It is pretty interesting. Not only can it tell us whether the number is divisible by 7 but also the exact remainder obtained if the number is divided by 7.

How this works is that each node is connected to a remainder by 7. That is each node has a unique number in [0..6] associated with it. And the rest is simple modular arithmetic.

To enumerate the associated numbers, the bottom-most node has 0 associated with it. From there, keep following the black arrows- you’ll obtain the node associated with the next number. As in the right-most node is associated with 1, the topmost with 2 and so on. You’ll notice that the leftmost node is associated with 6 which in turn leads to 0.

You should also notice that the white arrows lead up from nodes n mod 7 to nodes n∗10 mod 7. This is important.

Now let us consider a number N with digits N1,N2...Nm with Nm being the unit digit.

We begin applying the given algorithm:

1. We follow the black arrow N1 number of times and arrive at the node N1 mod 7.
2. Next we follow the white arrow to arrive at the node N1∗10 mod 7.
3. We repeat step 1 to follow the black arrow N2 number of times. Now we are at N1∗10+N2 mod 7.
4. We follow step 2 again to arrive at N1∗100+N2∗10 mod 7.
5. Applying the same procedure until the last digit, we finally arrive at the node N1∗10m−1+N2∗10m−2+..+Nm mod 7 which equals N mod 7. If we are at the bottommost node, then we conclude that N mod 7=0 and hence proving the divisibility of N by 7.

If anything is unclear, feel free to leave comments.

If you’re interested in the concept, type in the phrase “casting out nines” in your search engine of choice.
In the example you gave, the basic idea is that you know that X and Y have the same remainder when divided by 9 since they share the same digits1. Hence, the difference must be a multiple of 9. Since we know the sum of the digits of any multiple of 9 is itself a multiple of 9, you can deduce the last digit.
(In the case that the sum is itself a multiple of 9, you can use the fact that they hid a non-zero digit to deduce that the remaining digit is in fact a 9.)

1. Burn the rope 1 from one end and rope 2 from both ends
2. As soon as rope 2 burnt completely,
3. We have 30 minutes gone and 30 minute left in rope 1
4. Burn the other side of rope 2 and start counting (30 Minute left in rope 2)
5. Fire is from both end so, rope will be burnt in 15 minute and hence , we get 15 minutes

Her name is emerald June…

With, 20 years being emerald (anniversary name) and 6th month being June..
THUS … Emerald June.

For any given number, divide them into two group of equal number.
Say we have 10 Coins, and we have to divide them into 5-5’s Group,

First 5 would contain say x Heads, and ( 5- x ) tails.
while other group would contain ( 5- x ) heads and x Tails.

Just flip one Group totally, and it would be same as the other Group. in other words,
x heads and ( 5- x ) tails would convert into ( 5- x ) heads and x tails. which is similar to other group.

From: 6. Xera and Rir had never met before the tournament.

We know that these 2 are not from the same group.

From: 7. Xera will be visiting Pyi’s group when the Yomi go on a special excursion to that part of the planet.

We know that Xera is not part of the yomi, Pyi is.

From: 8. Pyi admires Teta’s colorful feathers, as well as her ability to soar, and once watched her and her teammate in the territory of the Grundi.

We know that Pyi is a yomi, so Teta cannot be. Teta is a Grundi bc she was in that territory and has never left home before now.

so..
Xera and Rir in diff group
Xera is not yomi
Pyi is yomi
Teta is grundi

thus Xera = Uti, Wora = yomi, teta = grundi
because Xera is not yomi and teta is already the grundi, so xera must be uti.

Now Pyi = yomi, Rir = grundi, vel = uti
because rir cannot be in the same group as Xera, and Pyi has already been named the yomi. Thus, vel is the uti.

The names of the Uti and the winners are:
Xera and Vel.

We Have-

4 Colors (Blue, Brown, Green, Red)
4 Martians: (Aken, Bal, Mun, Wora)
4 Groups: (Uti, Grundi, Yomi, Rafi)

1) Gurundi is Red
2) We know from the evidence that Wora is not Green, Wora is not in the Rafi, Mun is not in the Uti, and Bal is not in the Yomi.
3) The Yomi can’t be blue or brown (the argument), and can’t begreen, therefore Yomi = Green.
4) Bal isn’t Blue or Brown (the argument) and therefore can’t be in the Uti or Rafi (because Uti is blue or brown and Rafi is b lue or brown) and must be a red gurundi.
5) the Rafi and Brown agreed, thus the Rafi must be blue, the Uti must be brown.
6) Wora was caught agreeing with brown (whom Yomi disagreed with), therefore Wora is Blue.

Wora from Rafi has the blue feather.

If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So….

Number the coins 1 through 12.

1. Weigh coins 1,2,3,4 against coins 5,6,7,8.

1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).

1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.

1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don’t balance, you know that either 9 or 10 is light, so the top coin is the fake.

1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don’t balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.

1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.

1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.

1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.

1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.

1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.