Multiples / GCF question

[bronzie35]

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!

[Ian Stewart]

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.

[kanha81]

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.

[maihuna]

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