1. Number properties - Practice questions

2. 1. Find the least possible number which can be divided by 32, 36, 40
LCM of 32,36 and 40

3. 2. Find the number of numbers lying between 1 and 1000 which are divisible by each of 6, 7 and 15.
Lcm of 6,7 and 15 is 210 ,which is divisible by them. All the multiples of 210 lying between 1 and 1000 will be divisible i.e,210,420,630 & 840

4. 3. What is the least possible number which when divided by 24, 32 or 42 in each case it leaves the remainder 5?
LCM of (24,32 ,42)+5

5. 4. The least possible number of 3 digits when successively divided by 2,5,4,3 gives respective remainders of 1,1,3,1 is
372
275
273
193
Let N be the 3 digit number
N=2A+1=2(60D+36)+1=120D+73
A=5B+1=5(12D+7)+1=60D+36
B=4C+3=4(3D+1)+3=12D+7
C=3D+1
N=120+73=193

6. 5. A number when divided successively by 6, 7, 8, it leaves the respective remainders of 3, 5 and 4, what will be the last remainder when such a least possible number is divided successively by 8, 7, 6?

7. 6. Find the smallest positive number which is exactly divisible by 1/3, 1/2, 3/7, 4/11.

8. 7. Find the total number of factors for 10800

9. 8. Find the number of factors of 13!
2, 3, 5,7,11 are the prime no less than 12.
No of 2’s in 12! =10
No of 3’s =5
No of 5’s= 2
No of 7=1
No of 11=1
No of factors=(10+1)*(5+1)*(2+1)*(1+1)*(1+1)= 11*6*3*2*2=792

10. 9. Find the sum of factors of 270.
270=2^1 *3^3 *5^1
Sum of factors= (2^0 +2^1)* (3^0+3^1+3^2+3^3)*(5^0+5^1)=3*40*6=720

11. 10. Find the number of factors of 7056

12. 11. Find the highest power of 15 in 225!
No. of factors divided by 2

13. 12. Find the no of 26 in 200!
No of factors divided by 2

14. 13. Find the number of zeros in 133!
No of 5’s in 133! Are 32… so 32 zeros

15. 14. Find the highest power of 12 in 100!
12=2*2*3. No of 2’s in 100 is 97 so 4’s will be 48 and no of 3’s is also 48 . Total 48 12’s

16. 15. Number of zeros at the end of the following expression: (5. )5
No of zeros in 5! Is 1 and 5!=120 so der will be 120 zeors at end

17. 16. Find the last digit of
(6+4)=10, 0 will be the last digit

18. 17. Find the last digit of 3232^32 .
Last digit will be 6

19. 18. Find the last digit of the expression: 12 + 22 + 32 + 42 +…….+1002.
( )10=0 will be the last digit

20. 19. Find the unit digit of 11 + 22 + 33 + …..+1010.
1 = 1 2^2= 4 3^3 = 27 4^4 = unit digit will be 6 5^5= unit digit 5, 6^6 = unit digit 6 7^7 = unit digit 3 8^8 =unit digit 6
9^9 = unit digit 1
10^ 10= unit digit 0.
so( )=39, 9 will be the unit digit

21. 20. Find the unit digit of the expression: 888888. + 222222. + 333333
888! Will be having at least two zeros at the end so be comletely divisible by 4,so the answer will be ( )=4 unit digit

22. 21. The last digit of the following expression is: (1!)1 + (2!)2 + (3!)3 + ……+ (10!)10.
1! = 1,(2!)^2 = 4, (3! )^3= (3 × 2!)^3 = 216,(4!)^4 = (4 × 3! )^4= 13824, (5! )^5=( 5 × 4! )5= = 12 × ! = 6 × 5! = 720 = 72 × 10 7! = something × 10 and so on. So the last digit will be( )=5

23. 22. What is the remainder of 1421 * 1423 * 1425 when divided by 12 ?
1421, 1423 and 1425 gives 5, 7 and 9 as remainders respectively when divided by 12.
Remainder [ (1421 * 1423 * 1425 ) / 12 ] = Remainder [ (5 * 7 * 9) ] / 12, gives a remainder of 3.

24. 23. Find the reminder when 1. + 2. + 3. +. 99. + 100
23. Find the reminder when 1! + 2! + 3! ! + 100! is divided by the product of first 7 natural numbers
From 7! the remainder will be zero. Why ?  because 7! is nothing but product of first 7 natural numbers and all factorial after that will have 7! as one of the factor. so we are concerned only factorials till 7!, i.e, 1! + 2! + 3! + 4! + 5! + 6!
1! + 2! + 3! + 4! + 5! + 6! = 873 and as 7! > 873 our remainder will be 873

25. 24. What is the remainder when 444444^444 is divided by 7 ?
Remainder[ 444 444 ^ 444  / 7 ] = Remainder [ 3 444 ^ 444 / 7 ]
= Remainder [ ( 32  ) 222 ^ 444 / 7 ] = Remainder [ 2222 ^ 444 / 7 ] ( As Remainder [ 32 / 7 ] = 2 )
= Remainder [ ( 23 ) 74 ^ 444  / 7] =  Remainder [ 1 74 ^ 444  / 7 ] = 1 ( As Remainder [ 23 / 7 ] = 1

26. 25. What is the remainder when 49227 divided by 19?
49/19=R(11), 49^2/19=R(7), 49^3=R(1), remainder cycle is 11, 7,1 , so the answer is 7

27. 26. What is the remainder when 334334 is divided by 7?
334/7=R(5),334^2/7=r(4), 334^3/7=R(6), 334^4/7=R(2), 334^5/7=R(3), 334^6/7=R(1),So remainder cycle is 5, 4,6, 2, 3,1. so the remainder will be 4

28. 27. The average heights of three poles x,y,z is 45 cms
27. The average heights of three poles x,y,z is 45 cms. If average height of x and y is 40 cm and that of y and z is 43 cm, can you find the height of the second pole y.
334/7=R(5),334^2/7=r(4), 334^3/7=R(6), 334^4/7=R(2), 334^5/7=R(3), 334^6/7=R(1),So remainder cycle is 5, 4,6, 2, 3,1. so the remainder will be 4

29. 28. Find the average of first 10 odd numbers starting from 3.

30. 30. A boy scores an aggregate of 70% marks, scoring an average of 63% in 4 of the papers and 77% in the other papers. How many papers were there totally

31. 31. A batsman makes a score of 87 runs in the 17th innings and thus increases his average by 3. Find his average after 17 innings. 39

32. 32. 6 persons standing in queue are with different ages
32. 6 persons standing in queue are with different ages. After two years their average age will be 43. A seventh person joined with them and hence the current average age has become 45. Find the age of seventh person. 69

33. 33. ( ) = ?

34. 34. The average weight of two groups is 44 and 77
34. The average weight of two groups is 44 and 77. If a student of weight 66 from second group is shifted to the first, then which of the following is true
Average weight of first group decreases and second group decreases
Average weight of first group decreases and second group increases
Average weight of first group increases and second group decreases
Average weight of first group increases and second group increases