M K H T A

  Pic of question The answer is 204 squares.   This is because you have to calculate how many 1 x 1 squares, 2 x 2 square, 3 x 3 squares and so on that are on the chessboard. These numbers end up being the square numbers: 64, 49, 36, 25, 16, 9, 4, 1 respectively.  Hence, the total number of squares on the chessboard are 64+49+36+25+16+9+4+1 = 1²+2²+3²+4²+5²+6²+7²+8² = 8×9×17÷6 [∵ 1²+2²+3²+...+n² = n(n+1)(2n+1)/6 ]  i.e ., 204 Answer related pic

Pic of question.  We solve it by differentiate the triangles in the image into three types: small, medium, and large triangles.  Number of small triangles = 12 triangles (3 + 4 + 3 + 2) that are up facing and 12 facing down.  Sum = 24 triangles.  Number of medium triangles = 6 triangles (3 + 2 + 1) facing up and 6 facing down.  Sum = 12 triangles.  Number of large triangles = 2. One facing up and one facing down. Hence, the total number of triangles = 24 + 12 + 2 = 38 triangles .

Pic of question.  Given conditions are (i) exactly one of the statements is true. (ii) exactly one box contains a rat. By given conditions we solve it by three cases. Case I- If first statement is true i.e., the rat is in box 1, then other two statements are false [by condition (i)] i.e., the rat is in box 2 and the rat is in box 1. Which is a contradiction by condition (ii). [Since, exactly one box can contains a rat but here box 1 as well as box 2 contains rat. ] Case II- If second statement is true i.e., the rat is not in box 2, then other two statements are false [by condition (i)] i.e., the rat is not in box 1 and the rat is in box 1. Which is again a contradiction. Case III- If third statement is true i.e., the rat is not in box 1, then other two statements are false [by condition (i)] i.e., the rat is not in box 1 and the rat is in box 2. This case satisfied the condition (ii).  Thus, the rat is in box 2. [By case III] Hence, the answer is box 2 . ...

 The correct answer of this question is 5672111 . The pattern for the question is:  (A×B) (A×C) A×(C+B)-B= Answer (3×4) (4×5) 3×(4+5) -4=122023 (4×5) (5×6) 4×(5+6) -5=203039 (5×6) (6×7) 5×(6+7)-6=304259 (6×7) (7×8) 6×(7+8) -7=425683 Hence,  (7×8) (8×9) 7×(8+9)-8=5672111

 The Number of squares in this Visual engima are 40. Big block 1 Small blocks 16 Center squares 2 Inside the center squares 8 2×2 block squares 9 3×3 block squares 4

We have, 1 is not prime. 101=1+100 is prime. 10101=1+100+100² is not prime. [ Since, 3|10101] Now,  Let, N= 10101...1 and numbers of zero in N is n(say). Then N=1+100+100²+...+100ⁿ     .........(i)              =(100ⁿ⁺¹-1)/(100-1)              ={(10ⁿ⁺¹)²-1}/99  Therefore, 99=(10ⁿ⁺¹-1)(10ⁿ⁺¹+1)/N If possible, let N is prime then N|10ⁿ⁺¹-1 or, N|10ⁿ⁺¹+1 Now, N=1+100+100²+...+10²ⁿ    [By (i)]              >1+10²ⁿ              >1+10ⁿ⁺¹ [Since, 2n>n+1 ∀n≥2] Then it's can't happen that, N|10ⁿ⁺¹-1 or, N|10ⁿ⁺¹+1 Therefore our assumption N is prime is wrong. So, N can't be prime because n>1. Here only one possibility for N is prime which is if n=1. [Where n is the number/s of zero in N]

Mathematics tells us three of the saddest love story... Tangent- lines that had one chance to meet then parted forever. Parallel- lines that will never meant to meet. Asymptotes- lines who can get closer and closer but it will never be together.