The statements form an ironic combo. If ‘nobody goes there’, how can the place ‘always remain crowded‘?

T

The letters forming the circle (from top left,  clock-wise) are separated by the number equal to the numeric position of the letter at the centre. e.g. In the first one, the central letter is ‘A’, which is at numeric position 1 in the alphabet, and the letters forming the circle (F, G, H, I) are one letter apart. In the second image, the central letter is D (4th) and the  letters on the circle (G(7), K(11), O(15) and S(19)) are 4 places apart. In the third image, the central letter is ‘G‘ (No. 6) Accordingly, the circle should be composed of H (8), N (8 + 6 = 14), T  (14 + 6 = 20) and Z  (20 + 6 = 26)

Another repeat puzzle. Repeat of the ‘house on fire‘ puzzle.

1. P and Q cross together first – 2 min. (Even though P can cross in 1 min, he has to slow down because Q is slower than him AND both require the torch.)
2. P returns with the torch in 1 min. (because he’s alone and hence can move at his maximum speed)
3. R & S go together – 10 min.  (at the speed of S, the slower of the two)
4.  Q who is at the other end takes the torch and returns – 2 Min.
5. Both P and Q cross again – 2 Min.

Total Time: 2 + 1 + 10 + 2 + 2 = 17 Min.  The bridge collapses, but all have crossed safely.

Square No. 23, at the bottom left.  It is under square 17 (red), 22 (blue) and 24 (pink), and above square 18 (red)

31 Years ago, Krish was 14 (45 – 31 = 14) and Cristina was 42 (73 – 31 = 42). 14 x 3 = 42.

92

Needs a ‘trick’ to solve. (Otherwise, there is no solution because sum of three odd numbers can never be an even number.)

13 + 11 + 6 = 30 (use the ‘9’ upside down to make ‘6’)

126

62 squares of one unit each (excluding the two containing the pieces) +
41 Squares of 2 x 2 sides each +
18 Squares of 3 x 3 sides each +
4  Squares of 4 x 4 sides each +
1 Square of 5 x 5 (lower left corner)

Total: 62 + 41 + 18 + 4 + 1 = 126

Larger squares (6+ unit sides) are can’t be considered because any such square will contain at least one piece. There are 9 squares of 6 x 6, 4 squares of 7 x 7 and one 8 x 8 square (which is the entire board)

18 ? 12 ? 4 ? 5 = 59 => 18 * 12 / 4 + 5 = 59

The third wish’ should be to lose half his wealth. By the ‘double’ logic, the opponent would lose double. 🙂