# Numbers Questions

Q:

Find the remainder when 1234 x 1235 x 1237 is divided by 8.

 A) 4 B) 2 C) 0 D) 6

Explanation:

When we divide 1234 by 8, remainder is 2

When we divide 1235 by 8, remainder is 3

When we divide 1237 by 8, remainder is 5

$\inline \fn_jvn \small \Rightarrow 2\times 3\times 5=30$

As 30 will not be the remainder because it is greater than 8,

when 30 divided by 8, remainder = 6.

Filed Under: Numbers - Quantitative Aptitude - Arithmetic Ability
Exam Prep: CAT
Job Role: Bank PO

6 300
Q:

The difference between the squares of two consecutive odd integers is always divisible by ?

 A) 8 B) 2 C) 6 D) 4

Explanation:

Let the two consecutive odd integers be (2x + 1) and (2x + 3)

Then,

(2x + 3)2 - (2x + 1)2

= (2x + 3 + 2x + 1) (2x + 3 - 2x - 1)

= (4x + 4)(2)
= 8 (x + 1), which is always divisible by 8

4 274
Q:

What is the smallest number by which 2880 must be divided in order to make it into a perfect square ?

 A) 2 B) 3 C) 4 D) 5

Explanation:

By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.

4 270
Q:

What should be the maximum value of B in the following equation?
8A9 – 6B2 + 4C6 = 723

 A) 4 B) 1 C) 5 D) 7

Explanation:

1 1
8 A 9
+ 4 C 6
-  6 B 2
7 2 3
We may represent the given sum, as shown below:

1 + A + C - B = 12
A + C - B = 11

Now,giving the maximum values to A and C, i.e.

A = 9 and C = 9, we get B = 7.

4 240
Q:

How many zeros are there from 1 to 10000 ?

 A) 2893 B) 4528 C) 6587 D) 4875

Explanation:

For solving this problem first we would break the whole range in 5 sections

1) From 1 to 9
Total number of zero in this range = 0

2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)

3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180

4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893