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Director
Joined: 09 Aug 2006
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If integer k is equal to the sum of all even multiples of 15
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01 Jul 2007, 23:08
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifintegerkisequaltothesumofallevenmultiplesof125961.html
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Director
Joined: 09 Aug 2006
Posts: 730

1st even multiple of 15 b/w 295 & 615 is 300 (which is 15*20) and the last multiple of 15 is 600 (which is 15*40). Also, a multiple of 15 = 15*n, where n is even.
Therefore, from the above, finding the sum of all even multiples of 15 b/w 295 & 695 is equal to:
15(20)+15(22)+15(24)+....+15(40)
=15(20+22+24+....+40).....................1
Sum of n terms in an AP = n/2[2a+(n1)/2]
n=4020+1 = 21
a=20
d=2
sum of all terms = 21/2[2(40)+(211)2]=840.......2
Using 2 in 1 we get 15*840
Using prime factorization this is equal to 2^3*3^2*5^2*7
Therefore, the greatest prime factor is 7.
Answer should be B. But it's not!
Where am I going wrong???



Intern
Joined: 21 Mar 2007
Posts: 48

GK i see that while calculating the number of terms you have taken ther number as 21. i guess thats where u went haywire . it should be n= 11 .
by calculating taking n= 11, the highest prime would be 11.



Director
Joined: 13 Nov 2003
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Location: BULGARIA

HI, GK_Gmat,
The sum of all EVEN multiples, not ALL multiples of 15, which is 5*900+450=4950



Senior Manager
Joined: 18 Jul 2006
Posts: 482

C.
sum = (n/2)*(a+l) = (11/2)*(300+600)
Largest prime no = 11



Director
Joined: 09 Aug 2006
Posts: 730

forgmat wrote: GK i see that while calculating the number of terms you have taken ther number as 21. i guess thats where u went haywire . it should be n= 11 .
by calculating taking n= 11, the highest prime would be 11.
Thanks a bunch forgmat! i counted all multiples instead of only the even ones.



Manager
Joined: 09 Dec 2006
Posts: 90

PS integers
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23 Aug 2007, 12:58
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?
5
7
11
13
17



Current Student
Joined: 28 Dec 2004
Posts: 3222
Location: New York City
Schools: Wharton'11 HBS'12

can someone show the working...??
i got this one wrong i get 7..
here is how ...the first number should be 300...last number is 600
we are looking for the sum of even multiples..
300/15=20...615/15=41
number of even multiples (4120)=21/2 which means there 20/2 even multiples..
in other words the even multiples of 15 between 295 and 615 is like saying what are the 30s multiples...
so the number of even multiples is 10
sum=10/2 (300*2 + (101)15)
which is 5(600+9*15)=5(735) highest prime factor I get is 7...
what am i doing wrong????



Current Student
Joined: 28 Dec 2004
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Location: New York City
Schools: Wharton'11 HBS'12

OK..so taking a 30 minute break and I am back.......grrrrrrrrrr
last term=first term + (n1)d where d is the seperation..n is the number of terms
600=300 + (n1)30
we are looking for even multiples of 15..its like looking for multiples of 30...they will be 30 integers apart!!!!!
330=30n n=110
sum of these multiples = N/2 [2a + (n1)d]
=110/2 (2*300 + (1101)30)
110[300+109(15)]
110[15(20+109)]
110[15(129)]
prime factors 11 5, 43
technicall isnt 43 the highest prime???



Intern
Joined: 05 Jul 2007
Posts: 4

This is how I did it. There may be other ways to solve this problem. Took over a minute to solve.
K = 300 + 330 + 360 +.... + 600
K = 30 * (10 + 11 + 12+.....+ 20)
K = 30 * 165
K = (2 * 3 * 5) * (3 * 5 * 11)
So answer is 11



