Numbers

Question:
If n is a natural number, then (6n2 + 6n) is always divisible by:

[A].

6 only

[B].

6 and 12 both

[C].

12 only

[D].

by 18 only

Answer: Option B

Explanation:

(6n2 + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.

Question:
The difference of the squares of two consecutive odd integers is divisible by which of the following integers ?

[A].

3

[B].

6

[C].

7

[D].

8

Answer: Option D

Explanation:

Let the two consecutive odd integers be (2n + 1) and (2n + 3). Then,

(2n + 3)2 – (2n + 1)2 = (2n + 3 + 2n + 1) (2n + 3 – 2n – 1)

     = (4n + 4) x 2

     = 8(n + 1), which is divisible by 8.

Question:
The difference of the squares of two consecutive even integers is divisible by which of the following integers ?

[A].

3

[B].

4

[C].

6

[D].

7

Answer: Option B

Explanation:

Let the two consecutive even integers be 2n and (2n + 2). Then,

(2n + 2)2 = (2n + 2 + 2n)(2n + 2 – 2n)

     = 2(4n + 2)

     = 4(2n + 1), which is divisible by 4.

Question:
On multiplying a number by 7, the product is a number each of whose digits is 3. The smallest such number is:

[A].

47619

[B].

47719

[C].

48619

[D].

47649

Answer: Option A

Explanation:

By hit and trial, we find that

47619 x 7 = 333333.

Question: On dividing 2272 as well as 875 by 3-digit number N, we get the same remainder. The sum of the digits of N is:
[A].

10

[B].

11

[C].

12

[D].

13

Answer: Option A

Explanation:

Clearly, (2272 – 875) = 1397, is exactly divisible by N.

Now, 1397 = 11 x 127

The required 3-digit number is 127, the sum of whose digits is 10.

Question:
7589 – ? = 3434

[A].

4242

[B].

4155

[C].

1123

[D].

11023

Answer: Option B

Explanation:

Let 7589 -x = 3434

Then, x = 7589 – 3434 = 4155

Question:
The difference between the place value and the face value of 6 in the numeral 856973 is

[A].

973

[B].

6973

[C].

5994

[D].

None of these

Answer: Option C

Explanation:

(Place value of 6) – (Face value of 6) = (6000 – 6) = 5994