Official BOS Trial 2014 Thread (1 Viewer)

[Speed6]

No. Integration.

Trapizyzz.

 

[RealiseNothing]

Are you a gun at maths? If yes which one? And which topics do you like the most?

This sounds way too much like "are you a gun at maths, and if yes, which gun are you?"

 

[Speed6]

This sounds way too much like "are you a gun at maths, and if yes, which gun are you?"

I was expecting him to say something along the lines of 'Yes, MX2, Mechanics'.

Speed out.

 

[trapizi]

Trapizyzz.

I'll be somewhere in the bottom of this exam.

 

[Speed6]

I'll be somewhere in the bottom of this exam.

Do your best, there's nothing wrong with having a go

 

[anomalousdecay]

This sounds way too much like "are you a gun at engineering, and if yes, which gun are you?"

Rail gun life is engineering life.

Works for engineering haha.

 

[RealiseNothing]

Hope carrot doesn't mind me doing this.

We know how much carrot loves putting in questions which have very elegant and quick solutions. So here is a question which has a very elegant and quick solution in preparation for tomorrow's exam. I'll rep if anyone finds the shortest possible solution:



For example take the number 6547, it becomes 6+5+4+7=22, 21 becomes 2+1=3, etc.

 

[seventhroot]

Hope carrot doesn't mind me doing this.

We know how much carrot loves putting in questions which have very elegant and quick solutions. So here is a question which has a very elegant and quick solution in preparation for tomorrow's exam. I'll rep if anyone finds the shortest possible solution:



For example take the number 6547, it becomes 6+5+4+7=22, 21 becomes 2+1=3, etc.

can we write a computer program to do it?

 

[RealiseNothing]

can we write a computer program to do it?

You can try, maths will still be faster.

 

[seventhroot]

You can try, maths will still be faster.

alright; I'll write it in C

do you mean like

1 + 2 + 3 + 4 + ... + 321 + 322 + ... + 1,000,000

becomes

1 + 2 + 3 + 4 + ... + 3+ 2 + 1 + 3 + 2 + 2 + ... + 1 + 0 +...

?

 

[anomalousdecay]

can we write a computer program to do it?

It would be pretty easy to set a for loop to count it all up haha.

You can try, maths will still be faster.

This. In the end the computer will do a million times more calculations than you need to as a human. Think about all those poor electrons!

 

[Kurosaki]

Hope carrot doesn't mind me doing this.

We know how much carrot loves putting in questions which have very elegant and quick solutions. So here is a question which has a very elegant and quick solution in preparation for tomorrow's exam. I'll rep if anyone finds the shortest possible solution:



For example take the number 6547, it becomes 6+5+4+7=22, 21 becomes 2+1=3, etc.

Not sure if correct tho:

Consider 1 digit numbers. Sum is 1+2+3+...9
Consider 2 digit numbers. Sum is 10(1+2+....+9) + 10(0+1+2+3+...+9) (due to the two digits)
etc

then add the sum of all these?

 

[Speed6]

It would be pretty easy to set a for loop to count it all up haha.



This. In the end the computer will do a million times more calculations than you need to as a human. Think about all those poor electrons!

lol electrons are working very hard.

 

[Kurosaki]

alright; I'll write it in C

do you mean like

1 + 2 + 3 + 4 + ... + 321 + 322 + ... + 1,000,000

becomes

1 + 2 + 3 + 4 + ... + 3+ 2 + 1 + 3 + 2 + 2 + ... + 1 + 0 +...

?

he said digits, so the latter.

 

[RealiseNothing]

alright; I'll write it in C

do you mean like

1 + 2 + 3 + 4 + ... + 321 + 322 + ... + 1,000,000

becomes

1 + 2 + 3 + 4 + ... + 3+ 2 + 1 + 3 + 2 + 2 + ... + 1 + 0 +...

?

Exactly.

Not sure if correct tho:

Consider 1 digit numbers. Sum is 1+2+3+...9
Consider 2 digit numbers. Sum is 9(1+2+....+9) + 9(1+2+3+...+9) (due to the two digits)
etc

then add the sum of all these?

That works, but there is a faster method.

 

[seventhroot]

This. In the end the computer will do a million times more calculations than you need to as a human. Think about all those poor electrons!

I get a really big number that doesn't look right; I suspect it is a data overflow. what is the number around?

he said digits, so the latter.

soz; no engrish/10

but yeah assumed the latter

 

[RealiseNothing]

I get a really big number that doesn't look right; I suspect it is a data overflow. what is the number around?


soz; no engrish/10

but yeah assumed the latter

Answer is 27000001 iirc.

 

[Sy123]

Hope carrot doesn't mind me doing this.

We know how much carrot loves putting in questions which have very elegant and quick solutions. So here is a question which has a very elegant and quick solution in preparation for tomorrow's exam. I'll rep if anyone finds the shortest possible solution:



For example take the number 6547, it becomes 6+5+4+7=22, 21 becomes 2+1=3, etc.

Imagine writing all these numbers on top of each other, in columns, the 'units' column (the far right), every 10'th number is the same, so you have continuing cycles of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

So in this column, the sum is

The next column, the 'tens' column, has each digit appear in groups of 10 and cycles through all the digits, meaning the sum is:

which is just the same as the first sum

Similar arguments nets us the fact that all the sums for all the columns are the same

So the answer is

 

[seventhroot]

Imagine writing all these numbers on top of each other, in columns, the 'units' column (the far right), every 10'th number is the same, so you have continuing cycles of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

So in this column, the sum is

The next column, the 'tens' column, has each digit appear in groups of 10 and cycles through all the digits, meaning the sum is:

which is just the same as the first sum

Similar arguments nets us the fact that all the sums for all the columns are the same

So the answer is





Every number can be put into the notation of: This will do for all numbers from 0 to 1 000 000

spoiler alert m9; I am trying to code it

 

[Sy123]

Interpret 'all the numbers from 1 to 1 000 000' as every combination of digits in the 6 digit places

Each digit, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 can occupy it

Pick a random spot for these digits, it is found that if we fix a digit, say

1 xxx x2x

Where the x's are variable, there are 10^5 combinations for this, so the sum of all 2's in this column is 10^5 * 2

This is the same for every other digit, so you get 10^5(1+2+...+9)

The column is arbitrary, so 10^5(1+2+...+9)*6

And then add 1 at the end, getting 2700001

Is this more elegant?