Aptitude Discussion

Q. |
How many 3-digit numbers do not have an even digit or a zero? |

✖ A. |
60 |

✖ B. |
80 |

✔ C. |
125 |

✖ D. |
150 |

**Solution:**

Option(**C**) is correct

There are 5 digits that are not even or zero: $1, 3, 5, 7$ and $9$.

Now, let’s count all the three-digit numbers that can be formed from these five digits.

The first digit of the number can be filled in 5 ways with any one of the mentioned digits.

Similarly, the second and third digits of the number can be filled in 5 ways.

Hence, the total number of ways of forming the three-digit number is **125** $(= 5×5×5)$.

**Deeksha Verma**

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**Anand Gupta**

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Alternative interpretation: three digit numbers do not have even or zero : {111, 113, 115, 117, 119} {131, 133,135, 137, 139} . Basically 5 such digits in a group of 20 . Therefore, total digits will be 25 x 8 = 200. Above solution is true if digits are not allowed to be repeated.

As it's been asked all the no's should be odd(neither even nor zero).

1,3,5,7,9 are odd no's which can be used while framing odd numbers.

1,3,5,7,9 can once be used in units place, once in tense place and once in the hundreds place.

As 1,3,5,7,9 are 5 no's and can be used in 3 places.

so, 5^3

Which is equal to 125.

111,113,.......,119

131.....139

..............

........................ 911, 913.....919

931.....939 and so on.