**I take a 3 digit number with distinct digits. I can get 6 different 3 digit numbers by rearranging the digits of this number.**

__Question:__**Options:**

A. | True | |

B. | False | |

C. | 50 | |

D. | 51 |

**Answer: Option A**

: A

Let the3 digit number beabc.On rearrangingthedigits I get, abc, bac, cba, cab, acb, bca. Hence Iwill get 6 three digit numbers.

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## More Questions on This Topic :

- 1
- 5
- 3
- 2

**Answer: Option B**

: B

The sum of three A's is such anumber that hasA in its ones place. Therefore, the sum of two A's must be a number whose ones digit is 0. This happens only for A = 0 and A = 5 If A = 0, then the sum is 0 + 0 + 0 = 0, which makes B = 0.

Given that B is a natural number, thereforeB cannot be equal to 0. So we don't considerthis possibility.

Hence, A = 55+5+515 Therefore, A= 5 and B= 1

- 23
- 49
- 50
- 51

**Answer: Option C**

: C

From 1 to 4 there are 2 numbers divisible by 2, which are 2 and 4. From 1 to 10, there are 5 numbers, which are 2, 4, 6, 8, 10. Hence, the number of numbers divisible by 2 is half the number up to which we count. Half of 100 is 50. Hence 50 is the number of numbers divisible by 2 from 1 to 100.

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