Choose Your Goblet

For this problem you can use modular arithmetic .

The first 12 steps are A,B,C,D,E,F,G,F,E,D,C,B and then it repeats itself again. So, we just need to see how many sets of 12 steps there are, and then find the remainder number of steps.

Since $1000\quad \equiv \quad 4\quad (mod\quad 12)$ , therefore you should pick the $\boxed{4th}$ goblet

1. Observe the pattern repeats from A to G and back to B *
2. Realise that you count 12 times from A to G and back to B
3. 1000/12 = 83 r 4
4. Count 4 from A --> leading you to D, answer 4 :)

*Common error: you don't go back to A, or else you will have two A's i.e. A,B,C,D,E,F,G,F,E,D,C,B, A,A

if we write down the sequence (1,2,3,4,5,6,7,6,5,4,3,2,1,2,3,4,...) we notice that the first number repeated at the 13th number, therefore there is a cycle of 12 numbers repeating. So, 1000 ( mod ) 12 =4. as a general form , we can write 12(k)=0, 12(k)+1=1 , 12(k)+2=2 . . . 12(k)+12=2.

Layman solution:

Count 6 from the left. Then count 6 from the right. Then count 6 from the left. And so on. In this way you always count 6 but you land (after a count of 6) at one of two critical goblets every time (2 and 6).

$\frac{1000}{6}=166.6...$ You can only make 166 repeats of 6 counts, ie. $166 \times 6 = 996$ counts.

As 166 is an even number, the critical goblet will be 2. Try doing the 6 counts, you will find that after every oddth group of 6 counts you land on 6 and after eventh group of 6 counts you land on 2.

You have made 996 counts. You need 4 more. So starting from goblet 2, four more brings you to 1, 2, 3, then 4. The 1000th count lands on goblet 4.

Simplest possible way of solving: count the glasses back and forth, until you've counted to ten. To every power of ten you've counted, that glass, in this case, Glass D, will be the one you're on. Since 1000 is just 10 to the power of 3, D is the answer.

The way I did it is :: The first time you go across the row (A,B,C,D,E,F,G), there are 7 numbers. But, since you don't count A or G twice before going back across the row, the remaining times you go back and forth there are only six numbers.

1000-7 for the first row = 993 993/6 numbers per row = 165.5

This means that you go back and forth 165 times, and the gold goblet is halfway across the next row, or 3 goblets from the end goblet, or goblet D.

Wow, I forgot about the winding back and forth part and I did 1000 = x mod 6 but I still got x as 4. Is there a reason why its the same answer or is it just simple coincidence?

The numbers we can get for A are: 1, 13, 25 ... =1(mod12) the numbers we can get for B are: 2, 12, 14, 24 ... =2(mod12) or 0(mod12) the numbers we can get for C are: 3, 11, 15, 23 ... =3(mod12) or 11(mod12) the numbers we can get for D are: 4, 10, 16, 22 ... =4(mod12) or 10(mod12) the numbers we can get for E are: 5, 9, 17, 21 ... =5(mod12) or 9(mod12) the numbers we can get for F are: 6, 8, 18, 20 ... =6(mod12) or 8(mod12) the numbers we can get for G are: 7, 19, 31 ... =7(mod12) 1000= 12x83+4 = 4(mod12) the answer is D

A-------B-------C-------D-------E-------F-------G

1-------2--------3-------4-------5-------6-------7

13----12-------11-----10------9-------8-------X

X------14-------15-----16-----17-----18-----19

25-----24-------23-----22-----21-----20------X

..............

..............

subtract 1st 7 count => 1000-7 = 993 count remains.

after 1st row, each row contains 6 counts backward and forward alternatively.

=> 993/6 = 165.5 = 165 + 0.5

165 is odd

odd < 165.5 < even

=> backward complete and middle of forward counting.

now 0.5x6 = 3

3rd count while forward progress is D

=> D => 4

check:

D goblet count series is 4, 6, 16, 22, 28, .....

=> n-th term is 6n−2

=> 6n−2=1000 => n = 167

which is an integer => ok (Checked)

First you go to $7th glass$ . That is $G$ Now after $6$ step you will go to $A$ And again $6$ step ,again 7. Then $1000 = 7 +6k + something$ . Here $k = 165\rightarrow odd , and ,something = 3$
Which means $A$ and then it will go $3$ step which comes to $\rightarrow \boxed{D}$

The material used to make D is visibly different oops

Since I didn't know the formula pertinent to this problem, I counted for the case of 10 and got D. Then I counted for the case of 100 and got D again. I believe D would be the answer for all powers of 10 greater than 0.

