use algebraic manipulation to find the minimum sum-of-products expression for the function x1x2'x3' + x1x2x4+x1x2'x3x4'
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We need to find the number of integers between 100 and 500 that can be divided by 6, 8, or both. Now, to do this, we must as to how many are divisible by 6 and how many are multiples of 8.
The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have
We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.
That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.
Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.
Answer: 101
The closest number to 100 that is divisible by 6 is 102. 498 is the multiple of 6 closest to 500. To find the number of multiple of 6 from 102 to 498, we have
We can use the same approach, to find the number of integers that are divisible by 8 between 100 and 500.
That means there are 67 integers that are divisible by 6 and 50 integers divisible by 8. Remember that 6 and 8 share a common multiple of 24. That means the numbers 24, 48, 72, 96, etc are included in both lists. As shown below, there are 16 numbers that are multiples of 24.
Since we counted them twice, we subtract the number of integers that are divisible by 24 and have a final total of 67 + 50 - 16 = 101. Hence there are 101 integers that are divisible by 6, 8, or both.
Answer: 101