What the forget am I supposed to do here (math)

[kegeshan]

a little off topic, but it says:
"A solo dice game is played where, on each turn, a normal pair of dice is rolled."

so do you just like roll the dice two times and then you multiply it?

I think it means it's a game played by one person that uses a pair of dice that you multiply with.

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[SeventhSandwich]

^ Ok that's kind of an overcomplicated way to brute force it.

Yeah I know. I typed it up late at night and half of it isn't even necessary.

Right before I posted it I was like "I really hope Truce doesn't show up and point out how inefficient it is". 

[Illuminati]

a = 1st roll score
b = 2nd roll score
c = 3rd roll score
d = 4th roll score
e = 5th roll score

From the question, it is known that

b = a + 5
c = b - 6
d = c + 11
e = d - 8

Now, each score should be shown in terms of a, so

c = b - 6 = (a+5) - 6 = a - 1
d = c + 11 = (a-1) + 11 = a + 10
e = d - 8 = (a+10) - 8 = a + 2

Now we can obtain a formula for the total score.

a + b + c + d + e = total score = a + (a+5) + (a-1) + (a+10) + (a+2) = 5a + 16

Now, to find a suitable value of a, create a list of numbers from 1-36 (minimum score is 1 and maximum is 36) and remove impossible numbers. For example, 7 cannot be equal to a because no two whole numbers from 1-6 can multiply together to get 7.

Since c  = a - 1 and c must be greater than or equal to 1 (since the minimum score is 1), a must be greater than 1

1 2   3   4  5  6  7  8  9  10  11  12  13  14  15  16  17  18 19  20 21  22  23  24  25  26  27  28  29  30 31  32  33  34  35  36

Now just go through the remaining numbers and see if any of them fit the conditions previously stated.

a = 2
b =

...

a = 9 - Valid
b = a + 5 = 14 - impossible number

a = 10   - Valid
b = a + 5 = 15  - Valid
c = b - 6 = 9 - Valid
d = c + 11 = 20 - Valid
e = d - 8 = 12 - Valid

Since I assume this problem is designed to give one solution only, a = 10.

So the the total score is 10 + 15 + 9 + 20 + 12 = 66.

[beachbum111111]

Sorry to bump this thread but I'm really confused about this question:


I've always been terrible with these kinds of questions.

[Rusty]

1 in 1 O
1 in 2 X
1 in 3 O
1 in 4 X
1 in 5 O
1 in 6 X
ect...

1000/1 - 1000/2 + 1000/3 -1000/4 + 1000/5 - 1000/6... ect...

But that's not right because some doors are already in the correct state...

Sorry this is quite hard, maybe look into sigma notation or something...

1000/Σn = total doors opened by pupils
500/Σn = total doors closed by pupils

I just dont know how to find out which doors have been closed after being opened ect.

[tikitai]

calculate the total number of divisors of each locker number, if the number of divisors is odd, it will be open, if its even, it will be locked

... that would still take too long

theres probably a simpler way to solve it, but i have no idea how to

[Truce]

calculate the total number of divisors of each locker number, if the number of divisors is odd, it will be open, if its even, it will be locked

Yeah, this is on the right track. For example, locker 10 is visited by 1, 2, 5, and 10 - its factors, or divisors as tikitai was referring to. You'll notice it has an even number of factors. This is because they're two numbers you're multiplying together to get the result, so they come in pairs. The only time it wouldn't be even is if a factor is repeated. For example, 9 can be made from 1*9 and 3*3, but we only list 1, 3, 9. Therefore square numbers (1, 4, 9, 16, etc) are going to be the ones that are open. Unless staying up this late made me miss something.

[BluetoothBoy]

Pretty sure it'd be 499 or 500 doors. I thought about it long and hard, and realized that, once every other door is open, that's half of them open. Then, a strange thing happens when you increase one integer at a time. For every other door each consecutive student opens, they will close the next one that they interact with. For example, student number three with first close a door on the third door, then open one on the sixth. This pattern works as long as you increase one door at a time, which this problem does.

This means that each student will open as many doors as they close (give or take one for the odd numbers). As a result, at the end of each run, the number will basically be the same - it's like changing both sides of an algebra problem equally. And thus, you have 499 or 500 doors open at the end.

[Ipquarx]

There are exactly 500 doors open at the end.

Just look through the first few iterations of it:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 1
1 0 0 1 1 1 1 1 0 0 1 0 1 0 0 0

Here 1 means open and 0 means closed.
Note that from the second line on, exactly half are open and exactly half are closed. This will remain true for the entire pattern.

[BluetoothBoy]

Didn't I just say that? *confusion*

[Doomonkey]

I think Truce is right.

There are exactly 500 doors open at the end.

Just look through the first few iterations of it:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 1
1 0 0 1 1 1 1 1 0 0 1 0 1 0 0 0

Here 1 means open and 0 means closed.
Note that from the second line on, exactly half are open and exactly half are closed. This will remain true for the entire pattern.

Unfortunately, you have not expressed all of the factors of all of the first 16 positive integers. You have to go through as many iterations as there are lockers. Also, you screwed up a bit in your counting.

01 - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
02 - 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
03 - 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0
04 - 1 0 0 1 1 1 1 1 0 0 1 0 1 0 0 1
05 - 1 0 0 1 0 1 1 1 0 1 1 0 1 0 1 1
06 - 1 0 0 1 0 0 1 1 0 1 1 1 1 0 1 1
07 - 1 0 0 1 0 0 0 1 0 1 1 1 1 1 1 1
08 - 1 0 0 1 0 0 0 0 0 1 1 1 1 1 1 0
09 - 1 0 0 1 0 0 0 0 1 1 1 1 1 1 1 0
10 - 1 0 0 1 0 0 0 0 1 0 1 1 1 1 1 0
11 - 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 0
12 - 1 0 0 1 0 0 0 0 1 0 0 0 1 1 1 0
13 - 1 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0
14 - 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0
15 - 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0
16 - 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1

As predicted by Truce, square numbers 1 4 9 and 16 appear.

[BluetoothBoy]

I think Truce is right.
Unfortunately, you have not expressed all of the factors of all of the first 16 positive integers. You have to go through as many iterations as there are lockers. Also, you screwed up a bit in your counting.

01 - 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
02 - 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
03 - 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0
04 - 1 0 0 1 1 1 1 1 0 0 1 0 1 0 0 1
05 - 1 0 0 1 0 1 1 1 0 1 1 0 1 0 1 1
06 - 1 0 0 1 0 0 1 1 0 1 1 1 1 0 1 1
07 - 1 0 0 1 0 0 0 1 0 1 1 1 1 1 1 1
08 - 1 0 0 1 0 0 0 0 0 1 1 1 1 1 1 0
09 - 1 0 0 1 0 0 0 0 1 1 1 1 1 1 1 0
10 - 1 0 0 1 0 0 0 0 1 0 1 1 1 1 1 0
11 - 1 0 0 1 0 0 0 0 1 0 0 1 1 1 1 0
12 - 1 0 0 1 0 0 0 0 1 0 0 0 1 1 1 0
13 - 1 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0
14 - 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0
15 - 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0
16 - 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1

As predicted by Truce, square numbers 1 4 9 and 16 appear.

This is correct. Turns out I didn't do enough iterations myself. That makes the answer the square root of 1000, or ~31 doors.