yangsta8 wrote:
If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?
A. 5
B. 7
C. 11
D. 13
E. 17
Even multiples of 15 are integers which are multiples of both 2 and 15, which means these numbers should be a multiple of LCM(2, 15) = 30. Thus, the question is to determine the sum of multiples of 30 between 295 and 615. To do that, we will use the following formula:
Sum = Average * Number of terms
To determine the number of terms, we will use the following formula:
Number of multiples of 30 in an interval = 1 + [(the largest multiple of 30 in that interval - the smallest multiple in that interval)/30]
The smallest multiple of 30 between 295 and 615 is 300, and the largest multiple of 30 in the same interval is 600. Thus, the number of multiples of 30 between 295 and 615 is:
1 + [(600 - 300)/30] = 1 + 300/30 = 1 + 10 = 11
So, number of terms = 11.
Next, we determine the average of the multiples of 30 between 295 and 615. To do that, we will use the fact that multiples of 30 between 295 and 615 is a set of evenly spaced integers, and the average of a set of evenly set of integers is given by (largest term + smallest term)/2. Thus:
average = (largest multiple of 30 between 295 and 615 + smallest multiple of 30 between 295 and 615)/2
average = (600 + 300)/2 = 900/2 = 450
We can conclude that the sum of all multiples of 30 between 295 and 615 is 11 * 450. Since 450 = 2 * 3^2 * 5^2, the greatest prime factor of k = 11 * 2 * 3^2 * 5^2 is 11.
Answer: C _________________
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