Each of X alarm tolls at regular intervals. All of them tolls together twelve times a day. No two alarm at equal intervals of time. If each alarm tolls after a whole number of minutes, what is the maximum possible value of X?

Answer: A The alarm tolls together twelves times a day. Therefore, they toll together once every 2 hours (or 120 minutes). Since no two alarms toll at equal intervals of time, the total number of distinct factors of 120, including 1 and 120 itself = 2^{3}(3)(5) The number of factors = (3+1)(2)(2) = 16. The maximum value of X is 16.

Q. No. 2:

How many three digit numbers can be formed using the digits 1,2,3,4,5,6,7 and 8 without repeating the digits and such that the tens digit is greater than the hundreds digit and less than the units digit?

Answer: B Ten's digit = 7 => units digit =8 => Hundred's digit =1,2,3,4,5,6. => Number of ways =1(6) Ten's digit = 6 => units digit =7,8 => Hundred's digit =1,2,3,4,5. => Number of ways =2(5) Ten's digit = 5 => units digit =6,7,8 => Hundred's digit =1,2,3,4 => Number of ways =3(4) Ten's digit = 4 => units digit =5,6,7,8 => Hundred's digit =1,2,3. => Number of ways =4(3) Ten's digit = 3 => units digit =4,5,6,78 => Hundred's digit =1,2 => Number of ways =5(2) Ten's digit = 2 => units digit =3,4,5,6,7,8 => Hundred's digit =1 => Number of ways =6(1) Total number of ways = 6+10+12+12+10+6 =56.

Q. No. 3:

A set of S consists of i). all odd numbers from 1 to 55. ii). all even numbers from 56 to 150. What is the index of the highest power of 3 in the product of all the elements of the set S?

Answer: A The product of all element of S is {3, 5, 7, 56, 58, 60 9, 11, 13, 62, 64, 66 ................................ ............................... 51, 53, 55, 146, 148, 150 } => { 3, 3*3, 3*5, .................3*17 3*20, 3*22, 3*24, .................(3*50)M} => 3^{9}(3)(9)(3)(3^{16})(24)(30)(36)(42)(148)N => 3^{35}S

Q. No. 4:

A person use the base x for their number system, where x<=10. A student had to add two-digit numbers. None of the digits was 0. By oversight, he reversed the digits of both the numbers and added them. He later found that the difference between his answer and the correct answer was (84)_{10}. If this value is the maximum possible for x, then what is the value of x.

Answer: C The difference between a number pq with non-zero digits and its reverse qp in base x is |p-q|(x-1). This is maximum when {p,q}= {x-1, 1}. This difference is (x-2)(x-1).For the sum of two numbers and the sum of their reverses. It is 2(x-2)(x-1). 2(x-2)(x-1) = 84. => x =8.

Q. No. 5:

Let X be a four-digit number with exactly three consecutive digits being same and is a multiple of 9. How many such X’s are possible?

Answer: D Let the four digit number be 'aaab' or ‘baaa’ Since, the number has to be a multiple of 9, therefore 3a + b should be either 9, 18 or 27. Case I: 3a + b = 9 Possible cases are: (1116, 6111, 2223, 3222, 3330, 9000) Case II: 3a + b = 18 Possible cases are: (3339, 9333, 4446, 6444, 5553, 3555, 6660) Case III: 3a + b = 27 Possible cases are: (6669, 9666, 8883, 3888, 7776, 6777, 9990) Hence total number of cases 20.

Q. No. 6:

Find the sun of the digits of the least natural number P, such that the sum of the cubes of the four smallest distinct divisors of P equals 2P.

Answer: A Let the least number be P, 1 is its least divisor. Let 2nd, 3rd and 4th least divisors be x,y and z respectively. We consider the following values of divisor a and the corresponding values of a^{3}, from x,y and z exactly 1 or all 3 are odd.(P is even) a=1 : a^{3}=1 a=2 : a^{3}=8 a=3 : a^{3}=27 a=4 : a^{3}=64 a=5 : a^{3}=125 a=6 : a^{3}=216 For x, y and z = (2,3,4), 2P = 100 (i.e P=50). But 3 is not a divisor of 50. For x,y,z = (2,3,6), 2P = 252 (i.e P = 126) and the 1,2,3,6 are four least distinct divisor of 126. The required number is 126. The sum of digits is 9.