If you count one loop, you find that each cup occurs once every 12 counts. Using this, I would try goblet A, the resulting equation being 1+ 12*X with X being a random number. I try this until X = 83 results in a value of 997. I count the remaining B=998, C=999, so $\boxed{D=1000}$ (which is the golden goblet).

Oh boy, this is my chance to strike it rich! I don't have the time to count 1000 times, though, so I need another solution.

Luckily for me, after every 12th goblet counted, the counting restarts at goblet 1. The largest multiple of 12 that is at most 1000 is 996 , so we know number 997 will be goblet A . Therefore, I only need to count four goblets and run away with my prize!

And just like that, goblet D is revealed to be pure gold . Now, this better sell for some big money.

We can take here modulo 7 to make the solution.First see that if here was mentioned the 16th goblets we can write this as, 16=7 * 2+2.And the goblet will be (2+2)th or 4th goblet counting from the front side(taste it yourself).But if it was mentioned 9th goblet,where 9=7 * 1+2,the answer would be (1+2)th or 3th from the back side.Why?Because here the quotient is odd(1).So if K=7 * a+b where a is even, the K th goblet will be the (a+b)th goblet counting from the front side and when a is odd the Kth goblet will be the (a+b)th goblet counting from the back side. Now come to the point,1000=7 * 142+6.So the goblet will be the (142+6) or 148th goblet. We can do the same thing again,148=7 * 21+1.So the goblet will be the (21+1)th or 22th goblet.Note that here we don't need that the goblet is conting from where because we haven't reach the solution yet.Finally 22=7 * 3+1,so the goblet will be the (3+1)th or 4th goblet.The solution will remain same if you count it from front or back side. So the goblet will be the 4th goblet.

first and last goblet always follow the pattern 6x + 1, first goblet when x is even, while it's last goblet when x is odd. observing first integer dividable by 6 before 1000, it's 996. The x would be even. So the first goblet follows the same pattern and it's 997. It follows B = 998, C=999, D=1000

A more intuitive solution: First we write down the goblets: A B C D E F G The we write the numbers that ends up on each of them when we count them as the problem says A B C D E F G 1 2 3 4 5 6 7 13 12 11 10 9 8 14 15 16 17 18 19 We notice that goblets A, C, E and G will always have an odd number, and since 1000 is even we can rule them out. Then we can see that the numbers of goblet D increase by 6 every time. Since the starting number for goblet D is 4, we need 996 more to reach 1000, and since $996 \equiv 0 \pmod{6}$ We find out that goblet D will be the 1000th

Easy sol: This has a cycle 12 and it repeats this means that we can use modular arithmetic

$1000 \mod12$

we just need the remainder because its like we've counted many times to get only A and count from this to the remainder that is 4 , so it would be

A B C D E F G F E D C B

Since I haven't learned modular arithmetic, I solved for the maximum number divisible by 6 that is below 1000. That number is 996. And since the pattern repeats D F F D B B, any number divisible by 6 is the goblet B. So counting from B, it should be goblet D.

A B C D E F G

1 2 3 4 5 6 7

14 13 12 11 10 9 8

15 16 17 18 19 20 21

Let's choose any goblet, I chose D and the pattern of D is as follows: 4+, 10-, 16+, 22-, 28+,...........n the numbers form an arithmetic sequence, while the sign is positive when n is odd and negative when n is even

for 1000th term:

$4+6 \cdot (n-1)=1000$

$6 \cdot (n-1)=996$

$n= 167$

since n is an integer, D is already the gold goblet. We don't have to calculate the sign.

What if n is not an integer, what if the 1001st goblet is the golden one, which one would it be, how would we know whether it's C or E? That can be determined using the sign: n is odd so the sign is positive and E is the 1001st goblet.

You don't even need to use modular arithmetic. Just notice that you pattern in counting restarts every 12, and 12 divides 996, so when you have counted 996, you just restart at A. So 997 is A...1000 is D.

I just count to 1,000 which takes lesser than 5 minites if you are good at quick counting and get low human errors.

Just Kidding.I just count to 100 and it ended up in D.Repeat that 10x and it confirmes land in D.

Another way is to observe that counting 12 times will end up in B easily.So I took 1000/12 and the remainder would be 4.Count that from A and you end up in D.

Basic Java code to solve it:

    public class Goblet{
    //Since it's ABCDEFG (and then it goes back) FEDCB (and then it starts over again). So we         can use % to get the module, thus the answer.
    static int numberOfGoblets = 12;

    public static void main(String[] args){
        System.out.println(1000%numberOfGoblets);
    }
}

goblet A=25, B=50 C=75 D=100 thus with increment of every 25 we move to the next goblet hence goblet D is the 1000th. ie goblet 4.