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Anatoly

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Cool Math - And you won't finish it!!

NUMBER 1

You can split number 15 with this following numbers:

7+8
4+5+6
1+2+3+4+5

So you can make only this following combination. Which combinations are possible for 115? How to find the combinations for any N?

NUMBER 2

In a square one diagonal line will go trough some rect-corners. If you make a diagonal, which corners will be touched by the rect 199x991. How many blocks will this line touch (also if it won't touch the corner at all.) Please note that its in x² cubes, so of course you can't make the rect (199x991) with 19,9 x 99,1 cubes.

NUMBER 3

With coins with the numbers 3 and 5 you can make and combinations by making 3x + 5y. The first number (in this case) is 8, because after 8 there is a cycle which makes every number possible. X and Y can't be negative. ax+by.

a,b will be like 3 and 5
x, y will be theire count. So 3+5 = 8, 3+3=9, 5+5=10 and you can make after it every number. Since which number will it be possible by ax+by. Get the formula.

Another post (Check it out!)

Number 4

The numbers 4 and 9 are math cubes and have 3 dividers:

4 - 1, 2, 4
9 - 1, 3, 9

Is any number with 3 dividers a cube? Which types of number are existing for numbers with 4 dividers? Example: 1, 2, 3, 6 = 6.
5 dividers?
6?

Get the numbers for any natural numbers with X dividers, and their types, and how to find their dividers count without dividing the number?

Number 5

Lets get any square which was received by 2ⁿ x 2ⁿ.

Can you split the square into 1x3 lines, without bending this 1x3 lines?

Number 6

You are having a number, for example 112, then you have to split this number into 1³ + 1³ + 2³ because you will add the cube of its digits. Lets go on:

112 => 1³ + 1³ + 2³ => 10
10 => 1³ + 0³ => 1 (cycle 1 -> 1 -> 1 -> ....)

Lets check it with "3"

3 => 3³ => 27
27 => 2³ + 7³ => 351
351 => 3³ + 5³ + 1³ = 153 (cycle 351 -> 153 -> 153, because 3³ + 5³  + 1³ is the same as 1³ + 5³ + 3³)

And with number X?

Another post (Check it out) - Currently not existing

Number 7

Comming soon

TaskManager

Formerly maxi123

From: i really should update this

Joined: 2015-03-01

Posts: 9,465

Re: Cool Math - And you won't finish it!!

the answer is 42

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signature by HG, profile picture by bluecloud, thank!!
previous signature by drstereos

Koya

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From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

3+3 does not equal 9; assuming you meant 3×3=9 and not 3+3=9.

There is only 1 solution, the one you stated.

x = √(n+1)
y = n-x - (n+2)/2
These lines only cross in 1 place, (3,5)

I thought something was wrong but forgot to check basic calculation to validate the y value of my incorrect answer

Graph: https://www.desmos.com/calculator/peo6nhtoxg (drag the value of n)

Putting all equations together to make
n-2•√(n+1)-2=0
n = 8 only.

---

Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

Koya

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From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

New post for new problem.

For the first one there are 3 solutions to 115
7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 115
21 + 22 + 23 + 24 + 25 = 115
57 + 58 = 115

I made a little thing so you can try out any number and get all solutions: http://codepen.io/OfficialAntarctica/pe … itors=1010
There is at least one answer to every number unless the number is equal to 2^n where n is an integer, also when the number is greater than 1 as 1 = 1 is not a cumulative sum.

Edit: I actually really like this question, gave the calculator some styling because I am a fan.

---

Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

Koya

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From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

New post for new problem

For the second one there are no solutions as both numbers are prime numbers - therefore share no prime factors, 199 and 995 will work because they share the prime factor 199.

Thing I made so you can see other values, http://codepen.io/OfficialAntarctica/pe … itors=0010

Edit: I FINISHED IT!

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Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

hummerz5

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From: wait I'm not a secret mod huh

Joined: 2015-08-10

Posts: 5,854

Re: Cool Math - And you won't finish it!!

Edit: I FINISHED IT!

NUMBER 4

Comming soon...

ergo, you can never finish it!

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thanks HG, HG, HG, HG! anatoly, zumza! zoey!

Koya

Fabulous Member

From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

Edit: I FINISHED IT!

NUMBER 4

Comming soon...

ergo, you can never finish it!

