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Even multiples of 3 between 500 and 800

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by faraz_jeddah » Sat Aug 03, 2013 12:31 pm
What is the sum of the even multiples of 3 between 500 and 800 inclusive?

A - 193,050
B - 193,047
C - 32,550
D - 31,836
E - 31,830

OA - C

MY approach

Find relevant extremes - 501 & 798

All multiples of 3 = (798-501) / 3 = 99
Add 1 = 99 + 1 = 100

Half of these multiples are even and odd, so 50 even multiples.

Sum = # of terms * Average = 50 * (501+798)/2 = 32,475

Where am I going wrong?
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by satish_iitg » Sat Aug 03, 2013 1:17 pm
even multiple of 3 - 3*2k

k varies from 84 to 133 (504-798)

Sum = (s1 + sn)*n/2

s1 = 504
sn = 798
n = 50
ans 32550
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by GMATGuruNY » Sat Aug 03, 2013 2:09 pm
faraz_jeddah wrote:What is the sum of the even multiples of 3 between 500 and 800 inclusive?

A - 193,050
B - 193,047
C - 32,550
D - 31,836
E - 31,830

OA - C

MY approach

Find relevant extremes - 501 & 798

All multiples of 3 = (798-501) / 3 = 99
Add 1 = 99 + 1 = 100

Half of these multiples are even and odd, so 50 even multiples.

Sum = # of terms * Average = 50 * (501+798)/2 = 32,475

Where am I going wrong?
Sum = number * average.

You correctly determined that there are 50 even multiples of 3 between 500 and 800.
Thus, number = 50.

Average = (biggest + smallest)/2.
In your solution above, the value in red is incorrect.
501 is NOT even, so it is not the smallest value in the set.
The smallest even multiple of 3 between 500 and 800 is 504.
Thus:
Average = (798+504)/2 = 651.

Thus:
Sum = number * average = 50*651 = 32,550.

The correct answer is C.
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by Matt@VeritasPrep » Sat Aug 03, 2013 2:14 pm
Your extremes should be even: 504 <-> 798.

To count, I usually write them as multiples: 504 = 3 * 168 and 798 = 3 * 266. We start with an even and end with an even, and every other term is odd, so we have (266-168)/2 + 1 even multiples of 3, or 50.

Now we have 50 * (504+798)/2 = 50 * 651 = 32,550
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by faraz_jeddah » Sat Aug 03, 2013 10:56 pm
:shock: Cant believe I missed that.

Thanks.
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by freyesinsb » Mon Aug 05, 2013 9:32 am
Is this an effective way to calculate the number of multiples or did i just stumble into the right number?

500-800 inclusive

800-500= 300/2=150 to get the # of even multiples
150/3=50 to get the # of multiples of 3?
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by macattack » Mon Aug 05, 2013 11:39 pm
faraz_jeddah wrote:What is the sum of the even multiples of 3 between 500 and 800 inclusive?

A - 193,050
B - 193,047
C - 32,550
D - 31,836
E - 31,830

OA - C

MY approach

Find relevant extremes - 501 & 798

All multiples of 3 = (798-501) / 3 = 99
Add 1 = 99 + 1 = 100

Half of these multiples are even and odd, so 50 even multiples.

Sum = # of terms * Average = 50 * (501+798)/2 = 32,475

Where am I going wrong?
Hey why did you add 1 to 99 to get the number of multiples?
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by faraz_jeddah » Tue Aug 06, 2013 5:31 am
Thats the rule when you are finding multiples in a range.


For example find all the multiples of 2 between 1 and 13

First Method
- You find the relevant extremes i.e both should be multiples of 2 == > 2 and 12
- Subtract the extremes ==> 12-2 = 10
- Divide by the length (in this case 2) = 10/2 = 5
- Add 1 ==> 5+1 = 6

Second Method
List down the multiples of 2 in the interval = 2,4,6,8,10,12 which is a total of 6
But obviously you cannot use this second method when you have big interval like 500 and 800.

Hope that helps.
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