dtermine the smallest positive integer that is divisible by each of the first ten numbers (1,2,3,4,5,6,7,8,9,10).
10*9*8*7 should do it because
the rest have factors included in those
9*8*7*5 is enough. 8*5 has the 5 and the 2 needed for 10.
To do this problem in an organized way try facoring each number
1 = 1
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 2*3
7 = 7
8 = 2*2*2
9 = 3*3
10 = 5*2
so
I need three twos for 8
I need two threes for 9
I need a five for 10 (already have the 2)
I need a seven for seven
I already have the fctors for one through four.
so
2*2*2 * 3*3 * 5 * 7
= 8 * 9 * 5 * 7 = 2520
To determine the smallest positive integer that is divisible by each of the first ten numbers (1, 2, 3, 4, 5, 6, 7, 8, 9, 10), we need to find the least common multiple (LCM) of these numbers.
Here's how you can find the LCM step-by-step:
1. Start by listing the prime factorization of each number:
- 1 has no prime factors.
- 2 = 2^1
- 3 = 3^1
- 4 = 2^2
- 5 = 5^1
- 6 = 2^1 * 3^1
- 7 = 7^1
- 8 = 2^3
- 9 = 3^2
- 10 = 2^1 * 5^1
2. Identify the highest power of each prime factor that appears in any number:
- The highest power of 2 is 3.
- The highest power of 3 is 2.
- The highest power of 5 is 1.
- The highest power of 7 is 1.
3. Multiply the prime factors raised to their respective highest powers:
LCM = 2^3 * 3^2 * 5^1 * 7^1
4. Calculate the result:
LCM = 8 * 9 * 5 * 7 = 2,520
So, the smallest positive integer that is divisible by each of the first ten numbers is 2,520.