I shall be eagerly waiting for number 4.

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Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

Special symbols, credits here.

Announced: Here is more!

Number 4

The numbers 4 and 9 are math cubes and have 3 dividers:

4 - 1, 2, 4
9 - 1, 3, 9

Is any number with 3 dividers a cube? Which types of number are existing for numbers with 4 dividers? Example: 1, 2, 3, 6 = 6.
5 dividers?
6?

Get the numbers for any natural numbers with X dividers, and their types, and how to find their dividers count without dividing the number?

Number 5

Lets get any square which was received by 2ⁿ x 2ⁿ.

Can you split the square into 1x3 lines, without bending this 1x3 lines?

Number 6

You are having a number, for example 112, then you have to split this number into 1³ + 1³ + 2³ because you will add the cube of its digits. Lets go on:

112 => 1³ + 1³ + 2³ => 10
10 => 1³ + 0³ => 1 (cycle 1 -> 1 -> 1 -> ....)

Lets check it with "3"

3 => 3³ => 27
27 => 2³ + 7³ => 351
351 => 3³ + 5³ + 1³ = 153 (cycle 351 -> 153 -> 153, because 3³ + 5³  + 1³ is the same as 1³ + 5³ + 3³)

And with number X?

Number 7

Comming soon

Koya

Fabulous Member

From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

I don't understand number 4:
4 and 9 are squares not cubes, I don't know if I'm then looking for a cube or a square with 3, 4, 5, 6 factors.

For Number 5, the answer is any positive whole values of n
v^2=2^n×2^n
v^2=(2^n)^2
v = 2^n
You are only multiplying 2's together which has to be whole

I don't understand what you mean by the second part.

For number 6 this doesn't work with unnamed values of X
The first equation will go on indefinitely
v = (⌊x/(10^(⌊log(x)⌋-1))⌋^3
+ (⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)^3
+ (⌊(⌊(⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)/(10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1))⌋•10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1))/(10^(⌊log((⌊(⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)/(10^(⌊log(x)⌋-1))⌋•10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1)))⌋-1))⌋)^3
+ ... it goes on for ⌊log(x)⌋ times!! (the equation I gave works for numbers with 3 digits)

---

Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

I don't understand number 4:
4 and 9 are squares not cubes, I don't know if I'm then looking for a cube or a square with 3, 4, 5, 6 factors.

You are looking for a number N which has X amount of factors. Without dividing N into its factors how to get it's X?

For Number 5, the answer is any positive whole values of n
v^2=2^n×2^n
v^2=(2^n)^2
v = 2^n
You are only multiplying 2's together which has to be whole

I don't understand what you mean by the second part.

So you have for example a square 2^n x 2^n and you have to remove one of the corner blocks.

2ⁿ x 2ⁿ - 1 mod 3 = 1 then it will work.
so basically for (2ⁿ)^2 or better said 4ⁿ mod 3 = 1

It will work only for numbers which can be divided by 2.

For number 6 this doesn't work with unnamed values of X
The first equation will go on indefinitely
v = (⌊x/(10^(⌊log(x)⌋-1))⌋^3
+ (⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)^3
+ (⌊(⌊(⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)/(10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1))⌋•10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1))/(10^(⌊log((⌊(⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)/(10^(⌊log(x)⌋-1))⌋•10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1)))⌋-1))⌋)^3
+ ... it goes on for ⌊log(x)⌋ times!! (the equation I gave works for numbers with 3 digits)

Sorry, what?

Koya

Fabulous Member

From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

I don't understand number 4:
4 and 9 are squares not cubes, I don't know if I'm then looking for a cube or a square with 3, 4, 5, 6 factors.

You are looking for a number N which has X amount of factors. Without dividing N into its factors how to get it's X?

What?

For Number 5, the answer is any positive whole values of n
v^2=2^n×2^n
v^2=(2^n)^2
v = 2^n
You are only multiplying 2's together which has to be whole

I don't understand what you mean by the second part.

So you have for example a square 2^n x 2^n and you have to remove one of the corner blocks.

2ⁿ x 2ⁿ - 1 mod 3 = 1 then it will work.
so basically for (2ⁿ)^2 or better said 4ⁿ mod 3 = 1

It will work only for numbers which can be divided by 2.

What?

