| Q17) Find the unit digit in 1! + 2! + 3! + ..... + 49! 1) 3 2) 1 3) 0 4) 7 1! + 2! + 3! + 4! + (5! + ..... + 49!) n! ends with a 0 for n ≥ 5 ∴ 5! + ..... + 49! ends with a 0. 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 i.e. units place is 3 Hence, option 1. Q18) Which of the following numbers is divisible by 9. 1) 231897 2) 313243 3) 741082 4) 139275 A number is divisible by 9 if the sum of the digits of the number is divisible by 9. Sum of the digits of 139275 is 27 (27 is divisible by 9), so 139275 is divisible by 9. Hence, option 4. Q19) How many different numbers of the form 25y178x are divisible by 8. 1) 9 2) 10 3) 1 4) 0 A number is divisible by 8 if last three digits of a number are divisible by 8. The given number is divisible by 8 only when 'x' is 4. So, only 1 value of 'x' exists. As 'y' is a digit therefore, 'y' can take any value from 0 - 9. Thus, 10 values of 'y' exist. So, total numbers possible are 10. Hence, option 2. Q20) Find the units place of 1111×1212×1313×1414×1515 1) 0 2) 5 3) 1 4) 4 1111×1212×1313×1414×1515 contains a 12 and a 15. ∴ 12×15 = 180 Hence, option 1. |