) = 5

Does that help?

Cheers,
Brent

Brent, if you wouldn't mind providing an example that would be great because I am severely blocked on this question. Something I am not seeing or getting is really frustrating me here.

Thanks
gmat-admin's picture

No problem.

Let's say I tell you that integer K has 12 positive factors.

Since there are many way to get a PRODUCT of 12, there are many possible ways to write K.

K COULD = (p^2)(q^3) [where p and q are prime numbers].
When we apply our earlier RULE, the number of positive factors of K = (2 + 1)(3 + 1) =(3)(4) = 12. Works!

Or K COULD = (p^1)(q^5) [where p and q are prime numbers].
When we apply our earlier RULE, the number of positive factors of K = (1 + 1)(5 + 1) =(2)(6) = 12. Works!

Or K COULD = (p^11) [where p is a prime number].
When we apply our earlier RULE, the number of positive factors of K = (11 + 1) = 12. Works!

Since there are many ways to get a product of 12, there are several different ways to write K: K = (p^2)(q^3), K = (p^1)(q^5) and K = (p^11)

However, since there is only 1 way (using positive integers) to get a product of 13, then there is only 1 way to write N^3

If MUST be the case that N^3 = p^12 (where p is some prime number)

Does that help?

Cheers,
Brent

Hi Brent for the above explanation, how can we conclude that N=P?
gmat-admin's picture

Oops, my mistake.
That was supposed to read N = p^4
Thanks for catching that! I have edited my response accordingly.

Cheers,
Brent

Hi Brent,

720 = 2^4 * 3^2 * 5^1

While I understood the application of the fundamental counting principle to arrive at the number of divisors of 720, I want to know what is wrong in the below approach

There are three positions to be filled and we have 10 choices - 2^0, 2^1, 2^2, 2^3, 2^4, 3^0, 3^1, 3^2, 3^3, 5^0, 5^1, 5^2

Isn't this a combinatorics problem where I have to make a selection of 3 from 10 where the order does not matter (since 2^0 * 3^2 * 5^1 = 5^1 * 2^0 * 3^2) ?

10C3 is 120 which does not tally with the answer derived using the fundamental counting principle. Where am I going wrong?
gmat-admin's picture

There are two main problems with your solution:

1) Why are you selecting 3 of the 10 choices?
Selecting 1 choice (2 or choices) will also be a factor of 720

2) There's some duplication in the outcomes.
For example, one possible outcome is selecting 2^0, 3^1 and 5^2. which is the same as selecting 3^0, 3^1 and 5^2
Likewise, selecting 2^1, 2^3 and 5^1, is the same as selecting 2^4, 3^0 and 5^1

Does that help?

Cheers,
Brent

"Does that help?"

Yes, I get it. Thanks. Also, if we take 3^0, 3^1 and 3^2 then the product is 27 which is not a factor of 720.
gmat-admin's picture

Aha! Another reason it doesn't work!

Cheers,
Brent

A number N^2 has 15 factors. How many factors can N have?

A. 5 or 7 factors
B. 6 or 8 factors
C. 4 or 6 factors
D. 9 or 8 factors
E. 3 or 5 factors
gmat-admin's picture

Great question!

RULE: If K = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors/factors of K is equal to (a+1)(b+1)(c+1)...

GIVEN: N² has 15 factors

Case 1: If N² = p^14 (where p is a prime number), then the total number of positive divisors/factors of N = 14+1 = 15
So, N² = p^14 satisfies the condition of having 15 factors
If N² = p^14, then N = p^7, since (p^7)² = p^14
So, if N = p^7, the RULE tells us that 7+1 = the number of positive factors of N
In other words, N can have 8 factors
ELIMINATE A, C, E

Case 2: IF N² = (p^2)(q^4) (where p and q are prime), then the total number of positive divisors/factors of N = (2+1)(4+1) = 15
So, N² = (p^2)(q^4) satisfies the condition of having 15 factors
If N² = (p^2)(q^4), then N = (p)(q^2), since [(p)(q^2)]² = (p^2)(q^4)
So, if N = (p^1)(q^2), the RULE tells us that the number of positive factors of N = (1+1)(2+1) = 6
In other words, N can have 6 factors
ELIMINATE D

Answer: B

Here's a similar question to practice with: https://gmatclub.com/forum/a-number-n-2-has-35-factors-how-many-factors-...

Cheers,
Brent

Hi Brent,

Thank you for the lesson. Is there an easier way to calculate the prime factorization of 14000? (in the video you showed the factorization directly).
gmat-admin's picture

Here's my lesson on Prime Factorization.

Here's how I'd probably do it.
14,000 = 14 x 1000
= 14 x 10 x 10 x 10
= 2 x 7 x 2 x 5 x 2 x 5 x 2 x 5
= 2 x 2 x 2 x 2 x 5 x 5 x 5 x 7
= (2^4)(5^3)(7)

Cheers,
Brent

Hi Brent,

I am confused about solution for this question https://gmatclub.com/forum/if-x-and-y-are-prime-numbers-and-n-is-a-positive-integer-what-is-the-267973.html

I thought that when a statement is given to us it is a fact and we cannot question it, it is given to us that x*y=6 and that x,y are primes. Hense I thought that if we know that x*y=6 is a fact, than we can deduce that exponents are equal to 1, because if we will have higher exponents say 2^2*3^1=12, which is not equal to 6, so exponents have to be 1 and 1. What am I missing here?
gmat-admin's picture

Question link: https://gmatclub.com/forum/if-x-and-y-are-prime-numbers-and-n-is-a-posit...

I think you may be incorrectly reading Statement 1 as saying (x^n)(y^n) = 6

Statement 1 actually tells us that xy = 6
Since it is given that x and y are prime, we can conclude that one value (x or y) must be 2, and the other must be 3.

Does that help?

Cheers,
Brent

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