If xΩy is defined as the smallest integer of which both x and y are factors, then 10Ω32 is how much greater than 6Ω20?
xΩy = LCM(x,y)
10Ω32 = LCM(10,32)=160
6Ω20=LCM(6,20)=60
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please explain in detail If xΩy is a least integer of which x and y are both factors than 10Ω132 is how much greater than 6Ω20
the answer of this was little bit complicated please explain
To find the value of xΩy, we need to determine the smallest integer that is divisible by both x and y.
Let's start by calculating 10Ω32. To find the smallest integer divisible by both 10 and 32, we need to find their least common multiple (LCM).
Step 1: Find the prime factorization of both numbers:
- Prime factorization of 10: 2 × 5
- Prime factorization of 32: 2^5
Step 2: Identify the highest power of each prime factor:
- Highest power of 2: 2^5 (from 32)
- Highest power of 5: 5 (from 10)
Step 3: Multiply the highest powers of the prime factors:
- 2^5 × 5 = 10 × 32 = 320
Therefore, 10Ω32 is equal to 320.
Now let's calculate 6Ω20. Again, we need to find the least common multiple (LCM) of 6 and 20.
Step 1: Find the prime factorization of both numbers:
- Prime factorization of 6: 2 × 3
- Prime factorization of 20: 2^2 × 5
Step 2: Identify the highest power of each prime factor:
- Highest power of 2: 2^2 (from 20)
- Highest power of 3: 3 (from 6)
- Highest power of 5: 5 (from 20)
Step 3: Multiply the highest powers of the prime factors:
- 2^2 × 3 × 5 = 4 × 3 × 5 = 60
Therefore, 6Ω20 is equal to 60.
To find how much greater 10Ω32 is than 6Ω20, we subtract the value of 6Ω20 from 10Ω32:
10Ω32 - 6Ω20 = 320 - 60 = 260
Therefore, 10Ω32 is 260 greater than 6Ω20.