HOW MANY TIMES DO I GET TO BRAG ABOUT MY LATEX
[size=200]PART A[/size]
Wait. it's possible to have an odd number of divisors? For divisor, there is always another divisor that pairs it up! So that means it must be an even number!
Oh yeah?
What if I told you this....
What if the other divisor that pairs it up is the same number!
:what?:
How about this:
How many factors does $4$ have?
$$1,2,4$$
Oooooooh. So that means if the amount of divisors is an odd number, it must be a perfect square, right?
(end of conversation)
Now let's go look at some exceptions.
$$16$$
Now this fits the circumstances of being a perfect square, but does it fit the circumstances of having three positive divisors?
$$\text{divisors of 16}=1,2,4,8,16$$
FIVE?! AW CMON!
Let's look at this more in depth.
$$16$$
$$16=4^2$$
$$16=(2\cdot 2)^2$$
AHA! We know what happened! The square root was a composite number, therefore we could break it down more!
So that means The square root can't be a composite number......
This is because if we take a look at the formula that calculates the number of divisors based on the prime factorization, the only way we can get only 3 positive divisors is if we have some prime number squared.
If we have the number $x=p_{1}^{e_{1}}\cdot p_{2}^{e_{2}} \cdots p_{n}^{e_{n}}$ where $p_{n}$ is a prime number and $e_{n}$ is the exponent of the prime number, we can find the number of positive factors, which we can call $f$, through this formula: $f=(e_{1}+1)(e_{2}+1)\cdots (e_{n}+1)$
We know that $f = 3$, so the only way we can have this is by having one of the exponents equal to 2 and the rest equal to 1. If $(e_{n}+1)=1$, $e_n=0$, so $p_n^{e_n}=1$, which is not a prime factor. However, $e_1 + 1 = 3$, so $e_n = 2$, which means there is only one prime factor, which is squared, if a number has 3 positive divisors. An example would be $x = 25$, $x=49$, but not $x=36$.
So that means that all positive integers that have exactly 3 positive divisors are $\boxed{\bold{\mathbb{PERFECT\ SQUARES\ THAT'S\ SQUARE\ ROOT\ IS\ A\ PRIME\ NUMBER!}}}$!
[size=200]PART B[/size]
The way to start this problem is to first find the prime factorization of $8400$ so you can have a list of factors. Let's do that first!
$$8400$$
$$2\cdot 4200$$
$$2\cdot 2\cdot 2100$$
$$2\cdot 2\cdot 2\cdot 1050$$
$$2\cdot 2\cdot 2\cdot 2\cdot 525$$
$$2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 175$$
$$2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\cdot 35$$
$$2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\cdot 5\cdot 7$$
$$2^4\cdot 3^1\cdot 5^2\cdot 7^1$$
Now we can use the formula:
$$x^n\cdot y^m$$ where if we want to find the number of positive divisors, we will do:
$$(n+1)(m+1)$$
Using this formula we will get:
$$2^4\cdot 3^1\cdot 5^2\cdot 7^1$$
$$(4+1)(1+1)(2+1)(1+1)$$
$$(5)(2)(3)(2)$$
$$10\cdot 3\cdot 2$$
$$30\cdot 2$$
$$60$$
Positive integer divisors of $8400$.
Now let's ask ourselves some questions.
What do you call a number that has one positive divisor?
What do you call a number that has two positive divisors?
What do you call a number that has three positive divisors?
We ask these questions so that we can exclude these number since the amount of positive divisors are less than 4.
1:You may think that no integer has one positive divisor, but you forgot about one number: [bold]one[/bold]
2:Two positive divisors? Isn't that the definition of a [b]prime number[/b]
3:Using knowledge from Part A, we see that all numbers that have exactly three positive divisors are [b]perfect squares that their square root is a prime number[/b]
Remember, the prime factorization was $2^4\cdot 3^1\cdot 5^2\cdot 7^1$.
Now we can start eliminating!
From the answer of the first question we get only one divisor: one.
$$60-1=\boxed{59}$$
From the answer of the second question, we see that the answer is:prime numbers
From the prime factorization, $2^4\cdot 3^1\cdot 5^2\cdot 7^1$, we can clearly see that there are only four prime numbers that are divisible of $8400$.
$$59-4=\boxed{55}$$
Now let's do the last question: perfect squares that their square root is a prime number.
From the prime factorization, $2^4\cdot 3^1\cdot 5^2\cdot 7^1$, we see that the only two perfect squares are $2^2$ and $5^2$
$$55-2=\boxed{53}$$
Therefore, there are only $\boxed{\bold{53}}$ divisors if $8400$ that have four or more divisors.
