If X is equal to the sum of the even integers...

[Guest]

If X is equal to the sum of the even integers from 40 to 60, inclusive, and Y is the number of even integers from 40 to 60, inclusive, what is the value of X+Y?

550
551
560
561*
572

I know there is an equation to use for consectutive integers but is it supposed to be used for patterns?

[Sudhan]

I just used the brute force method because the range of numbers is very minimal.

X= Sum(Even Intg from 40-60 Incl)
Y= # of Even Intg from 40-60 Incl)

X= 40+42+44+46+48+...60 =550
Y= (40 to 60 Incl )= 60-40 = 20/2=10+1 Since incl ==> 11

X+Y=561

Thanks

[rfernandez]

If X is equal to the sum of the even integers from 40 to 60, inclusive, and Y is the number of even integers from 40 to 60, inclusive, what is the value of X+Y?

550
551
560
561*
572

First, to find the number of terms in the set, you can think of 40 as the 20th even integer and 60 as the 30th even integer. How many even integers are there? 30 - 20 = 10. But you add one because the set is inclusive of 20 and 30, so there are 11.

The way to find a sum of consecutive integers is to multiply the mean of the set by the number of integers in the set. Here, we don't want the sum of all of the integers, just the even ones. Well, the method still works:

mean is 50 and number of terms is 11
sum = 50*11 = 550

So X = 550 and Y = 11, so X + Y = 561.

Why does the sum of consecutive integers formula work in this context? Well, it's a small enough set to list out the values to see why it works here:

40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60
Notice that the first and last terms, 40 and 60, sum to 100, which is twice the mean. Also, the second and tenth terms, 42 and 58, also sum to 100, also twice the mean. And so on, until you get to the "middle" number 50. So, how many multiples of the mean are there in total? 11, which equals the number of terms that are in the set. So we can shortcut the process by multiplying 50 by 11.

Rey

[mynameisdong]

Hi Rey,

Just to build on top of this -

if the question asked to find the sum of all odd integers from 40 - 60... would it be:

# of odd integers in the set 40 -> 60
[ (59-41) / 2 ]+ 1 = 10 ?

then, [ (59+41) / 2 ] * 10 ?

I'm trying to understand the alternative question they could've asked.

thank you!

[tim]

looks good!

[mynameisdong]

looks good!

thanks!

[RonPurewal]

well played