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Joined: 20 Feb 2017
Posts: 2506
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If integer k is equal to the sum of all even multiples of 15 between 2
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18 May 2021, 01:05
18 May 2021, 01:05
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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
A. 5
B. 7
C. 11
D. 13
E. 17
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Location: India
WE:Education (Education)
If integer k is equal to the sum of all even multiples of 15 between 2
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19 May 2021, 11:37
19 May 2021, 11:37
1
GeminiHeat 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
A. 5
B. 7
C. 11
D. 13
E. 17
Even multiple of 15 between 295 and 615: 300, 330, 360, 390, 420, 450, 480, 510, 540, 570, 600
Sum = K = Average x number of terms = \(\frac{(600+300)}{2}(11) = (450)(11) = (2)(3^2)(5^2)(11)\)
Hence, option C
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Re: If integer k is equal to the sum of all even multiples of 15 between 2
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21 Jul 2021, 06:22
21 Jul 2021, 06:22
2
GeminiHeat 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
A. 5
B. 7
C. 11
D. 13
E. 17
Multiples of 15: 15, 30, 45, 60, 75, 90, 105, etc
EVEN multiples of 15: 30, 60, 90, 120, ....
So k = 300 + 330 + 360 + ... + 570 + 600
Since each number is a multiple of 30, let's rewrite each value as the product of 30 and some integer:
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)
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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