For number 6 this doesn't work with unnamed values of X
The first equation will go on indefinitely
v = (⌊x/(10^(⌊log(x)⌋-1))⌋^3
+ (⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)^3
+ (⌊(⌊(⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)/(10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1))⌋•10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1))/(10^(⌊log((⌊(⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1)/(10^(⌊log(x)⌋-1))⌋•10^(⌊log((⌊(⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1))/(10^(⌊log((⌊x/(10^(⌊log(x)⌋-1))⌋•10^(⌊log(x)⌋-1)))⌋-1))⌋)•10^(⌊log(x)⌋-1))⌋-1)))⌋-1))⌋)^3
+ ... it goes on for ⌊log(x)⌋ times!! (the equation I gave works for numbers with 3 digits)

Sorry, what?

Splitting up a number with maths is long, you can use the number above to split X into each digit and cubing them and adding them.

You can use that for a number like 241 to split it into 2, 4, 1 by finding the exponent of the 2 in 241 (2) and then splitting it from the group 2x10^2 to leave you with 41 and you do the same process with the modified number - the modifications to the numbers stack up with each digit; if you plug in a 2 digit number into the equation above it will work and the third part will just be 0 so you would need to generate this equation for the longest number; I can't be bothered to make the equation to then try and find a loop, might just do it programatically later.

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Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

What?

What?

What exactly you don't understand

Splitting up a number with maths is long, you can use the number above to split X into each digit and cubing them and adding them.

You can use that for a number like 241 to split it into 2, 4, 1 by finding the exponent of the 2 in 241 (2) and then splitting it from the group 2x10^2 to leave you with 41 and you do the same process with the modified number - the modifications to the numbers stack up with each digit; if you plug in a 2 digit number into the equation above it will work and the third part will just be 0 so you would need to generate this equation for the longest number; I can't be bothered to make the equation to then try and find a loop, might just do it programatically later.

I will read it later.

Stagecrew

Member

Joined: 2015-02-15

Posts: 289

Re: Cool Math - And you won't finish it!!

Nice problems, but your wording was a bit difficult to figure out on a few.

▼Hidden text

1. If n is your number, the answer is σ(n/v(n))-1 where σ(n) is the divisor function which gives the number of divisors of n and v(n) is the greatest power of two dividing n. For example, 115 = 5x23 gives σ(115)=4 (v(n)=1) so the answer is 3, like Koya found.

This is equivalent to finding when n is a difference of two non consecutive triangular numbers. These are of the form 1/2(i)(i+1), so we want to solve for i>j

2n=i^2+i-j^2-j = (i-j)(j+i+1)

One of these factors is odd, while the other is even. In fact, for any pair of factors (a,b) with opposite parities we can solve for i,j. So we're really interested in the number of distinct factor pairs (a,b) with opposite parities of 2n. WLOG, let a be the even one: then a=2*v(n)*(odd number), while b is another odd number. Therefore it's only dependent on the odd numbers, which gives σ(n/v(n)) ways. Subtract one for the case of just 'n' as our list of numbers.

2. See later post.

3. This is best done with modular arithmetic. Let a>b with gcd(a,b)=1, since if gcd(a,b)>1 then we can't make numbers not divisible by their gcd. Once we have achieved all possible residues modulo b using only multiples of a, we know that we can create any number after the greatest multiple.Given that a,b are relatively prime, a modulo b is relatively prime to b and so a*(b-1) will cover the last residue modulo b. However, we can create b-1 numbers before a*(b-1) using previously obtained residues, so we really can obtain every value that is at last (a-1)*(b-1).

4. (I'm taking this to mean find the set of numbers with x divisors) This actually relates a bit back to the first one: we want to find the set S_x of all n such that σ(n)=x. First, make a list of all ways to get a product of x with integers>1. If the prime factorization of n is n = p_1^e_1 * p_2^e_2 ... , then σ(n)=(e_1+1)(e_2+1)... . For each way to express x, we subtract 1 from each term in the product to get the e_i. Then substitute any set of primes in to get a value for n.

As an example: take x=4. We have 4=2x2, 4 (only ways), so S_x=p_1*p_2 or p_1^3.

