Numbers

Question:
A number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. The number is:

[A].

214

[B].

476

[C].

954

[D].

1908

Answer: Option A

Explanation:

4 | x z = 6 x 1 + 4 = 10
———–
5 | y -2 y = 5 x z + 3 = 5 x 10 + 3 = 53
———–
6 | z – 3 x = 4 x y + 2 = 4 x 53 + 2 = 214
———–
| 1 – 4

Hence, required number = 214.

Question:
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be

[A].

1, 2

[B].

2, 3

[C].

3, 2

[D].

4, 1

Answer: Option B

Explanation:

4 | x y = (5 x 1 + 4) = 9
——–
5 | y -1 x = (4 x y + 1) = (4 x 9 + 1) = 37
——–
| 1 -4

Now, 37 when divided successively by 5 and 4, we get

5 | 37
———
4 | 7 – 2
———
| 1 – 3

Respective remainders are 2 and 3.

Question: A number when divide by 6 leaves a remainder 3. When the square of the number is divided by 6, the remainder is:
[A].

0 

[B].

1

[C].

2

[D].

3

Answer: Option D

Explanation:

Let x = 6q + 3.

Then, x2 = (6q + 3)2

   = 36q2 + 36q + 9

   = 6(6q2 + 6q + 1) + 3

Thus, when x2 is divided by 6, then remainder = 3.

Question:
3251 + 587 + 369 – ? = 3007

[A].

1250

[B].

1300

[C].

1375

[D].

1200

Answer: Option D

Explanation:

3251 Let 4207 – x = 3007
+ 587 Then, x = 4207 – 3007 = 1200
+ 369
—-
4207
—-

Question: How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336
[A].

4

[B].

5

[C].

6

[D].

7

Answer: Option A

Explanation:

132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264 11,3,4 (/)

396 11,3,4 (/)

462 11,3 (X)

792 11,3,4 (/)

968 11,4 (X)

2178 11,3 (X)

5184 3,4 (X)

6336 11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.

Question:
Which of the following numbers is divisible by each one of 3, 7, 9 and 11 ?

[A].

639

[B].

2079

[C].

3791

[D].

37911

Answer: Option B

Explanation:

639 is not divisible by 7

2079 is divisible by each of 3, 7, 9, 11.

Question: How many of the following numbers are divisible by 3 but not by 9 ?
2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276
[A].

5

[B].

6

[C].

7

[D].

None of these

Answer: Option B

Explanation:

Marking (/) those which are are divisible by 3 by not by 9 and the others by (X), by taking the sum of digits, we get:s

2133 9 (X)

2343 12 (/)

3474 18 (X)

4131 9 (X)

5286 21 (/)

5340 12 (/)

6336 18 (X)

7347 21 (/)

8115 15 (/)

9276 24 (/)

Required number of numbers = 6.

Question:
If the number 653 xy is divisible by 90, then (x + y) = ?

[A].

2

[B].

3

[C].

4

[D].

6

Answer: Option C

Explanation:

90 = 10 x 9

Clearly, 653xy is divisible by 10, so y = 0

Now, 653×0 is divisible by 9.

So, (6 + 5 + 3 + x + 0) = (14 + x) is divisible by 9. So, x = 4.

Hence, (x + y) = (4 + 0) = 4.

Question:
If x and y are positive integers such that (3x + 7y) is a multiple of 11, then which of the following will be divisible by 11 ?

[A].

4x + 6y

[B].

x + y + 4

[C].

9x + 4y

[D].

4x – 9y

Answer: Option D

Explanation:

By hit and trial, we put x = 5 and y = 1 so that (3x + 7y) = (3 x 5 + 7 x 1) = 22, which is divisible by 11.

(4x + 6y) = ( 4 x 5 + 6 x 1) = 26, which is not divisible by 11;

(x + y + 4 ) = (5 + 1 + 4) = 10, which is not divisible by 11;

(9x + 4y) = (9 x 5 + 4 x 1) = 49, which is not divisible by 11;

(4x – 9y) = (4 x 5 – 9 x 1) = 11, which is divisible by 11.

Question: A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?

[A].

10

[B].

11

[C].

12

[D].

15

Answer: Option A

Explanation:

4 a 3 |
9 8 4 } ==> a + 8 = b ==> b – a = 8
13 b 7 |

Also, 13 b7 is divisible by 11      (7 + 3) – (b + 1) = (9 – b)

  (9 – b) = 0

  b = 9

(b = 9 and a = 1)     (a + b) = 10.