Answer:
To determine the number of 4-digit numbers that are multiples of at least one of the numbers 2 and 5, we can use the principle of inclusion-exclusion.
First, let's find the number of 4-digit numbers that are multiples of 2. The first 4-digit multiple of 2 is 1000, and the last 4-digit multiple of 2 is 9998. To find the count of multiples, we can divide the difference between these two numbers by 2 and add 1 (since we're including both endpoints):
Number of multiples of 2 = (9998 - 1000) / 2 + 1 = 4500
Next, let's find the number of 4-digit numbers that are multiples of 5. The first 4-digit multiple of 5 is 1000, and the last 4-digit multiple of 5 is 9995. Similar to before, we divide the difference between these two numbers by 5 and add 1:
Number of multiples of 5 = (9995 - 1000) / 5 + 1 = 1800
However, if we simply add these two counts together, we will be counting the multiples of 10 twice (since 10 is a multiple of both 2 and 5). To correct this, we need to subtract the number of 4-digit multiples of 10.
To find the number of 4-digit multiples of 10, we divide the difference between the first and last 4-digit multiples of 10 (1000 and 9990) by 10 and add 1:
Number of multiples of 10 = (9990 - 1000) / 10 + 1 = 900
Now we can use the principle of inclusion-exclusion to calculate the final count:
Number of 4-digit numbers that are multiples of at least one of 2 or 5 = Number of multiples of 2 + Number of multiples of 5 - Number of multiples of 10
= 4500 + 1800 - 900 = 5400
Therefore, there are 5400 4-digit numbers that are multiples of at least one of the numbers 2 and 5.