Teacher trying to flunk me with these math questions! Help

the factorial of 768 should be the answer to your second question- enter 768! into a calculator
1

Walt:

Given that f(x)+2f(8-x)=4 for all real x values find f(2).

What is the smallest positive integer with exactly 768 divisors.

Guest's answer.
the factorial of 768 should be the answer to your second question- enter 768! into a calculator

Our guest had a good theory, unfortunately
4 is the smallest positive integer with 3 divisors and it is not 1x2x3 so maybe the theory does not hold up under testing

I've had a quick look at both your questions and you could be right.
maybe your teacher does want to flunk you, but i think it is much more likely that your teacher wants to really extend you.
I assume you are one smart person.

I haven't succeeded yet. They both appear to be very difficult. I haven't had much time today. Maybe I'll get a chance to have a better look tomorrow.

What is the smallest positive integer with exactly 768 divisors.

This question is driving me nuts.

What is the smallest positive integer with exactly 768 divisors.

Well, it is less than 2^767. I am convinced of that.

Oh yes, 2^767 does have 768 distinct factors.

Consider 2^4. It has 5 different factors.

1,2,4,8,16 see.

I know there are smaller answers.

It would be some multiple of powers of prime numbers. Like

2 ^a x 3 ^b x 5 ^c x 7 ^d x 11 ^e etc

that's the best i can do so far.

Maybe your teacher does want to flunk you, but i think it is much more likely that your teacher wants to really extend you.

Hi Walt.
I have asked my friends to take a look at you questions (next time it would be better if you put your questions as 2 different posts.
Since we have made no headway with the other one yet maybe you could repost it by itself. Just a thought.

I was going to give you a summary version of their answers but I think it might be better if you see the thought processes that everyone is havin.
This was my original question.

Hi everybody,
What is the smallest positive integer with exactly 768 divisors?

I think that 2^767 might have 768 divisors, but I think there would be a smaller one.
I'm thinking that it would be some product of prime numbers.

Like 2^x * 3^y * 5^z etc

Anyone got any ideas.
Thanks.

1st 'answer' from Romsek

I would think it's the product of the first 768 primes... 2*3*5*7*11*13.... *5489

You do mean unique divisors I assume. Otherwise i guess it would be 2^768.

My response (Melody)

I do mean unique.

Think about 2^5. By my logic this should have 6 distict divisors
1*2*2*2*2*2

1,2,4,8,16,32 See it has 6 distict divisors.

The reason I know that this is not the smallest one is demonstrated as follows
2^3=8, this has 4 factors, 1,2,4,and 8
but
6 is a smaller number with 4 factors, that is 1,2,3, and 6

2nd response from Bob

If you were to use the suggested 2^{x}.3^{y}.5^{z}, a divisor would be the product of 0,1,2,... or x 2's, 0,1,2,... or y 3's and 0,1,2,... or z 5's.
That would give you a possible (x+1)(y+1)(z+1) divisors. You would need that to equal 768.

I think though it's necessary to include higher primes.

If we assume that the divisors are of the form 2^{a}.3^{b}.5^{c}.7^{d}...... then we would need

(a+1)(b+1)(c+1)(d+1) ..... = 768.

Since 768=2^{8}.3, the best that I can come up with for the moment, subject to further investigation, is,

a=2, b=c=d=e=f=g=h=I=1 which produces the number 2^{2}.3.5.7.11.13.17.19.23=446185740.

My (Melody) response
Thanks Bob,
I wasn't suggesting that 5 would be the highest prime, (I actually put, etc, in my question)
I've got to think about your answer. I don't understand the logic you have used. I'm tired now. It might make more sense tomorrow.
Does your list really have 768 distinct combinations. (I hope i got my wording right that time) I did a quick calculation of my own and it could do. Still, I thought there would be more low prime numbers than that.
I'll be back

3rd response from ebaines

Building from BobP's good work, I think a better mix of values for a, b, c, etc are:

a=3, b=3, c= 2, d=1, e=1, f=1, and g=1. This gives:

2^3 x 3^3 x 5^2 x 7 x 11 x 13 x 17 = 91891800.

with total number of factors = 4 x 4 x 3 x 2 x 2 x 2 x 2 = 768.

The strategy is to use higher powers for lower primes, as long as the prime raised to that value is less than the next highest prime that would otherwise be needed. For example 2^3 = 8 is smaller than adding more primes (19, 23, etc). It would be nice to be ableto use 2^4, but then (a+1) = (4+1) = 5, and that does not divide into 768, so we/re limited to 2^3. Notice that one of the divisors of 768 is 3, so one of the primes is going to have to have a power of 2. So we need to see which works better: 3^2 and include 19 or 3^3, 5^2 and not include 19. I foind the latter gives a lower value.

4th response from Bob (again)

2^{5} x 3^{3} x 5 x 7 x 11 x 13 x 17 = 73513440,

has

6 x 4 x 2 x 2 x 2 x 2 x 2 = 768

divisors.

Any further offers ?

Walt wrote: What is the smallest positive integer with exactly 768 divisors.
******************

To restate in this form: "What number (N) is divisible by 768 (unique) Divisors that have a remainder of zero (0)."

