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
OA C
Source: Manhattan GMAT
If integer k is equal to the sum of all even multiples of 15
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- Jay@ManhattanReview
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The first even multiples of 15 between 295 and 615 is 300 and the last one is 600.BTGmoderatorDC 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
OA C
Source: Manhattan GMAT
Thus, the terms are 300, 330, 360, ... 600.
Total number of terms = [(600 - 300)/30] + 1 = 11 terms
Sum of terms = Number of terms*Average of the first and the last term = 11*[(300 + 600)/2] = 11*450
Taking prime factors of 11*450, we have 11*450 = 11*9*5*10 = 11*3^2*5^2*2
The greatest prime is 11.
The correct answer: C
Hope this helps!
-Jay
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Multiples of 15: 15, 30, 45, 60, 75, 90, 105, etcBTGmoderatorDC 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: 30, 60, 90, 120, ....
So k = 300 + 330 + 360 + ... + 570 + 600
300 = 30(10)
330 = 30(11)
360 = 30(12)
390 = 30(13)
.
.
.
570 = 30(19)
600 = 30(20)
So k = 30(10 + 11 + 12 + ... + 19 + 20)
------------------------------------------------------
Let's examine this sum: 10 + 11 + 12 + ... + 19 + 20
Since 20 - 10 + 1 = 11, we know there are 11 numbers to add together.
Since these red numbers are equally spaced (consecutive integers), their sum = (# of values)(average of first and last values)
= [11][(10+20)/2]
= [11][15]
= (11)(15)
-------------------------------------------------
So, k = 30(10 + 11 + 12 + ... + 19 + 20)
= 30(11)(15)
= (2)(3)(5)(11)(3)(5)
We can see that 11 is the greatest prime factor of k
Answer:C
Cheers,
Brent
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The smallest even multiple of 15 between 295 and 615 is 15(20) = 300, and the largest is 15(40) = 600. Therefore, k = 300 + 330 + 360 + ... + 600 = (300 + 600)/2 x 11 = 450 x 11 = 45 x 10 x 11 = 3^2 x 5 x 2 x 5 x 11. So the largest prime factor of k is 11.BTGmoderatorDC 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
Answer: C
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