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## Comment on

Counting the Divisors of Large Numbers## great tip! thank you so much!

## Should we consider 1 as the

## 1 is a divisor of all

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

## I cover this a 1:50 in the

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

I got 20 for 3375 and 36 for 100 000. Where am I going wrong? Hmm.

## 3375 = (3)(3)(3)(5)(5)(5) =

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?

## Question link: https:/

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

Thanks

## No problem.

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

## Oops, my mistake.

Oops, my mistake.

That was supposed to read N = p^4

Thanks for catching that! I have edited my response accordingly.

Cheers,

Brent

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