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multiples of 15

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by [email protected] » Mon Sep 08, 2014 11:14 pm
Hi j_shreyans,

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

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by Brent@GMATPrepNow » Tue Sep 09, 2014 6:49 am
j_shreyans 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

OAC
In other words, we're looking for multiples of 30

So, k = 300 + 330 + 360 + ... + 570 + 600

Let's examine some terms in this series. . . .

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)

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

Now, let's examine this sum: 10 + 11 + 12 + ... + 19 + 20
Since 20 - 10 + 1 = 11, we know there are 11 numbers to add together.
Aside: A nice rule says: the number of integers from x to y inclusive equals y - x + 1

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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Brent
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by j_shreyans » Fri Oct 31, 2014 9:31 am
Aside: A nice rule says: the number of integers from x to y inclusive equals y - x + 1

Hi Brent ,

One thing if the number of terms from x to y (x and y are not inclusive), then how can we find the number of terms?
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by Matt@VeritasPrep » Fri Oct 31, 2014 9:55 am
j_shreyans wrote:Aside: A nice rule says: the number of integers from x to y inclusive equals y - x + 1

Hi Brent ,

One thing if the number of terms from x to y (x and y are not inclusive), then how can we find the number of terms?
(y - x - 1), presuming that we're dealing with consecutive integers.

You can see this by examining some small sets.

The number of integers BETWEEN 5 and 9 is (9 - 5 - 1):
5 {6 7 8} 9

The number of integers BETWEEN 5 and 9, inclusive, is (9 - 5 + 1):
{5 6 7 8 9}
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by GMATGuruNY » Fri Oct 31, 2014 10:35 am
j_shreyans wrote:Aside: A nice rule says: the number of integers from x to y inclusive equals y - x + 1

Hi Brent ,

One thing if the number of terms from x to y (x and y are not inclusive), then how can we find the number of terms?
For any set of consecutive integers, the number of integers = biggest - smallest + 1.

If x=23 and y=47, and z is equal to the number of consecutive integers between x and y, excluding x and y, what is the value of z?
Implication:
z is equal to the number of consecutive integers between 24 and 46, inclusive.
Thus:
z = biggest- smallest + 1 = 46 - 24 + 1 = 23.
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by Brent@GMATPrepNow » Fri Oct 31, 2014 11:18 am
j_shreyans wrote:Aside: A nice rule says: the number of integers from x to y inclusive equals y - x + 1

Hi Brent ,

One thing if the number of terms from x to y (x and y are not inclusive), then how can we find the number of terms?

In the original question, I took a question that where the values were NOT inclusive and turned it into one where the values ARE inclusive.
You can always do the same thing.
Example, let's say we want to find the number of integers BETWEEN (i.e., not inclusive) 4 and 10.
This is same as finding the number of integers from 5 to 9 INCLUSIVE.
At this point we can apply the formula to get...
The number of integers = 9 - 5 + 1 = 5

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by j_shreyans » Tue Feb 17, 2015 8:45 am
Hi All ,

One confusion, in the it is given that sum of even multiple of 15 between 295 and 615.

so the multiple would be 300,330,360.......600

now we have to find the number of term.

So a rule is if x&y are multiple of k then the number of k from x to y inclusive = (y-x)/k+1

so the number of term will be (600-300)/15+1

Number of term is 21

sum = (600+300)21/2

which is 9450

so the prime factor of 9450= 2X5X5X3X3X3X7

and the greatest is 7, but the OA is 11

please advise and correct .

Thanks,

Shreyans
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by GMATGuruNY » Tue Feb 17, 2015 8:52 am
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?

1)5
2)7
3)11
4)13
5)17
For any EVENLY SPACED SET:
Number of terms = (biggest-smallest)/interval + 1, where the interval is the distance between successive terms.
Average = median = (biggest + smallest)/2.
Sum = (number of terms)(average).

The even multiples of 15 between 295 and 615 are the MULTIPLES OF 30 between 295 and 615.
Thus:
k = 300 + 330 + 360 + ... + 600.
Here, the interval between successive terms is 30.
Thus:
Number of terms = (600-300)/30 + 1 = 11.
Average = (600+300)/2 = 450.
Sum = 11*450.

Prime-factoring k = (11)(450), we get:
(11)(2*3*3*5*5).
Thus, the greatest prime factor of k is 11.

The correct answer is C.
j_shreyans wrote: so the number of term will be (600-300)/15+1
The value in red does not accurately represent the interval between successive terms.
Since we are counting only the EVEN multiples of 15 -- in other words, the MULTIPLES OF 30 -- the interval between successive terms is not 15 but 30.
Thus:
Number of terms = (600-300)/30 + 1 = 11.
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