Double Divisibility

Let n denote an arbitrary integer between $102$ and $999$ . If $2|n$ and $5|n$ , then $10|n$ .

First, consider the set of numbers between $100$ and $999$ . Since we want to count the number divisible by $10$ , this forces ones digit to be $0$ . In this case,

• For hundreds digit, there are $9$ possible choices from $1$ to $9$ .
• For tens digit, there are $10$ possible choices from $0$ to $9$ .
• Ones digit is fixed.

So there are $9 \cdot 10 = 90$ numbers divisible by $10$ in the hundred range. Thus, for the problem, since $100$ does not occur between $101$ and $999$ , we have $90 - 1 = \boxed{89}$ numbers.

If a number is divisible by 2 and 5, then it is divisible by 10. A number that way would be of the form 10n. We want to know how many numbers of the form 10n are between 10 * 10.1 and 10 * 99.9. We want natural numbers, so the actual range is 10 * 11 and 10 * 99. The answer is the amount of numbers from 11 to 99, which is 89.

• We can solve this using AP
• For a number to be divisible by both 2 and 5 it must be divisible by 10.
• So we can use the formula a+(n-1)d
• The n th term will be 990 as it is the last number before 999 which is divisible by 10.
• a=110
• Substituting we get,
• $n=89$