multiples of 15

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?

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}

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.

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

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.