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# If integer k is equal to the sum of all even multiples of 15

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If integer k is equal to the sum of all even multiples of 15 [#permalink]

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11 Nov 2009, 05:27
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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
[Reveal] Spoiler: OA
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Posts: 44588
Re: Even multiples of 15 [#permalink]

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11 Nov 2009, 05:45
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yangsta8 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

[Reveal] Spoiler:
ANS = C

Even multiple of 15 is of a form 15*2n=30n, hence it's is multiple of 30.

First multiple of 30 in the range 295-615 is obviously 300 and the last 600. so we have:

300+330+360+...+600=30*(10+11+...+20)

10+11+..+20 sum of 11 consecutive integers =$$\frac{first \ term+last \ term}{2}*number \ of \ terms=\frac{10+20}{2}*11=15*11$$

$$300+330+360+...+600=30*(10+11+...+20)=30*15*11=2*3^2*5^2*11$$

Greatest prime 11.

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Re: Even multiples of 15 [#permalink]

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11 Nov 2009, 05:46
2
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C.

You can find the sum by taking the average of the highest and lowest numbers that satisfy your criteria multiplied by how many numbers there are. Find out how many numbers there are by subtracting the highest number that satisfies your condition minus the lowest number that satisfies your condiction, divide by the interval (here that's 30), then add 1.

600 - 300 = 300; 300 / 30 = 10, 10 +1 = 11

Now the average of 600 and 300 is 450. so 450 * 11 = 4950. Notice also that 11 is prime and the question calls for the largest prime number. Because we already used 11, we know C is possible, and 13 and 17 are not factors.

4950 / 13 = 380 and 10/13 - not a factor

4950 / 17 = 291 and 3/17 = not a factor.

For methods on finding divisibility by the primes up to 50, check out this link

http://home.egge.net/~savory/maths1.htm

yangsta8 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

[Reveal] Spoiler:
ANS = C

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Senior Manager
Joined: 31 Aug 2009
Posts: 406
Location: Sydney, Australia
Re: Even multiples of 15 [#permalink]

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11 Nov 2009, 05:55
awesome responses.
+1 to both of you.
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Joined: 28 Oct 2009
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17 May 2010, 16:46
Thanks, this was a difficult problem for me.
Manager
Joined: 25 Jul 2010
Posts: 129
Re: even multiples of 15 between 295 and 615 [#permalink]

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27 Sep 2010, 14:31
super. many thanks.
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Joined: 16 Jun 2010
Posts: 167
Re: even multiples of 15 between 295 and 615 [#permalink]

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27 Sep 2010, 15:19
Bunnel --- Amazing solution.
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Re: If integer k is equal to the sum of all even multiples of 15 [#permalink]

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12 Jan 2012, 05:44
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300+330+.........+600
600=300+(n-1)*30
n=11

sum=k=11/2[2*300+10*30]
=11*450= 11*5^2*3^2*2
the largest prime = 11
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Re: If integer k is equal to the sum of all even multiples of 15 [#permalink]

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11 Feb 2012, 17:22
Those are great answers... i especially like the step of factoring out the 30.
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11 Feb 2012, 17:25
even integers of 15 can only be multiples of 30. so find all the multiples of 30 in 300-600. then you can factor out and find that the total is 4950 and there is a factor of 11 in there which makes it largest prime. factor.
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Re: If integer k is equal to the sum of all even multiples of 15 [#permalink]

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26 Nov 2016, 08:36
yangsta8 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

Slightly different approach.

We have $$15*20 + 15*22 + 15*24 + ….. + 15*40 = 15*(20 + 22 + 24 + … + 40)$$ =

= 15* [(sum of even numbers from 1 to 40) – (sum of even numbers from 1 to 18)]

# of even numbers from 1 to 40 = $$\frac{(40-2)}{2} + 1 = 20$$

# of even numbers from 1 to 18 = $$\frac{(18-2)}{2} + 1 = 9$$

Sum of consecutive even numbers = $$n(n+1)$$

$$= 15*(20*21 - 9*10) = 15*(420 – 90) = 15*330=15*30*11$$

Greatest prime factor is $$11$$.
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Re: If integer k is equal to the sum of all even multiples of 15 [#permalink]

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02 Dec 2016, 10:08
Great question to make you think!!

multiples of 15 within the range --> 300 (15 x 20) ----- 615 (15 x 41)

41-20+1 = 22 multiples total
22/2 = 11 multiples that will be even

median of 11 = 6

20+5 (we're including 20, so you'll have 6 values added together) = 25
15x25 = 375 (KEEP TO THE SIDE)

sum/# = avg --> this nifty formula can be applied here!
sum/11 = 375 --> sum = 11x375

Largest prime value is already in front of us --> 11!

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Re: If integer k is equal to the sum of all even multiples of 15 [#permalink]

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26 Feb 2018, 12:33
Hi All,

The "math" behind this question can be done in a variety of ways, but you will eventually have to use what's called "prime factorization" to answer this question.

First, we need to figure out what we're summing: the even multiples of 15 between 295 and 615. That will be all the even multiples of 15 from 300 to 600, inclusive.

300 + 330 + 360…..+600

Since these numbers are all multiples of 30, we have….

30(10 + 11 + 12…..+20)

Using "bunching" rules, this gives us….

30(165)

Prime factoring this gives us…

(3x2x5)(11x3x5)

So the greatest prime factor is 11

[Reveal] Spoiler:
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Re: If integer k is equal to the sum of all even multiples of 15   [#permalink] 26 Feb 2018, 12:33
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