5. No. In the end, we want 4^n = 1 (mod 3) squares. However, at every step there are 0 mod 3 squares.

6. We get a cycle when we hit a number which is equal to the sum of the cubes of its digits. However, we can easily put a bound on this: there are floor(log(n)) digits (base 10 log here), each which is at most 9^3. When 9^3ceil(log(n))<n, we can't have such a number anymore. This happens around n=3000. (Although they become increasingly improbable way before this)

A quick brute force search gives 1, 153, 370, 371, 407 as the only such numbers. For the next bit, let f(n) be the sum of the cubes of the digits of n.

The specific fixed point is dependent on the starting number modulo 3: a number is a multiple of 3 plus k if the sum its digits is a multiple of 3 plus k, and cubing preserves values of numbers modulo 3, so repeated application of f on some number n does not change its value modulo 3. if 3|n, then f(n) 153. If n is a multiple of 3 plus one, it goes to 1 or 370 if there isn't a cycle. If n is a multiple of 3 plus two, it goes to 371 or 407. These aren't too tricky to prove, since f(n)<n for n>3000, so we only need to check that all n in this range end up in a cycle or a fixed point as claimed. It does turn out that all multiples of 3 end up at 153, and all multiples of 3 plus two end up at 371 or 407. Unfortunately, this fails for multiples of 3 plus one because you can end up in a cycle with a number like 133.

I really liked this last one, I had no clue that there was any sort of pattern in this sort of stuff. 3000 numbers isn't a lot to search, but if anybody can lower the number of checks you make or eliminate them completely that would be nice.

Koya

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From: The island with those Brits

Joined: 2015-02-18

Posts: 6,310

Re: Cool Math - And you won't finish it!!

2 means imagine a graph y=991x/199, which points are both x and y whole numbers

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Thank you eleizibeth ^

I stack my signatures rather than delete them so I don't lose them

Stagecrew

Member

Joined: 2015-02-15

Posts: 289

Re: Cool Math - And you won't finish it!!

2 means imagine a graph y=991x/199, which points are both x and y whole numbers

Oh, okay. That makes much more sense. I was put off by diagonal of square, because that is definitely not diagonal of a square.

EDIT: Solved it. If we have a AxB rectangle, the answer is A+B-gcd(A,B). The number of boxes you pass through increases by 1 each time x passes an integer or y passes an integer, which would lead you to believe that the answer is A+B. However, sometimes we count the same box twice when the lines passes through a lattice point: this happens at gcd(A,B) points.

(Also Koya I think you should read the question again, it's asking how many boxes the line passes through, not lattice points.)

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

Another one:
This one is my own, not from the teachers!

Let's get from start! These are the combinations for some numbers:

2 -> 12 -> 30 -> 56 -> 90

We have the following possible change explanations:

1 *2 -> 3 * 4 -> 5 * 6 -> 7 * 8 -> 9 * 10 -> 11 * 12 -> ....
AND
2 -> +10 -> +18 -> +26 -> Every time a number for 8 bigger. So the next numbers will be +34 +42 +50 .....

What means

1*2 == 2
3*4 == 2 + 10 = 12
5*6 == 12 + 18 = 30
7*8 == 30 + (18 + 8) = 56
9*10 == 56 + (18 + 8 + 8) = 90

Explain this, if it will happen always, and explain why it shouldn't be possible.

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

I will start numbers from new. Geometrical now!

1

A geometric shape. N angles. Purely random.

This 7-angle got 31 dots (the orange ones). How to find them + formula?

2

Clue that all 3 medians will  meet in one point.

3

3 Points are given, these are the middles of every of the triangles' sides. Construct a triangle, from these dots. Same with n-angle.

4 (My one)

Find and create your own theorems about a

- equal-angled 6-angle
- equal-sized 6-angle
- written in circle N-ANGLE

5

Create theorems for a 6-angle where every second angle/size is same.

6

A paper is given with the size 1 :  √2 (a normal paper), per "origami" you can split it into ½ parts, not a question. How about 1/14, 1/91 ??

7

Given is a circle and two points on it. A, B. C is "alive". If it will move around on the circle, how will "special points" of the triangle move? (centre of medians...)

8

I don't get it so idk.

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

Xfrogman43

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From: need to find a new home

Joined: 2015-02-15

Posts: 4,174

Re: Cool Math - And you won't finish it!!

http://www.coolmath-games.com/

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thanks zoey aaaaaaaaaaaand thanks latif for the avatar

Anatoly

Guest

Re: Cool Math - And you won't finish it!!

Locking, people should make their own topic if they want to share something.