‘Z’ is a two digit number in which the unit place digit is ‘P’ and the tens place digit is ‘Q’. The value of ‘P’ is smaller than the value of ‘Q’. The product of P and Q is the multiple of three. The unit place digit of (P) ²  ,(P) ³ and (P) ⁴ is the same. The difference between the values of P and Q is 4. If the unit and tens place digit of the number ‘Z’ are interchanged with each other, then which of the following is the nearest prime number greater than the newly formed two digit number (after interchanging digits).

Solution

‘Z’ is a two digit number in which the unit place digit is ‘P’ and the tens place digit is ‘Q’. So the number is ‘QP’.    Eq.(i) The value of ‘P’ is smaller than the value of ‘Q’. The product of P and Q is the multiple of three. PxQ = multiple of three    Eq.(iii) The unit place digit of (P) ²  ,(P) ³ and (P) ⁴ is the same. When P = 1, 5 and 9 , then only the above given condition will be true. The difference between the values of P and Q is 4. From Eq.(ii) and the above given condition, (Q-P) = 4 When P = 1, then (Q-1) = 4 Q = 4+1 = 5 When P = 5, then (Q-5) = 4 Q = 4+5 = 9 When P = 9, then (Q-9) = 4 [This is not possible. Because in this case ‘Q’ will be a two digit number.] So Z = QP ⇒ 51 and 95 Because of Eq.(iii), (Z = 51) will not be possible. So Z = 95 If the unit and tens place digit of the number ‘Z’ are interchanged with each other, then the newly formed number = 59 The nearest prime number greater than the newly formed two digit number (after interchanging digits) = 61