Intern
Joined: 11 Aug 2007
Posts: 2

fresinha12 wrote: OK..so taking a 30 minute break and I am back.......grrrrrrrrrr
last term=first term + (n1)d where d is the seperation..n is the number of terms
600=300 + (n1)30
we are looking for even multiples of 15..its like looking for multiples of 30...they will be 30 integers apart!!!!!
330=30n n=110
sum of these multiples = N/2 [2a + (n1)d]
=110/2 (2*300 + (1101)30)
110[300+109(15)]
110[15(20+109)] 110[15(129)]
prime factors 11 5, 43
technicall isnt 43 the highest prime???
330 = 30n
n=11
11/2(600+300)
11(450)
so answer is 11



Current Student
Joined: 28 Dec 2004
Posts: 3222
Location: New York City
Schools: Wharton'11 HBS'12

DNT wrote: fresinha12 wrote: OK..so taking a 30 minute break and I am back.......grrrrrrrrrr
last term=first term + (n1)d where d is the seperation..n is the number of terms
600=300 + (n1)30
we are looking for even multiples of 15..its like looking for multiples of 30...they will be 30 integers apart!!!!!
330=30n n=110
sum of these multiples = N/2 [2a + (n1)d]
=110/2 (2*300 + (1101)30)
110[300+109(15)]
110[15(20+109)] 110[15(129)]
prime factors 11 5, 43
technicall isnt 43 the highest prime??? 330 = 30n n=11 11/2(600+300) 11(450) so answer is 11



Manager
Joined: 19 Aug 2007
Posts: 161

PS greatest prime factor
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30 Oct 2007, 09:54
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?
5
7
11
13
17



Director
Joined: 11 Jun 2007
Posts: 865

Re: PS greatest prime factor
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30 Oct 2007, 10:04
jimjohn 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?
5 7 11 13 17
i get 11
if we break down what the stem is asking what is the sum of all mult of 30 between 300 and 600.
using arithmetic progression to find n : 600 = 300 + (n  1) 30
300+ 30 = 30n
330 = 3n => n = 11
the sum would be: 11* mean
mean = [600 + 300] / 2 = 450
11*450 = 4950
not necessary to find but you can see 11 would be the largest prime factor (try 13 or 17.. both doesn't work)



Manager
Joined: 02 Aug 2007
Posts: 142

Why did you use 600 & 300 and 30 as distance?



Manager
Joined: 19 Aug 2007
Posts: 161

hey beckee just wondering where do you get that equation from:
600 = 300 + (n1) *30
i did get the same answer 11, but did it the much longer way



Director
Joined: 11 Jun 2007
Posts: 865

jimjohn wrote: hey beckee just wondering where do you get that equation from: 600 = 300 + (n1) *30
i did get the same answer 11, but did it the much longer way
it is the arthmetic progression equation
tn = a + (n1) * d
tn = last term of series
a = first term of the series
n = # of terms in the series
d = difference between each term



Manager
Joined: 16 Sep 2010
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Location: India
Schools: Terry, Georgia Tech
WE 1: Working for Wipro Technologies since April 2010

Re: even multiples of 15 between 295 and 615
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27 Sep 2010, 21:13
Orange08 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? 5 7 11 13 17 i followed a more conventional path.... even multiples mean the common difference is 30. therefore number of terms = (600 300)/30 + 1 = 11 Sum = 11 ( 600 + 10 * 30 )/2 ie n[2a + {n1}d]/2 = 4950 = 2 * 5^2 * 3^2 * 11 Thus, 11 ANS
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Manager
Joined: 15 Apr 2010
Posts: 114

Re: Even multiples of 15
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04 Oct 2010, 07:21
The even multiples of 15 in the given range would be 300, 330, 360, 390.....600. Since this is an arithmetic progression, the sum i.e., k can be calculated using the following formula: (n/2)*(first term+last term)
In order to calculate n we can use the following formula: a[n] = a[1] + (n1)*d 600 = 300 + (n1)*30 n = 11
Substituting the value of n in the formula to calculate sum, we get k = 11*450 and if you factorize 450, you will get all the factors less than 11. Hence the greatest prime factor of k is 11.



Senior Manager
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K is the sum of multiples of 30 between 295 and 615.
K= 30 (10+11 …+20) 10+11+…20 = 11/2(10+20)=15*11 K=30*15*11
11 is the greatest prime factor







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