Peter Y. answered • 05/08/14
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Starting with 56, the multiples of 56 (the product if 8&7) are:
56,112,168,224,280,336, and 392. Of those 7 only 168 and 336 are divisible by 6. Thus 169 and 336 are the only numbers we need to exclude from our count of multiples of 6.
Now for the multiples of 6. We use the formula for an arithmetic sequence: an=a1+d(n-1)
The first number divisible by 6 is 6, the last is 396. So plugging in: 396=6+6(n-1). Solving for n:
390=6(n-1)
65=n-1
66=n
So there are 66 numbers between 1 and 400 divisible by 6. Subtracting the two divisible by 8&7 our final answer is 64
56,112,168,224,280,336, and 392. Of those 7 only 168 and 336 are divisible by 6. Thus 169 and 336 are the only numbers we need to exclude from our count of multiples of 6.
Now for the multiples of 6. We use the formula for an arithmetic sequence: an=a1+d(n-1)
The first number divisible by 6 is 6, the last is 396. So plugging in: 396=6+6(n-1). Solving for n:
390=6(n-1)
65=n-1
66=n
So there are 66 numbers between 1 and 400 divisible by 6. Subtracting the two divisible by 8&7 our final answer is 64
Peter Y.
Anytime! Glad I could help.
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