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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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New post 11 Nov 2009, 04:27
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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
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Re: Even multiples of 15  [#permalink]

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New post 11 Nov 2009, 04:45
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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

ANS = C


Please explain how you arrive at your answer.


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.

Answer: C (11)
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Re: Even multiples of 15  [#permalink]

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New post 11 Nov 2009, 04:46
2
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.

So C is the answer.

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

ANS = C


Please explain how you arrive at your answer.

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Re: Even multiples of 15  [#permalink]

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New post 11 Nov 2009, 04:55
awesome responses.
+1 to both of you.
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Re: Number Properties  [#permalink]

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New post 17 May 2010, 15:46
Thanks, this was a difficult problem for me.
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Re: even multiples of 15 between 295 and 615  [#permalink]

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New post 27 Sep 2010, 13:31
super. many thanks.
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Re: even multiples of 15 between 295 and 615  [#permalink]

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New post 27 Sep 2010, 14:19
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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New post 12 Jan 2012, 04:44
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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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New post 11 Feb 2012, 16:22
Those are great answers... i especially like the step of factoring out the 30.
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New post 11 Feb 2012, 16:25
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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New post 26 Nov 2016, 07:36
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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New post 02 Dec 2016, 09:08
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 between 2  [#permalink]

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New post 29 Apr 2017, 19:20
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dflorez 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


Even multiples of 15 means multiple of 30...
295 to 615 clearly shows 300-600..
So we are looking at 300+330+....+570+600...
\(300+330+360+....+570+600=30(10+11+12+....+19+20)\)
Now 10+11+12+.....+20= average of lowest and greatest * number of items=\(\frac{10+20}{2}*11=15*11\)..
So sum is 30*15*11....
Clearly 11 is the biggest prime factor.

C
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Re: If integer k is equal to the sum of all even multiples of 15 between 2  [#permalink]

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New post 29 Apr 2017, 19:40
dflorez 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



sum = mean *no. of terms
since all are in sequence , mean = 300+600/2= 450
no of terms = 15*2 (10,11,12....20) , total 11 nos.
thus Sum = 450*11 = 9*5*5*2*11
clearly largest prime is 11

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Re: If integer k is equal to the sum of all even multiples of 15 between 2  [#permalink]

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New post 29 Apr 2017, 20:07
Even multiple of 15 basically means multiples of 30

Series => 300,330....600
Mean => 900/2=450
Number of terms => 600-300/30 +1 => 10+1 => 11

Sum => 11*450 => 11*5^2*3^2*2
Hence k=> 2*3^2*5^2*11

Greatest Prime => 11

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

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New post 26 Feb 2018, 11:33
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

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Re: If integer k is equal to the sum of all even multiples of 15 between 2  [#permalink]

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New post 05 Aug 2018, 07:26
dflorez 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


OA: C

Even multiples of \(15\) between \(295\) and \(615\) are \(300(20*15),330(22*15),360(24*15),................................,600(40*15)\)

\(K =\) sum of even multiples of \(15\) between \(295\) and \(615\) \(= 300+ 330+360+...........................+600\)

\(K=15( 20+22+24+26+....................+40)\)
\(20+22+24+26+....................+40\) is sum of an A.P series with \(a=20\) and \(d=2\)
to find the number of terms in A.P, we will use formula \(a_n=a+(n-1)d\)
\(40 =20+(n-1)2\)
\(n-1=\frac{20}{2}\)
\(n=11\)
Sum of A.P series \(= \frac{n}{2}\)[First Term + Last Term]\(=\frac{11}{2}[20+40]=11*30\)
\(K =15*11*30=2^1*3^2*5^2*11^1\)

Greatest prime factor of \(K =11\)
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Re: If integer k is equal to the sum of all even multiples of 15  [#permalink]

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New post 06 Aug 2018, 08:17
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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


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)

------------------------------------------------------
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

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

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New post 21 Oct 2018, 05:12
Since any even number must be divisible by 2, any even multiple of 15 must be divisible by 2 and by 15, or in other words, must be divisible by 30. As a result, finding the sum of even multiples of 15 is equivalent to finding the sum of multiples of 30. By observation, the first multiple of 30 greater than 295 will be equal to 300 and the last multiple of 30 smaller than 615 will be equal to 600.

Thus, since there are no multiples of 30 between 295 and 299 and between 601 and 615, finding the sum of all multiples of 30 between 295 and 615, inclusive, is equivalent to finding the sum of all multiples of 30 between 300 and 600, inclusive. Therefore, we can rephrase the question: “What is the greatest prime factor of the sum of all multiples of 30 between 300 and 600, inclusive?”

The sum of a set = (the mean of the set) × (the number of terms in the set)

Since 300 is the 10th multiple of 30, and 600 is the 20th multiple of 30, we need to count all multiples of 30 between the 10th and the 20th multiples of 30, inclusive.
There are 11 terms in the set: 20th – 10th + 1 = 10 + 1 = 11
The mean of the set = (the first term + the last term) divided by 2: (300 + 600) / 2 = 450
k = the sum of this set = 450 × 11

Note, that since we need to find the greatest prime factor of k, we do not need to compute the actual value of k, but can simply break the product of 450 and 11 into its prime factors:
k = 450 × 11 = 2 × 3 × 3 × 5 × 5 × 11

Therefore, the largest prime factor of k is 11.

The correct answer is C.
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