Expected CAT question

EXPECTED CAT QUESTION

Q1:. Let O and E represent odd and even digits respectively. ∴ S can have digits of the form O _ O _ E or O _ E _ O or E _ O _ O
Case 1: O _ O _ E The first digit can be chosen in 3 ways out of 1, 3 and 5 The third can be chosen in 2 ways. The fifth digit can be chosen in 2 ways after which the second and fourth digits can be chosen in 2 ways. ∴ There are 3 × 2 × 2 × 3 = 24 ways in which this number can be written. 12 out of these ways will have 2 in the rightmost position and 12 will have 4 in the rightmost position. ∴ The sum of the rightmost digits in Case 1 = (12 × 2) + (12 × 4) = 72
Case 2: O _ E _ O This number can again be written in 24 ways. 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position. Thus the sum of the rightmost digits in Case 2 = (8 × 1) + (8 × 3) + (8 × 5) = 72
Case 3: E _ O _ O This number can also be written in 24 ways. As in Case 2, 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position. ∴ The sum of the rightmost digits in Case 3 = (8 × 1) + (8 × 3) + (8 × 5) = 72 ∴ The sum of the digits in the rightmost position of the numbers in S = 72 + 72 + 72 = 216 Hence, option 2.

 

Q2:. 302720 = 32720 × 102720 The rightmost non zero digit of 302720 will be the digit in the unit’s place of 32720. 3’s power cycle is 3, 9, 7, 1 and cyclicity is 4. 2720 = 680 × 4 ∴ The digit in the unit’s place of 32720 is 1. ∴ The rightmost non-zero digit of 302720 is 1. Hence, option 1.