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by GMATGuruNY » Fri Feb 08, 2013 10:09 am
An integer is a multiple of 3 if the sum of its digits is a multiple of 3.
Thus:
The smallest multiple of 3 between 500 and 800 is 501, since the sum of the digits = 5+0+1 = 6, which is a multiple of 3.
The biggest multiple of 3 between 500 and 800 is 798, since the sum of the digits = 7+9+8 = 24, which is a multiple of 3.

Question rephrased: What is the sum of the multiples of 3 between 501 and 798, inclusive?

The multiples of 3 constitute an EVENLY SPACED SET, since the distance between each term and the next is always 3.
I explain how to determine the sum of an evenly spaced set here:

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by Ian Stewart » Fri Feb 08, 2013 10:33 am
This kind of question is more interesting if we can see the answer choices. It obviously could involve a tedious calculation, so I'd at least glance at the answer choices to see if there are any shortcuts available. If the answers are far apart, we could do a rough estimate: if you add all the numbers between 500 and 800, you are summing roughly 300 numbers (301 to be precise, but we're estimating) which average to 650, so the sum would be approximately 300*650. We're only adding 1/3 of those numbers, since only 1/3 of the numbers in that range will be multiples of 3, so the sum should be very close to 300*650/3 = 65,000. That might be all you need to do if the answers are far apart (the exact answer turns out to be 64,950, so the estimate is good). We're also adding multiples of 3 here, so the answer must be a multiple of 3. So summing the digits of each answer choice might let you get to the right answer as well, if we're fortunate enough to find that only one answer is divisible by 3.

If the answers are close together, and if several of them are divisible by 3, then you would not have any alternative but to carry out the calculation as Mitch outlined above. But in my experience, it is very often possible to find a shortcut in most real GMAT questions that potentially involve lengthly calculations.
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by Tommy Wallach » Fri Feb 08, 2013 11:08 am
And just to finish up this trinity of expert advice pieces, I'll just put the evenly spaced set stuff right here:

Equation for Sum of an Evenly Spaced Set: Average * # of terms

Equation for Average of an Evenly Spaced Set: (First Term + Last Term) / 2

Equation for # of terms in an Evenly Spaced Set: ((Last Term - First Term) / Interval) + 1

So, in this case:

Average = 501 + 798 / 2 = (1299 / 2) = 649.5
# of terms = (( 798 - 501 ) / 3 + 1 = (297 / 3) + 1 = 99 + 1 = 100

Sum = Average * # 0f terms = 649.5 * 100 = 64950

Hope that helps!

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by Soumita Ghosh » Sat Feb 09, 2013 11:43 am
What is the sum of even multiples of 3 between 500-800 inclusive?
A)193050
B)193047
C)32550
D)31836
E)31830

is even divisible is not same as even multiples?? will I not follow the same method that GMATGuruNY and Tommy mentioned?
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by The Iceman » Sat Feb 09, 2013 12:00 pm
Soumita Ghosh wrote:What is the sum of even multiples of 3 between 500-800 inclusive?
A)193050
B)193047
C)32550
D)31836
E)31830

is even divisible is not same as even multiples?? will I not follow the same method that GMATGuruNY and Tommy mentioned?
Evenly divisible by 3 means that the number is divisible by 3; however even multiple of 3 means that the number is only a multiple of 6. So, you will have to exclude all such numbers that are not multiples of 6 (or that are odd multiples of 3).

So, the two boundary numbers divisible by 6 are 504 and 798. The number of terms are 1+(798-504)/6 =50

The required sum = (504+798)*50/2 =32550
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