Peter Y. answered 05/08/14
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Starting with 105, the multiples of 35 (the product if 5&7) are:
105,140, and 175. Of those three only 105 is divisible by three. Thus 105 is the only number we need to exclude from our count of multiples of three.
Now for the multiples of three. We use the formula for an arithmetic sequence: an=a1+d(n-1)
105,140, and 175. Of those three only 105 is divisible by three. Thus 105 is the only number we need to exclude from our count of multiples of three.
Now for the multiples of three. We use the formula for an arithmetic sequence: an=a1+d(n-1)
the first number divisible by three is 102, the last is 189. So plugging in: 189=102+3(n-1). Solving for n:
87=3(n-1)
29=n-1
30=n
So there are 30 numbers between 100 and 200 divisible by three. Subtracting the one divisible by 5&7 our final answer is 29
Nantu B.
08/29/18