Trick for The Lightneing Fast Addition!

Why lightning fast speed needed?

Gauss attached 0 to the series and made pairs of numbers having addition 100.

100 + 0 = 100

99 + 1 = 100

98 + 2 = 100

97 + 3 = 100

96 + 4 = 100

95 + 5 = 100
..
..
..
..
..
..
51 + 49 = 100

This way he got 50 pair of integers (ranging in between 1-100) having sum equal to 100.

So sum of these 50 pairs = 100x50 = 5000.

 
And the number 50 left added to above total to get sum of integers 1 - 100 as 5000 + 50 = 5050

Story of 7 Generous Dwarfs

The Seven Dwarfs are having breakfast, and Snow White has just poured them some milk. Before drinking, the dwarfs have a ritual. First, Dwarf #1 splits his milk equally among his brothers' mugs (leaving himself with nothing). Then Dwarf #2 does the same with his milk, etc. The process continues around the table, until Dwarf #7 has distributed his milk in this way. (Note that Dwarf #7 is named Dopey!) At the end, each dwarf has exactly the same amount of milk as he started with!

 

How much milk does each cup contain, if there were 42 ounces of milk altogether?

Finding difficult? Click here for answer! 

Behind the Story of 7 Generous Dwarfs

What was the story?

First thing is very clear that Dwarf 7 must have 0 ounces of milk at the start and end. Let's assume that 'a' be the maximum amount of milk (in ounces) that any dwarf has in his mug at any point of time. 

For a moment, let's assume Dwarf 1 himself has this 'a' amount of milk.

Now, when D1 distributes his 'a' amount of milk among 6 others, D7 receives 'a/6' amount of milk. At this point of time somebody else will be having maximum amount of milk 'a'. Let D2 be that person now having milk 'a'.

Next is D2's turn where he gives a/6 to all. So now D1 has a/6, D7 has 2a/6 and somebody else say D3 has maximum a. Continuing in this way, for each Dwarf's turn gives - 

Now, when we assumed D2 has maximum milk amount a after receiving a/6 from D1, then it's clear that he must had earlier 5a/6. Similarly, D3 had maximum amount of milk a after receiving a/6 from D1 and D2 indicates that he had 4a/6 milk initially. Continuing in this way, we can find the amount of milk each had initially like below.

So at any point of time, the milk distribution is like a, 5a/6, 4a/6, 3a/6, 2a/6, a/6, 0 where amounts are distributed among 7 dwarfs in cyclic order. But the total amount of milk available is 42 ounces.Hence, 

a + 5a/6 + 4a/6 + 3a/6 + 2a/6 + a/6 +  0 = 42

a + 15a/6 = 42

a + 5a/2 = 42

7a = 84

a = 12.

That means at any point of time the maximum amount of milk that one dwarf can have is equal to 12 ounces. And then others would have 10, 8, 6, 4, 2, 0.

To conclude, the 7 dwarfs had 12, 10, 8, 6, 4, 2, 0 ounces of milk initially.  

A Determined Cat on a Ladder!

A ladder is leaning against a wall. On the center rung is a cat. She must be a very determined cat, because she remains on that rung as we draw the foot of the ladder away from the tree until the ladder is lying flat on the ground. What path does the cat describe as she undergoes this indignity?

She follows this path!

A Path Followed by Determined Cat

What was the question?

 Interestingly, the cat follows the circular path whose center is at the foot of tree. 

Actually, as ladder is drawn out a series of right triangles with the same hypotenuse (the ladder) are created with respect to the foot of tree.

The point of hypotenuses where cat is sitting determinedly will be always at the same distance from all 3 vertices. So if all such point are joined then we get a circular path having center at the foot of tree. 

(Figure is for illustration purpose only & may not have accurate measurements.

Navigation Paths Between Two Points

Consider a rectangular grid of 4×3 with lower left corner named as A and upper right corner named B. Suppose that starting point is A and you can move one step up(U) or one step right(R) only. This is continued until B is reached. 

How many different paths from A to B possible ? 

Here is calculation of total number of paths. 

Combinations of Naviagation Paths

What is the question?

If the right move is represented as R and up move as U then, RRRUU is the one path to reach at B.

UURRR is one more path between points A and B.

URRRU is another way to reach at B.

Further, one can reach at B via RURRU.

So number of such paths are possible.

However, if all paths above are observed, we can conclude that total 5 moves are needed to reach from point A to B. Out of those 5, 3 have to be RIGHT and 2 have to be UP. 

That is, any combination having 3R and 2U in 5 moves will give a valid path to reach at B.

Now, number of ways 3R can be placed in 5 moves can be calculated as - 

C(5,3) = 5!/(5-3)! * 2! = 10.

To sum up, there are 10 paths available between points A and B.