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Multiples / GCF question

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Re: Multiples / GCF question

by Ian Stewart » Fri May 01, 2009 5:55 pm
bronzie35 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 is c

Thank you for your help!
An even multiple of 15 is just a multiple of 30, so we want to find the largest prime factor of:

300 + 330 + 360 + 390 + ... + 570 + 600
= 30(10 + 11 + 12 + ... + 19 + 20)

There are a lot of ways to compute the sum of consecutive integers; for example, we can use the fact that the average of any 'equally spaced' list is always equal to the average of the smallest and largest numbers in that list. So the average of {10, 11, 12, ..., 19, 20} is just (10 + 20)/2 = 15. Since there are 11 numbers in the list, using:

avg = sum/n
sum = n*avg

the sum is equal to 11*15. So

300 + 330 + 360 + 390 + ... + 570 + 600
= 30(10 + 11 + 12 + ... + 19 + 20)
= 30*11*15
= 2*3*5*11*3*5
= 2*(3^2)*(5^2)*11

and the largest prime factor is 11.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Multiples / GCF question

by kanha81 » Fri May 01, 2009 10:02 pm
Ian Stewart wrote:
bronzie35 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 is c

Thank you for your help!


An even multiple of 15 is just a multiple of 30, so we want to find the largest prime factor of:

300 + 330 + 360 + 390 + ... + 570 + 600
= 30(10 + 11 + 12 + ... + 19 + 20)

There are a lot of ways to compute the sum of consecutive integers; for example, we can use the fact that the average of any 'equally spaced' list is always equal to the average of the smallest and largest numbers in that list. So the average of {10, 11, 12, ..., 19, 20} is just (10 + 20)/2 = 15. Since there are 11 numbers in the list, using:

avg = sum/n
sum = n*avg

the sum is equal to 11*15. So

300 + 330 + 360 + 390 + ... + 570 + 600
= 30(10 + 11 + 12 + ... + 19 + 20)
= 30*11*15
= 2*3*5*11*3*5
= 2*(3^2)*(5^2)*11

and the largest prime factor is 11.
Wow! How do you do it Ian? I kept trying to solve this problem with brute force for 15-20 mins., but couldn't capitalize.
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by maihuna » Fri May 01, 2009 11:16 pm
Its goes like this: 15*20 = 300
15*40 = 600

So K= 15*(20 + 22 + 24 + ... + 40)
= 15*2(10+11+12+.......+20)
= 15*2*(11/2)*(10+20)
= 15*11*30
= 3*5*2*3*5*11
greatest prime factor is : 11
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