Answer:
2500 numbers
Step-by-step explanation:
Problem;
Solving for the number of 5-digit numbers divisible either by 45 or 60 but not divisible by 90;
Let us approach this in a step wise manner;
5 -digits numbers ranges from 10,000 to 99,999
To solve this problem, let us find the number of digits that are divisible by 45;
the first number divisible by 45 within this range is 10035
the last number divisible by 45 in this range is 99990
increment is 45
The total number in this range is:
[tex]\frac{last number - first number}{increment}[/tex] + 1 = [tex]\frac{99990 - 10035}{45}[/tex] + 1
= 2000 numbers
To solve this problem, let us find the number of digits that are divisible by 60;
the first number divisible by 60 within this range is 10020
the last number divisible by in this range is 99960
increment is 60
[tex]\frac{99990 - 10080 }{90}[/tex]
The total number in this range is:
[tex]\frac{last number - first number}{increment}[/tex] + 1 = [tex]\frac{99960 - 10020}{60}[/tex] + 1
= 1500 numbers
To solve this problem, let us find the number of digits that are divisible by 90;
the first number divisible by 90 within this range is 10080
the last number divisible by in this range is 99990
increment is 90
The total number in this range is:
[tex]\frac{last number - first number}{increment}[/tex] + 1 = [tex]\frac{99990 - 10080}{90}[/tex] + 1
= 1000 numbers
Now solution:
Number of 5 digits = number of 45 + number of 60 - number of 90
= 2000 + 1500 - 1000
= 2500 numbers
