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

Question

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

Bunuel · Accepted 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)

jallenmorris · Answer

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

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

yangsta8 · Answer

awesome responses.
+1 to both of you.

marcusaurelius · Answer

Thanks, this was a difficult problem for me.

Orange08 · Answer

super. many thanks.

devashish · Answer

Bunnel --- Amazing solution.

BDSunDevil · Answer

300+330+.........+600
600=300+(n-1)*30 
n=11

sum=k=11/2 
 =11*450= 11*5^2*3^2*2
the largest prime = 11

calreg11 · Answer

Those are great answers... i especially like the step of factoring out the 30.

chetan2u · Answer

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

EMPOWERgmatRichC · Answer

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

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

BrentGMATPrepNow · Answer

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 ] 
=  11 ] 
=  (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

Cheers,
Brent

GMATYoda · Answer

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.

joeljohndaniel · Answer

A Fast and Efficient approach could be,

Counting..

Since 295 is not a multiple of 15, the next multiple of 15 is 300.

From 300 to 600 there are 300 Numbers, of which every 15th Number After 300 will be a multiple of 15.

Therefore, Total multiples of 15 between 300 and 600 (excluding 300) is equal to  300/15 . Which is Equal to   20   --- (i)

Now count 300 as the first multiple =   1   --- (ii) and 615 as the last multiple =   1   ----(iii)

Adding the counts of (i), (ii), and (iii);

We get,   20+ 1+ 1= 22

Now, The question stem asks for the count of even multiples of 15, by virtue of Odd-Even parity, which will be  1/2  of the total Count.

Thus,   22/2 = 11  . Fortunately this is also the highest prime number.

This is the answer. 11

Prasannathawait · Answer

Fast method would be counting. 
As the question says even multiples of 15, that means first number would be 300 and then add 30 upto 600.
11 numbers are there and average will be the median of 11 i.e. 450
450x11 will be the sum.
Upon prime factorizing, we will get 11 as the largest prime factor.

Kinshook · Answer

Given:  Integer k is equal to the sum of all even multiples of 15 between 295 and 615.

Asked:  What is the greatest prime factor of k?

Integer k is equal to the sum of all even multiples of 15 between 295 and 615.

k = 300 + 330 + .... + 600 
600 = 300 + (n-1)30
n = 300/30 + 1 = 11

k = 11 * 450 = 2*3^2*5^2*11

Greatest prime factor of k = 11

IMO C

exc4libur · Answer

multiples15(295…615)=m15(300…600)=(600-300/15)+1=21=(even,odd)=(10,11) 
 average15(300…600)=(600+300)/2=450 
 average•n=sum…450*11=sum 
 gprimef(450*11)=(45*100*11)=(5,3^2,5^2,2^2,11)=11

Answer (C)

gagan543 · Answer

First, let us simplify the problem by rephrasing the question. 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:

Schachfreizeit · Answer

"Even multiple of 15 is of a form 15*2n=30n, hence it's is multiple of 30."
Can someone elaborate on this? what would we have to do, if the questions asks for the sum of all odd multiples?

ScottTargetTestPrep · Answer

Even multiples of 15 are integers which are multiples of both 2 and 15, which means these numbers should be a multiple of LCM(2, 15) = 30. Thus, the question is to determine the sum of multiples of 30 between 295 and 615. To do that, we will use the following formula:

Sum = Average * Number of terms

To determine the number of terms, we will use the following formula:

Number of multiples of 30 in an interval = 1 +

The smallest multiple of 30 between 295 and 615 is 300, and the largest multiple of 30 in the same interval is 600. Thus, the number of multiples of 30 between 295 and 615 is:

1 +   = 1 + 300/30 = 1 + 10 = 11

So, number of terms = 11.

Next, we determine the average of the multiples of 30 between 295 and 615. To do that, we will use the fact that multiples of 30 between 295 and 615 is a set of evenly spaced integers, and the average of a set of evenly set of integers is given by (largest term + smallest term)/2. Thus:

average = (largest multiple of 30 between 295 and 615 + smallest multiple of 30 between 295 and 615)/2

average = (600 + 300)/2 = 900/2 = 450

We can conclude that the sum of all multiples of 30 between 295 and 615 is 11 * 450. Since 450 = 2 * 3^2 * 5^2, the greatest prime factor of k = 11 * 2 * 3^2 * 5^2 is 11.

Answer: C

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

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