Hallo happy people!

Question concerns problem “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?

5
7
11
13
17

CAT’s 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:
k = 450 × 11 = 2 × 3 × 3 × 5 × 5 × 11

Therefore, the largest prime factor of k is 11.

The correct answer is C.

Problem: there are two options : first – 615 is included in the set (it should be stated! and it is not, so I assume it is not inclusive. Because problem “51 to 107” says “How many integers are there between 51 and 107, inclusive?”, so i assume that if it stated that numbers are inclusive in the set we count them, if it is not stated we only count numbers “in between”). If 615 is included in the set - everything is alright with solution, but formulation of problems is not consistent, it is not clear when numbers are included and when not?

Second - 615 is not included in the set. then we have 21 multiplier of 15 and the correct answer should be 7.

Question: what is the policy of using 'inclusive' and assuming 'inclusive' in the number sets, basically an the answer depends on that...

I agree that the solution should not say "between 295 and 615, inclusive" because, according to the wording of the problem, neither 295 nor 615 are included in the set. Thanks for the catch.

This does not, however, change the solution. The solution says that "integer k is equal to the sum of all even multiples of 15" and 615 is not an even multiple of anything - it is not an even number. :)

For the official test: if you see "between [one number] and [another number]" those two numbers are NOT included in the range. You have to have "inclusive" to include those two.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

thanks.
i fell into the trap, since i did not notice the word even in the problem :(

you are not alone...
:)

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

A bit late posting here based the original post date(s) of this thread. Anyhow, there are three formulas that can calculate a variation of the Sums: (Even, Odd, All).

n = number of terms between 600 and 300 inclusive multiples of 15: 300/15 = 20 + 1 = 21 = n
Even: n(n + 1)
Odd: n^2
All: (n(n + 2) / 2)

Even: n(n + 1) ==> (21)X (22) = 462
Primes of 462 = 2 X 3 X 7 X 11 ==> 11

as a warning to our students, these formulas are not all correct. as always, be careful - not everything you read on the internet is true! :)

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Tim Sanders
Manhattan GMAT Instructor