Wait. it's possible to have an odd number of divisors? For divisor, there is always another divisor that pairs it up! So that means it must be an even number!
Oh yeah?
What if I told you this....
What if the other divisor that pairs it up is the same number!
:what?:
How about this:
How many factors does $4$ have?
$$1,2,4$$
Oooooooh. So that means if the amount of divisors is an odd number, it must be a perfect square, right?
(end of conversation)
Now let's go look at some exceptions.
$$16$$
Now this fits the circumstances of being a perfect square, but does it fit the circumstances of having three positive divisors?
$$\text{divisors of 16}=1,2,4,8,16$$
FIVE?! AW CMON!
Let's look at this more in depth.
$$16$$
$$16=4^2$$
$$16=(2\cdot 2)^2$$
AHA! We know what happened! The square root was a composite number, therefore we could break it down more!
So that means The square root can't be a composite number......
This is because if we take a look at the formula that calculates the number of divisors based on the prime factorization, the only way we can get only 3 positive divisors is if we have some prime number squared.
If we have the number $x=p_{1}^{e_{1}}\cdot p_{2}^{e_{2}} \cdots p_{n}^{e_{n}}$ where $p_{n}$ is a prime number and $e_{n}$ is the exponent of the prime number, we can find the number of positive factors, which we can call $f$, through this formula: $f=(e_{1}+1)(e_{2}+1)\cdots (e_{n}+1)$
We know that $f = 3$, so the only way we can have this is by having one of the exponents equal to 2 and the rest equal to 1. If $(e_{n}+1)=1$, $e_n=0$, so $p_n^{e_n}=1$, which is not a prime factor. However, $e_1 + 1 = 3$, so $e_n = 2$, which means there is only one prime factor, which is squared, if a number has 3 positive divisors. An example would be $x = 25$, $x=49$, but not $x=36$.
So that means that all positive integers that have exactly 3 positive divisors are $\boxed{\bold{\mathbb{PERFECT\ SQUARES\ THAT'S\ SQUARE\ ROOT\ IS\ A\ PRIME\ NUMBER!}}}$!
[size=200]PART B[/size]
The way to start this problem is to first find the prime factorization of $8400$ so you can have a list of factors. Let's do that first!
$$8400$$
$$2\cdot 4200$$
$$2\cdot 2\cdot 2100$$
$$2\cdot 2\cdot 2\cdot 1050$$
$$2\cdot 2\cdot 2\cdot 2\cdot 525$$
$$2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 175$$
$$2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\cdot 35$$
$$2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\cdot 5\cdot 7$$
$$2^4\cdot 3^1\cdot 5^2\cdot 7^1$$
Now we can use the formula:
$$x^n\cdot y^m$$ where if we want to find the number of positive divisors, we will do:
$$(n+1)(m+1)$$
Using this formula we will get:
$$2^4\cdot 3^1\cdot 5^2\cdot 7^1$$
$$(4+1)(1+1)(2+1)(1+1)$$
$$(5)(2)(3)(2)$$
$$10\cdot 3\cdot 2$$
$$30\cdot 2$$
$$60$$
Positive integer divisors of $8400$.
Now let's ask ourselves some questions.
What do you call a number that has one positive divisor?
What do you call a number that has two positive divisors?
What do you call a number that has three positive divisors?
We ask these questions so that we can exclude these number since the amount of positive divisors are less than 4.
1:You may think that no integer has one positive divisor, but you forgot about one number: [bold]one[/bold]
2:Two positive divisors? Isn't that the definition of a [b]prime number[/b]
3:Using knowledge from Part A, we see that all numbers that have exactly three positive divisors are [b]perfect squares that their square root is a prime number[/b]
Remember, the prime factorization was $2^4\cdot 3^1\cdot 5^2\cdot 7^1$.
Now we can start eliminating!
From the answer of the first question we get only one divisor: one.
$$60-1=\boxed{59}$$
From the answer of the second question, we see that the answer is:prime numbers
From the prime factorization, $2^4\cdot 3^1\cdot 5^2\cdot 7^1$, we can clearly see that there are only four prime numbers that are divisible of $8400$.
$$59-4=\boxed{55}$$
Now let's do the last question: perfect squares that their square root is a prime number.
From the prime factorization, $2^4\cdot 3^1\cdot 5^2\cdot 7^1$, we see that the only two perfect squares are $2^2$ and $5^2$
$$55-2=\boxed{53}$$
Therefore, there are only $\boxed{\bold{53}}$ divisors if $8400$ that have four or more divisors.
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