Thought its application is used in analytic number theory and is related to (NOT the same as) Euler's Totient, and Sigma functions. (Eurler's Totient is the number of co-primes contained in a number). Theorem 329 of Wright and Hardy delves into the relation of the sum of divisors to the totient. A. Olofsson also broaches this.

A simple, but incomplete, solution is to use the number of divisors as an exponent of 2 -- (2 ⁷⁶⁸). Though true, it's not the smallest number -- not even close.

The primary algorithm for solving this is the prime number decomposition theorem. That is to decompose a number into primes then reassemble them into a composite number. This is relatively simple to do. The difficult part is reducing the number to the lowest value.

This is a simplified statement of the prime number decomposition theorem:
Every integer N is the product of powers of prime numbers: If N is a power of a prime, N = P ^x, then it has x + 1 factors. Note: Every value >1 has at least two factors (divisors) --itself and 1. This only happens if the number is prime.

A more advanced restatement of how the (total) number of divisors of a number correlates to the number itself:
IF a positive integer N has the prime factorization of N = (P ₁) ^(X1)* (P ₂) ^(X2) * ... * (P _k ^(Xk) for distinct primes P _i and exponent of X _k. Then each factor of N equals (X ₁ + 1)(X ₂ + 1) ... (X _k + 1), which is the product of the factors.

(OK. Enough with the formal BS).

Eurler's Totient for a prime (P) is simply P-1. If a number is composite then factor out the primes, subtract 1 from each, then multiply the result of each P-1 together to find the number of divisors.

Note that many numbers can have exactly the same number of divisors -- observable in the example. The desired value is the lowest value that satisfies this condition.

Starting with a simple example: What is the smallest positive integer with exactly 10 divisors?

First, prime factor 10: P(10)= 2*5. We have two factors. From each of these subtract (1): 2 - 1 = 1 and 5 - 1 = 4. These will become the exponents of 2 primes that are currently unknown. We want the smallest value of N so start with the lowest primes. 2 and 3. P ₁ ¹ * P ₂ ² = 2 ¹ * =3 ⁴ = 162. Though this is true because 162 has 10 integers that divide into it, it is not the smallest. The smallest are 3 ¹ * 2 ⁴ = 48. You have to "play" with them to find the smallest.

It becomes more difficult for larger numbers and particularly for for primes with large exponents. For the record the 10 divisors for 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. For 162 they are 1, 2, 3, 6, 9, 18, 27, 54, 81, 162. Note: This formula tells how many divisors there are, but not what they are. Wolframalpha will calculate for a given value, the number of divisors and list them. Search for "divisor function". -- Note it will not solve this problem; however, after you solve for a number, it will tell you the number of divisors for that number, but not if it is the lowest that meets the criterion. That being said it seems a trivial matter to to write a program to do so.

To the actual question: What is the smallest positive integer with exactly 768 divisors?

Factor 768 = (2 ⁸)*3 : 256 and 3. Subtract 1 for 63 and 2.
768= 256 * 3 ---> (P ₁ ²⁵⁵ * P ₂ ²) = 2 ²⁵⁵ * 3 ² = N --- A massive number
The best approach is to "upper bound" the number by factoring 256 ---> 2 ⁸ so we have 8 powers of 2 and of course the 3.
(Remember to subtract 1 from each) so we have a 2 and eight 1's as the exponents. Set them as exponents for this sequence.
(Generally the largest exponent for the smallest prime)
2 ² * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23. Note: Exponents of 1 are not shown).

Attempt to reduce again by eliminating the highest primes. As a general process "elevate the lowest primes such that it does not exceed the value of the highest primes.
One way to keep track is to make a list of factors and exponents.

768 = 256 * 3
-------------------------------Exponent list
Factor 768 = 3*2*2*2*2*2*2*2*2--->2,1,1,1,1,1,1,1,1
Factor 768 = 3*4*2*2*2*2*2*2 --->2,3,1,1,1,1,1,1
Factor 768 = 3*4*4*2*2*2*2 --->2,3,3,1,1,1,1
Factor 768 = 6*4*4*2*2*2 --->5,3,3,1,1,1
Factor 768 = 6*4*2*2*2*2*2 --->5,3,1,1,1,1,1 <----This "looks" optimal


2 ⁵ * 3 ³ * 5 * 7 * 11 * 13 * 17 = 73513440

Many have said you can't add apples and oranges, but apples and oranges are related.

~~D~~

Edited for clarity

Melody:

6 x 4 x 2 x 2 x 2 x 2 x 2 = 768 divisors.
Any further offers ?

Nope. That looks pretty much to be optimal.

bump

bump

First one is fairly simple, I think:
f(x) + 2*f(8-x) = 4
So let x = 2
=> f(2) + 2*f(6) = 4 (let this be equation A)
then let x = 6
=> f(6) + 2*f(2) = 4 (let this be equation B)

equating A and B we have:
f(2) + 2*f(6) = f(6) + 2*f(2)
=> f(2) = f(6) (subtracting f(2) + f(6) from both sides)

plugging this back into equation A:
f(2) + 2*f(2) = 4
=3*f(2) = 4
=> f(2) = 4/3