Number System 2

 

Find the total number of factors of 15120

50

60

70

80

Ans:  80

 

The greatest among the numbers 3√2, 3√7, 6√5, 2√20 is

6√5

3√7

2√20

None of these

Ans:  6√5

 

 

A 4-digit number is formed by repeating a 2-digit number such as 2525, 3232, etc., Any number of this form is always exactly divisible by 

Smallest 3-digit prime number

7

11

13

Ans:  Smallest 3-digit prime number

 

 

Find the no. of prime factors in the product of 25^12 × 10^7 x 14^7

50

68

54

52

Ans: 52

25^12× 10^7 x 14^7 

=(5*5)^12 × (2*5)^7 ×(2*7)^7 

=5^12 × 5^12 × 2^7 × 5^7 × 2^7 × 7^7 

=5^31 × 2^14 × 7^7 

Now add the power to find the number of prime factors  31+14+7 =52 is option D.

 

 

How many of the following numbers are divisible by 3 but not by 9. 

4320, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276

5

6

7

None of these

Ans: 5

 

A 3-digit number 4p3 is added to another 3-digit number 984 to give the four-digit number 13q7, which is divisible by 11. Then, (p + q) is :

10

11

12

15

Ans: 10

(1+q)-(3+7)=0 

q=9

q=8+p

p=1

 

 

What should be the maximum value of Q in the following equation? 5P9 – 7Q2 + 9R6 = 823

5

6

7

9

Ans:  7

⇒ 5P9 - 7Q2 + 9R6 = 823
⇒ (500 + 10P + 9) - (700 + 10Q + 2) + (900 + 10R + 6) = 823
⇒ (500 + 900 - 700) + 10 (P + R - Q) + (9 + 6 - 2) = 823
⇒ 700 + 10 (P + R - Q) = 810
⇒ 700 + 10 (P + R - Q) = 700 + 110
⇒ 10 (P + R - Q) = 110
⇒ P + R - Q = 11
⇒ Q = (P + R - 11)
To get maximum value of Q we take P = 9 and R = 9
This gives Q = (9 + 9 - 11) = 7
Hence, the maximum value of Q is 7

 

 

What is the number in the unit place in (729)^59?

1

6

7

9

Ans: 9

729^58 * 729 = 729^59

729^58 gets unit place is 1 and then multiply again 729 gets unit place is 9.

 

 

The digit in unit’s place of the product 71 × 72 × .......× 79 is

0

2

6

8

Ans: 0

To calculate it consider only unit digits i.e. 1 * 2 * 3 * 4 *5 *6 *7 *8 *9 = 24 * 30 * 56 * 9

 

 

Find the remainder when 67^99 is divided by 7

4

6

1

2

Ans: 4

1. It doesn't matter how much power is there on that particular number.
2. Now divide 67 by 7, you will find that remainder is 4.
3. Now if you multiple 67, 99 times then also remainder will be 4
   

 

 

A number when successively divided by 3, 4 and 9 leaves respective remainders of 2, 3 and 7. What will be the remainders if the original number is divided successively by 3, 5 and 7?

2,2,6

6,1,2

2,4,6

6,4,2

Ans: 

 

 

When a number divided by 342 gives remainder 47. When the same number is divided by 19, what would be the remainder?

5

9

4

1

Ans: 9

On dividing the given number by 342 let k be the quotient and 47 as remainder. Then number – 342 k + 47 = 19 x 18k + 19 x 2 + 9 = 19 18k + 2 + 9. The given number when divided by 19 gives 18k + 2 as quotient and 9 as remainder.

 

 

How many keystrokes are needed to type numbers from 1 to 1000 on a standard keyboard?

3001

2893

2704

2890

Ans:  2893

• We need to 10 key strokes to type numbers from 0–9
• Now, if we need to type numbers from 10–19 we’ll need 10 (for typing 1’s)+10 (for typing 0–9) = 20 keystrokes
• Similarly if we need to type numbers from 20–29 we’ll need 10 (for typing 2’s) + 10 (for typing 0–9) = 20 keystrokes
• Pattern follows up to 99
• So to type numbers from 0–99, we’ll need to type 20*9 (for typing numbers from 10–99)+10 (for typing numbers from 0–9) = 180 + 10 = 190
• Now to type numbers from 100–199, we’ll need 100 (for typing 1’s) + 200 (for typing 00–99, here we are considering 0’s before 0–9 so 10 extra ) = 300
• Similarly for typing numbers from 200–299, we’ll need 100(for typing 2’s) + 200 (for typing 00–99) = 300
• Therefore to type 0–999 we’ll need 300*9(to type 100–999) + 190 (to type 0–99) = 2890 keystrokes.
• Typing 1000 will take 4 keystroke, So to type 0–1000 we’ll need 2894 keystrokes
• Therefore, to type 1–1000 we’ll need 2894–1=2893 keystrokes.
  

 

 

The numerator of a certain fraction is 8 less than the denominator. If 3 is added to the numerator and 3 is subtracted from the denominator, the fraction becomes 3/4. Find the original fraction?

2/5

7/9

3/11

8/5

Ans: 3/11

 

 

A number when divided by 3 leaves a remainder 1. When the quotient is divided by 2, it leaves a remainder 1. What will be the remainder when the number is divided by 6?

2

3

4

5

Ans:  4

3q+1 = p

2 x +1 = q

In the above equation q is equal to 2x+1 which is the form of an odd number so we can say that q is an odd number. Hence, q can be any odd number which is 3, 5, 7, 9 etc.

3(3)+1 = 10

10/6 remainder 4.

 

The difference between the local value and face value of 7 in the numeral 657903 is :

7896

6993

903

1662

Ans: 6993

 

 

It is being given that (2^32 + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?

2^16 + 1

2^16 - 1

7 x 2^23

2^96 + 1

Ans:  2^96 + 1

Let 2³² = x. Then, (2³² + 1) = (x + 1).

Let (x + 1) be completely divisible by the natural number N. Then,

(2⁹⁶ + 1) = [(2³²)³ + 1] = (x³ + 1) = (x + 1)(x² - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

 

 

Which of the following numbers will completely divide (4^61 + 4^62 + 4^63 + 4^64) ?

3

10

11

13

 Ans: 10

(4⁶¹ + 4⁶² + 4⁶³ + 4⁶⁴) = 4⁶¹ x (1 + 4 + 4² + 4³) = 4⁶¹ x 85

     = 4⁶⁰ x (4 x 85)

     = (4⁶⁰ x 340), which is divisible by 10.

 

 

 

(51+ 52 + 53 + ... + 100) = ?

2525

2975

3225

3775

Ans:  3775

51 + 52 + 53 + ...........+ 100
 = (1 + 2 + 3 + .... + 100) - (1 + 2 + 3 + ...... + 50)
 = It is in the form of n(n+1)2 series summationn(n+1)2 series summation
= n1 = 100 , n2 = 50
=[100(100+1)/2] − [50(50+1)/2] =10100/2 - 2550/2
 =  (5050 - 1275) = 3775

 

 

If the number 97215 * 6 is completely divisible by 11, then the smallest whole number in place of * will be:

1

2

3

5

 Ans: 3 

22 - (8+*) = 0 or multiple of 11
14 - * = 0 or multiple of 11
Here multiple of '11' is the best option which is '11' itself
14 - * = 11
Hence * = 3
Hence '3' is the smallest number that should replace * so that the number 97215*6 is divisible by 11.