To solve this problem, we need to find numbers that are divisible by both 20 and 60.
### Step 1: Calculate the Least Common Multiple (LCM) of 20 and 60.
The LCM of two numbers is the smallest number that is a multiple of both of them.
1. First, find the prime factorization of both numbers:
- [tex]\( 20 = 2^2 \times 5 \)[/tex]
- [tex]\( 60 = 2^2 \times 3 \times 5 \)[/tex]
2. To find the LCM, we take the highest power of each prime that appears in the factorizations:
- The highest power of 2 is [tex]\( 2^2 \)[/tex]
- The highest power of 3 is [tex]\( 3 \)[/tex]
- The highest power of 5 is [tex]\( 5 \)[/tex]
3. Thus, the LCM is:
[tex]\[
\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60
\][/tex]
### Step 2: Find the smallest 4-digit number divisible by 60.
A 4-digit number ranges from 1000 to 9999.
1. To find the smallest 4-digit number divisible by 60, divide 1000 by 60 and round up to the nearest whole number.
[tex]\[
\frac{1000}{60} \approx 16.6667
\][/tex]
Rounding up to the next whole number, we get 17.
2. Multiply 17 by 60 to find the smallest 4-digit number divisible by 60:
[tex]\[
17 \times 60 = 1020
\][/tex]
Thus, the smallest 4-digit number divisible by 60 is 1020.
### Step 3: Find the largest number less than 3000 that is divisible by 60.
1. To find the largest number less than 3000 divisible by 60, divide 3000 by 60 and round down to the nearest whole number.
[tex]\[
\frac{3000}{60} = 50
\][/tex]
2. Multiply 50 by 60 to find the largest number less than or equal to 3000 that is divisible by 60:
[tex]\[
50 \times 60 = 3000
\][/tex]
Since 3000 itself is divisible by 60, it is already less than or equal to itself, satisfying the problem condition.
### Summary
- The smallest 4-digit number divisible by both 20 and 60 is 1020.
- The number closest to and less than 3000 that is divisible by both 20 and 60 is 3000.
So, the final answers are:
- 1020 (smallest 4-digit number)
- 3000 (closest number less than or equal to 3000)
