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?

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

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.

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

Cant believe I missed that. Thanks.

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?

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?

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.