Using the digits from 0 to 9, how many 3-digit numbers can be constructed such that the number must be odd and greater than 500 and digits may be repeated?

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Answer 1 · 2017-11-11T06:16:16

$250$ $250$ numbers

If the number is $A B C$ $ABC$ , then:

For $A$ $A$ , there are $9$ $9$ possibilities: $5, 6, 7, 8, 9$ $5,6,7,8,9$

For $B$ $B$ , all the digits are possible. There are $10$ $10$

For $C$ $C$ , there are $5$ $5$ possibilities. $1, 3, 5, 7, 9$ $1,3,5,7,9$

So the total number of $3$ $3$ -digit numbers is:

$5 \times 10 \times 5 = 250$ $5xx10xx5 = 250$

This can also be explained as:

There are $1000, 3$ $1000,3$ -digit numbers from $000 \to 999$ $000 to 999$

Half of them are from $500 \to 999$ $500 to 999$ which means $500$ $500$ .

Of those, half are odd and half are even.

Hence, $250$ $250$ numbers.

Answer 2 · 2017-11-11T06:26:36

250 numbers

The 1st digit must be greater than or equal to 5 for the number to be greater than 500. There are 5 possibilities (5, 6, 7, 8, 9).

The 2nd digit has no restriction on it. There are 10 possibilities (0-9).

The 3rd digit must be odd in order for the number to be odd. There are 5 possibilities (1, 3, 5, 7, 9).

$5 \cdot 10 \cdot 5 = 250$ $5*10*5=250$ numbers