Random math questions

[FlorianK]

Starting with the fraction 7 over 8, without simplifying the new fractions, I do one of two operations: I either add 8 to the numerator or I add 7 to the denominator. After multiple operations the fraction is 7 over 8 again. How many operations are it at least?

[kamil9876]

So in other words you want to know the smallest non-trivial solution (x,y) in the non-negative integers such that:



Which is equivalent to:



But then since and are relatively prime this means divides and divides . So the smallest non-trivial solution must be and

[FlorianK]

thx

got another one

Let z be the smallest possible number of a*b*c. a, b and c are all integers and a²=2b³=3c⁵.
By how many integers is z divisible including 1 and z?

[kamil9876]

Translation error from German?

Is supposed to be the smallest element of the set ? Also should it be positive? (i.e should we also add the constraints ) ?

[kamil9876]

1+0=1 which is not 0

[FlorianK]

nah no translation error :/

z=a*b*c
a,b, and c are positive integers
a² = 2b³ = 3c⁵

By how many numbers is z divisible including 1 and z.

The answer is 60, but I'm not sure why :/

[Deleted User]

1+0=1 which is not 0

Now before you judge me, I was half-awake when I posted that question. 

[Deleted User]

For proofs, do we have to use the same proof techniques that our lecturer taught us to do our assignment? If we use our own methods, would we still get the marks?

[kamil9876]

nah no translation error :/

z=a*b*c
a,b, and c are positive integers
a² = 2b³ = 3c⁵

By how many numbers is z divisible including 1 and z.

The answer is 60, but I'm not sure why :/

Well it seems that the answer could vary as there are many possibilities for but combining this data with your previous post it seems my original formulation of the problem is close to what you want, let me state it:

Problem: Let be the minimum element of the set . Find the number of divisors of .

Now why don't we just find and then count the divisors ?

It is clear that will only have prime divisors and (why?) Hence we can write , and for lack of better notation.

So we we really need to solve is the following system in the non-negative integers:





So what are the smallest possible non-negative integer solutions to these (i.e such that is a minimum) ? Well for the first one, one can see that we need to be a multiple of and to be a multiple of . But hey, so  is a minimal solution.

As for the second one, we need s to be a multiple of , and to be a multiple of . What do you know... hence we can take

So in fact:



Now the number of divisors is as there are 11 choices of a multiple of 2 at most 2^10 and 7 choices for a multiple of 3 at most 3^6

(i.e the number of divisors of is and not which gives the supposed correct answer of )

This seems like a very contrived problem, where is it from?

[FlorianK]

It's from an old paper from a Multiple Choice Math Competition that I'm attending next week. They give out the questions and answers, but want me to pay for the worked solutions. Getting in the 99.x%tiles atm

77 was an answer as well. Maybe it was just an error. Thx

So or my understanding:
If we have n=2^x * 3^y * 5^z the number of divisors is (x+1)*(y+1)*(z+1) ?

From which 'part' of math did you learn this? discrete mathematics?

[kamil9876]

So or my understanding:
If we have n=2x * 3y * 5z the number of divisors is (x+1)*(y+1)*(z+1) ?

From which 'part' of math did you learn this? discrete mathematics?

Yeah it could count(pun!) as discrete mathematics I guess. I guess it is mentioned in some elementary number theory text books. That's really how the proof works, by number theory. A divisor of    must be of the form such that . There are choices for such an , namely (perhaps forgetting is where this company made a mistake) and hence different choices.

Of course this needs some justification, (how do you know every divisor is of such a form? how do you know different choices of f_k give different numbers (i.e no repetition) ?) All of this can be justified with unique prime factorization in the integers, which can be found in any elementary course/book/free website on number theory I guess.

[FlorianK]

Ok got another one
Let Z be the amount of 8 digit numbers of which all 8 digits are different from each other and from 0.
How many of those numbers are divisible by 9?

Z/8  |   Z/3   |   Z/9   |   8Z/9   |   7Z/8   |

and

a₀=1, a₁=2, a_n+2=a_n+ a²_n+1 for n>=0

What is a₂₀₀₉ mod 7 ?


and

You have a 3*3 square (like a 9th of a soduku). In the middle, right field is a 47 and in the middle, bottom field is a 63. The sum of the numbers in each row, collum and diagonal is the same.
Which number is in the top, left field?


and

this one if probs like the one in the post where 77 was the answer:
z has 2 digits
t(z) is the number that is created when you put each divisor of z in a row including 1 and z
So t(14)=12714
How many digits has the largest value of t(z)?

[kamil9876]

Let Z be the amount of 8 digit numbers of which all 8 digits are different from each other and from 0.
How many of those numbers are divisible by 9?

Z/8  |   Z/3   |   Z/9   |   8Z/9   |   7Z/8   |

There is a criterion: a number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Hence we wish to find the number of 8 element sequences of distinct numbers in such that their sum is a multiple of 9. Now the sum of all such elements in D is . The only number from D that we can remove so that the sum is a multiple of 9 is 9 itself (removing any other number will make the sum not divisible by 9). Hence in fact we are counting permutations of numbers between 1-8. So there are 8! such 8-digit numbers. So what is Z? Well it is in fact (every such number can be uniquely obtained by permuting the 9 digits and removing the last one). So Z/9 is the ans.

[Deleted User]

How to prove this by induction?

2^n > n^3

[kamil9876]

Is supposed to be a positive integer? (You can actually prove it for all integers but I am assuming this is what you mean).

You can verify it for small positive integers manually (these will be your base case). Then the following inductive step works whenever is large enough such that





So the point is: Find such that for all we have . Then verify the inequality for