# Number Divisibility

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Question 1 |

The sum of all 3 digit numbers divisible by 3 is:

165150 | |

164380 | |

168420 | |

165250 |

**Arithmetic Aptitude 2**

**Number Divisibility**

**Discuss it**

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 ?

546702 | |

556824 | |

367312 | |

467536 |

**Arithmetic Aptitude 2**

**Number Divisibility**

**Discuss it**

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?

6 | |

5 | |

4 | |

0 |

**Arithmetic Aptitude 2**

**Number Divisibility**

**Discuss it**

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:

1200 | |

800 | |

1000 | |

900 |

**Arithmetic Aptitude 2**

**Number Divisibility**

**Discuss it**

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. = 3^{2}x 2^{2}x 5^{2}= 900

Question 5 |

How many numbers between 20 and 451 are divisible by 9?

44 | |

48 | |

50 | |

52 |

**Arithmetic Aptitude 4**

**Number Divisibility**

**Discuss it**

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?

4, 0 | |

4, 5 | |

1, 0 | |

8, 0 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

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?

12 | |

14 | |

15 | |

13 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

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?

10 | |

2 | |

7 | |

9 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

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:

9980 | |

9870 | |

9540 | |

9640 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

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……..

44 | |

47 | |

48 | |

49 |

**Arithmetic Aptitude 6**

**Number Divisibility**

**Discuss it**

Question 10 Explanation:

Pattern is (2^{2}- 1), (3^{2}- – 1), (4^{2}- – 1), (5^{2}- – 1), .... Next number will be = 7^{2}- – 1 = 49 – 1 = 48

There are 16 questions to complete.