Discussion
Home ‣ Aptitude ‣ Number System Comments
- Question
- Which one of the following number is divisible by 11?
Options- A. 45678940
- B. 54857266
- C. 87524398
- D. 93455120
- Correct Answer
- 93455120
ExplanationWe know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11.
For 93455120
Sum of even places = 3 + 5 + 1 + 0 = 9
Sum of odd places = 9 + 4 + 5 + 2 = 20
There difference = 20 - 9 = 11
which is divisible by 11 so 93455120 is divisible by 11. - 1. How many numbers between -11 and 11 are multiple of 2 or 3
Options- A. 11
- B. 14
- C. 15
- D. None of these Discuss
- 2. The number 58129745812974 is divisible by
Options- A. 11
- B. 9
- C. 4
- D. None of these. Discuss
- 3. A number when divided by a divisor leaves remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
Options- A. 13
- B. 59
- C. 35
- D. 37 Discuss
- 4. A number when divided by 5, leaves 3 as a remainder. What will be the remainder when the square of this number is divided by 5.
Options- A. 3
- B. 4
- C. 5
- D. 0 Discuss
- 5. On dividing a certain number by 357, the remainder is 39. On dividing the same number by 17, what will be the remainder?
Options- A. 5
- B. 3
- C. 7
- D. 6 Discuss
- 6. When 17 200 divided by 18, find the remainder.
Options- A. 1
- B. 4
- C. 5
- D. 3 Discuss
- 7. What is the remainder when 4 1000 is divisible by 7?
Options- A. 1
- B. 2
- C. 4
- D. None of these. Discuss
- 8. A common factor of ( 41 43 + 43 43 ) and ( 41 41 + 43 41 ) is ...
Options- A. ( 43 - 41 )
- B. ( 4141 + 4341 )
- C. ( 4143 + 4343 )
- D. ( 43 + 41 ) Discuss
- 9. The remainder when 9 19 + 6 is divided by 8 is
Options- A. 2
- B. 3
- C. 5
- D. 7 Discuss
- 10. What will be remainder when 19 100 is divided by 20?
Options- A. 19
- B. 29
- C. 3
- D. 1 Discuss
Number System problems
Search Results
Correct Answer: 15
Explanation:
Number between 0 and 11 which are multiple of 2 or 3
= 11/2 + 11/3 - 11/6 = 5 + 3 - 1
= 7
Number between 0 and -11 which are multiple of 2 or 3
= 11/2 + 11/3 - 11/6 = 5 + 3 - 1
= 7
? Number be 15, including 0
Correct Answer: 11
Explanation:
We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11.
So number is 58129745812974
Sum of digit at odd places = 4 + 9 + 1 + 5 + 7 + 2 + 8 = 36
Sum of digit at even places = 7 + 2 + 8 + 4 + 9 + 1 + 5 = 36
So, required difference = 36 - 36 = 0
? number is divisible by 11.
Correct Answer: 37
Explanation:
Let the divisor be x and quotient be y.
Then, the number = xy + 24
Twice the number = 2xy + 48
Now 2xy is completely divisible by x.
On dividing 48 by x remainder is x
? x = 48 - 11
= 37
Correct Answer: 4
Explanation:
Let the number be N.
Therefore N = 5k + 3
On squaring both sides, we get
N2 = ( 5k + 3)2
= 5( 5k2 + 6k +1 ) + 4
? On dividing x2 by 5, the remainder is 4
Correct Answer: 5
Explanation:
Let the given number be N
Therefore N = 357k + 39
Then,
357 k + 39 = ( 17 x 21k ) + ( 17 x 2 ) + 5
? Required remainder = 5
Correct Answer: 1
Explanation:
We know that, ( xm - am ) is divisible by ( x + a ), for even value of m.
? 17200 - 1200 is divisible by 17 +1
? 17200 - 1 is divisible by 18.
Hence when 17200 divided by 18 the remainder is 1.
Correct Answer: 4
Explanation:
Remainder of 41000/7 = (42)500/7 = (16)500/7
= 2500/7
= Remainder of 4 x 8166/7
= 4
Correct Answer: ( 43 + 41 )
Explanation:
We know that when m is a odd number ( xm + am ) is divisible by ( x + a ).
? Each one is divisible by (41 + 43).
So Common factor = (41 + 43).
Correct Answer: 7
Explanation:
(919 + 6 ) / 8 = [( 8 + 1 )19 + 6] / 8
? Required remainder = ( 119 + 6 ) / 8
? Remainder = 7
Correct Answer: 1
Explanation:
19100 = ( 20 -1 )100
? ( -1 )100 / 20
? 1 / 20
? Required Remainder = 1
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.