How many 3 digit number that are divisible by 5 can be formed using the numbers 0 2?

Answer

- Hint: Use the fundamental principle of counting to find the number of three digit numbers divisible by 5. Use the property that if a number is divisible by 5, then there are 0 or 5 at unit place. Complete step-by-step solution -

The fundamental principle of counting: According to the fundamental theorem of counting, if a task can be done in n ways and another task can be done in m ways, then the number of ways in which both the tasks can be done is mn, and the number of ways in which either of the tasks can be done is m+n.
In the case of forming numbers, we have to fill the decimal places. When we create a three digit number, we need to fill the units places, the tens place and the hundreds place.
Filling the hundreds place: The hundreds place cannot be filled by 0. Hence the number of ways in which units can be filled is 9.
Filling the units place: The units place can be filled with only two digits 0 or 5. This is because a number is divisible by 5 if and only if the digit at units place is 0 or 5.
Filling the tens place: The tens place can be filled in 10 ways.
Hence the total number of three digit numbers divisible by 5 is $9\times 2\times 10=180$.
Hence there are 180 three digit numbers divisible by 5.

Note: Alternative Solution:
We know that the number of numbers from 1-n divisible by k is $\left[ \dfrac{n}{k} \right]$, where [.] denotes the greatest integer function.
Hence the number of numbers from 1-99 divisible by 5 $=\left[ \dfrac{99}{5} \right]=19$
The number of numbers from 1-999 divisible by 5 $=\left[ \dfrac{999}{5} \right]=199$
Hence the number of three digit numbers divisible by 5 is $199-19=180$, which is the same as obtained above.

How many 3-digit numbers divisible by 5, can be formed using the digits 2 3 5 6 7 and 9, without repetition of digits?

1. 216
2. 20
3. 120
4. 24

Answer (Detailed Solution Below)

Option 2 : 20

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NIMCET 2020 Official Paper

120 Questions 480 Marks 120 Mins

Given,

Numbers that can be used = {2,3,5,6,7,9} = 6

The three digits number has to be formed without repetition

∴ Let the three digit number be abc

Now c is the unit digit and for the number to be divisible by 5, the unit digit can be 5

∴ c = {5} = 1 way

b is the tenth digit and can be formed using numbers 2,3,6,7 and 9

∴ b = {2,3,6,7,9} = 5 or b = 5C1 = 5 

c is the hundredth digit and can be formed using remaining 4 numbers has there can be no repetition of digits

∴ c = 4C1 = 4

∴ Possiblity of a three digit number divisible by 5 = 4 × 5 × 1 = 20 ways

Last updated on Sep 21, 2022

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