Top 25 Funny Brain Teasers

Top 25 Funny Brain Teasers

Watch Top 25 Funny Brain Teasers.
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32 thoughts on “Top 25 Funny Brain Teasers

  1. This brain training game“nonu amazing only” (Google it) will keep your mind energetic. While I`m still young, I want to exercise my brain by doing numerous things so I could lessen the chances of getting brain problems brought by retirement years. You can enhance your concentration capacity and memory by expending a couple of minutes a day in the game.

  2. Josephus problem. A group of n people are standing in a circle, numbered consecutively
    clockwise from 1 to n. Starting with person no. 2, we remove every other person, proceeding
    clockwise. For example, if n = 6, the people are removed in the order 2, 4, 6, 3, 1, and the
    last person remaining is no. 5. Let j(n) denote the last person remaining. Find some simple
    way to compute j(n) for any positive integer n > 1.
    Solution. Note that the problem does not ask for a ”simple mathematical formula” for j(n),
    because the solution is not quite easy to express using only ordinary mathematical symbols,
    however there is a very simple way to compute j(n) using binary notation:
    j(n) = left rotation of the binary digits of n
    This means that if n = x1x2x3 . . . xn, where the xk are the digits of the binary representation
    of n (with x1 6= 0) then
    j(n) = x2x3 . . . xnx1
    For instance if n = 366 (base 10) = 101101110 (in base 2), then we take the ’1’ in the left
    and move it to the right: 011011101, so j(366) = 011011101 (base 2) = 111 (base 10).
    For instance for n = 6 = 110 (base 2) the last person was j(6) = 5 = 101 (base 2).
    The answer also can be expressed as j(n) = 2m + 1, where m = n − 2
    , 2k = maximum
    power of 2 not exceeding n, i.e., 2k ≤ n < 2
    1 This can be proved in the following way:
    (1) First check that if n = power of 2, say n = 2k
    , then the last person remaining is
    always no. 1. This can be proved by induction on k. For k = 1 there are only two
    people, no 2 is removed and no. 1 remains. Then assume that the statement is true
    for a given k. Assume that n = 2k+1. Then people no. 2, 4, 6, . . . are removed. After
    removals all even numbered people will have been removed, leaving us with exactly
    the 2k odd numbered people no. 1, 3, 5, . . . By induction hypothesis we know that in
    this case the first person (i.e. no. 1) remains, so the statement (that no. 1 remains if
    n = power of 2) is also true for k + 1. This completes the induction and proves the
    statement for every k ≥ 1.
    (2) Finally, if n = 2k + m, where 0 ≤ m < 2
    k ≤ n, we start by removing the m people
    numbered 2, 4, 6, . . . , 2m. Now we have a circle with 2k people, and the ”first one”
    (which will remain at the end) at that point is no. 2m + 1.

  3. why is the police efforts not foucused on rapists and murders?

  4. 2:05 actually an elephant doesn’t walk on hands it walks on digitigrades. so actually you would be talking about a person with one hand. A one handed person probably can’t pick up a elephant. 🐘

  5. that was the worst thing that I’ve ever seen 👎👎👎👎👎👎👎👎👎👎👎👎👎👎👎👎👎

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