Number Divisibility

12
Question 1
The sum of all 3 digit numbers divisible by 3 is:
A
165150
B
164380
C
168420
D
165250
Arithmetic Aptitude 2    Number Divisibility    
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Question 1 Explanation: 
All 3 digit numbers divisible by 3 are :
102, 105, 108, 111, ..., 999.

This is an A.P. with first element 'a' as 
102 and difference  'd' as 3.

Let it contains n terms. Then,
102 + (n - 1) x 3 = 999 
102 + 3n-3 = 999
3n = 900 or n = 300
Sum of AP = n/2 [2*a  + (n-1)*d]
Required sum = 300/2[2*102 + 299*3] = 165150. 
Question 2
Which of Following is not divisible from 4 ?
A
546702
B
556824
C
367312
D
467536
Arithmetic Aptitude 2    Number Divisibility    
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Question 2 Explanation: 
If last 2 digits are divisible by 4, then number is also divisible by 4.
In case of 546702 last 2 digits are 02 (x)
In case of 556824 last 2 digits are 24 (/)
In case of 367312 last 2 digits are 12 (/)
In case of 467536 last 2 digits are 36 (/) 
Question 3
What should be the value of * in 985*865, if number is divisible by 9?
A
6
B
5
C
4
D
0
Arithmetic Aptitude 2    Number Divisibility    
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Question 3 Explanation: 
If a number is divisible by 9, the sum of digits is also divisible by 9
9 + 8 + 5 + * + 8 + 6 + 5 = 9x
41 + * = 9x
Nearest value of 9x must be 45
41 + * = 45
* = 4
Question 4
The least perfect square, which is divisible by each of 15, 20 and 36 is:
A
1200
B
800
C
1000
D
900
Arithmetic Aptitude 2    Number Divisibility    
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Question 4 Explanation: 
LCM of 15, 20 and 36 is 180

Now 180 = 3 x 3 x 2 x 2 x 5

To make it perfect square, it must
be multiplied from 5.

So required no. = 32 x 22 x 52 = 900
Question 5
How many numbers between 20 and 451 are divisible by 9?
A
44
B
48
C
50
D
52
Arithmetic Aptitude 4    Number Divisibility    
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Question 5 Explanation: 
The required numbers are 27, 36, 45……450.

This is an A.P. with a = 27 and d = 9

Let it has n terms.

Then Tn = 450 = 27 + (n-1) x9

∴ 450 = 27+ 9n - 9

∴ 9n = 432

∴ n = 48
Question 6
Which digits should come in place of * and $ if the number 4675*2$ is divisible by both 5 & 8?
A
4, 0
B
4, 5
C
1, 0
D
8, 0
Arithmetic Aptitude 6    Number Divisibility    
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Question 6 Explanation: 
Since the given number is divisible by 5, 0 or 5 must come in place of $. But a number ending with 5 is never divisible with 8 because a divisible number must be even. Therefore 0 will replace $. If the number is divisible by 8, the number formed by last three digits must be divisible by 8. The number *20 must be divisible by 8. Among all 4 options, only placing 1 in place of * makes it divisible by 8.
Question 7
What least number must be added to 4000 to obtain a number exactly divisible by 17?
A
12
B
14
C
15
D
13
Arithmetic Aptitude 6    Number Divisibility    
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Question 7 Explanation: 
Let us divide 4000 by 17.

17) 4000 (235
       34
       ….
         60
         51
        …..
           90
           85
          …..
             5

We get 5 as remainder.
Number to be added = 17-5 = 12
Question 8
What least number must be subtracted from 5000 to get a number exactly divisible by 23?
A
10
B
2
C
7
D
9
Arithmetic Aptitude 6    Number Divisibility    
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Question 8 Explanation: 
Let us divide 5000 by 23.

23) 5000 (217
      46
      ….
        40
        23
        ….
         170
         161
         …..
            9

We get 9 as remainder.

Therefore required number to be subtracted = 9
Question 9
The largest 4 digit number exactly divisible by 5, 6 and 7 is:
A
9980
B
9870
C
9540
D
9640
Arithmetic Aptitude 6    Number Divisibility    
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Question 9 Explanation: 
The required number must be divisible by L.C.M. of 5,6 and 7.
L.C.M. of 5, 6 and 7 = 5 x 6 x 7 = 210

Let us divide 9999 by 210.

210) 9999 (47
      840
     ----
      1599
      1470
      ----
       129

Required number = 9999 – 129 = 9870
Question 10
34. What should be the next number in below series? 3, 8, 15, 24, 35……..
A
44
B
47
C
48
D
49
Arithmetic Aptitude 6    Number Divisibility    
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Question 10 Explanation: 
Pattern is (22- 1), (32- – 1), (42- – 1), (52- – 1), ....

Next number will be = 72- – 1 = 49 – 1 = 48
There are 16 questions to complete.
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