Each angle is 180(p-2)/p. 180-{360}/{p} = k So 360/p has to be an integer. 360 = 2^3 * 3^2 * 5^1 So there are 4 * 3 * 2 = 24 possibilities, but we exclude 1 and 2, because p > = 3 So , 24 -2 = 22 Hence, choice (c) is the right answer

Warning - There might be an easier method! f(1) + f(6) = 0f(1) + f(4) + 24 = 0f(1) + f(2) + 8 + 24 = 0f(1) + f(0) = -32 f(1) + f(2) + f(3) + f(4) + f(5) + f(6)= f(1) + f(6) + f(0) + f(1) + 3 + f(2) + 8 + f(3) + 15= 0 + f(0) + f(1) + 3 + f(0) + 8 + f(1) + 3 + 15= 2 (f(0) + f(1)) + 29= 2 (-32) + 29= -64 + 29 = -35Option D

Let the two digit number be represented as ab. a^2 – b^2 = (a – b)(a + b) Now, |a^2 – b^2| will be a prime number only when |a – b| = 1 and (a + b) comes out to be a prime number Two digit numbers in which |a – b| = 1 are 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87,89, 98 From these numbers, the numbers in the form of 'ab' in which (a + b) comes out to be a prime number are 12, 21, 23, 32, 34, 43, 56, 65, 67, 76, 89 and 98. There are 12 two digit numbers such that absolute difference of the square of the two digits results in a prime number.

Practice Problem - 5 : Seven children A, B, C, D, E, F and G started walking from the same point at the same time, with speeds in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 respectively and they are running around a circular park. Each of them carry flags of different colours and whenever two or more children meet, they place their respective flag at that point. However nobody places more than 1 flag at a same point. They are running in anti-clockwise direction. How many flags will be there in total, when there will be no scope of putting more flags? Solution : When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be |a – b| A and B will meet at |1 - 2| = 1 point.A and C will meet at |1 - 3| = 2 pointsA and D will meet at |1 - 4| = 3 pointsA and E will meet at |1 - 5| = 4 pointsA and F will meet at |1 - 6| = 5 pointsA and G will meet at |1 - 7| = 6 pointsSo A will put 1 + 2 + 3 + 4 + 5 + 6 = 21 flags. similarly B and C will meet at |2 - 3| = 1 pointB and D will meet at |2 - 4| = 2 pointsB and E will meet at |2 - 5| = 3 pointsB and F will meet at |2 - 6| = 4 pointsB and G will meet at |2 - 7| = 5 pointsSo B will put 1 + 2 + 3 + 4 + 5 = 15 flags Similarly find for C, D, E and F. We will get 21 + 15 + 10 + 6 + 3 + 1 = 56 flags

1000 = 2^3 * 5^ When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases .. Case -1 ; ( 1, 2^3, 5^3) ..now here ordered arrangements = 3^2*3! = 54 Case - 2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)now total arrangement of these cases is 3!/2! ( 15) = 45 case -3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1 So finally 54 + 45 + 1 = 100

@sibanand_pattnaik number of scores possible = 251 - 15 + 1 ...= 237 Could anyone explain this last step ?

@sibanand_pattnaik b18081

Q30)