number of number divisible by 6=1000-6/6= 165

average=sum/165
sum= 165/2(6+996)
=165*501

average=165*501/165= 501

don't know if this is the fastest way but...

sum of multiples of 6 until 1000 = 1*6 + 2*6 + 3*6... + 166*6 (which is 966, the largest multiple smaller than 1000)

so that = 6*(1+2+3+...+166)

so the average = 6*(1+2+3+...166) / 166

average of a series is equal to the median of the series.
median of series 1+2+3...166 = (83+84)/2 which is = 83.5

so final average = 6*83.5 = 501

edit: Ok, tohellandback's way is a lot faster, lol

according to tohellandback's method, the short cut is just to simply do: largest multiple + 6 then divide by 2: (996+6)/2 = 501. this is ofcourse after fact, but is there a logical reason for this?

tohellandback,

please explain the part where,

sum= 165/2(6+996)

thanks...

godemol wrote:tohellandback,

please explain the part where,

sum= 165/2(6+996)

thanks...

oh let me rewrite..may be i got you confused..
it is actually..165/2 * (6+996)
sum of n terms in AP is n/2* ( first term + last term)

capnx wrote:don't know if this is the fastest way but...

sum of multiples of 6 until 1000 = 1*6 + 2*6 + 3*6... + 166*6 (which is 966, the largest multiple smaller than 1000)

so that = 6*(1+2+3+...+166)

so the average = 6*(1+2+3+...166) / 166

average of a series is equal to the median of the series.
median of series 1+2+3...166 = (83+84)/2 which is = 83.5

so final average = 6*83.5 = 501

edit: Ok, tohellandback's way is a lot faster, lol

according to tohellandback's method, the short cut is just to simply do: largest multiple + 6 then divide by 2: (996+6)/2 = 501. this is ofcourse after fact, but is there a logical reason for this?

well thats not exactly what i have done..
sum of n terms in AP is n/2 * (1st term + last term)
thats why sum is 165/2 * (6+996)
and that gives you
165 * 501
when you take the average you divide by the total numbers of terms i.e. 165

sum of n terms in AP is n/2 * (1st term + last term)

thanks for ur quick response...one more thing, AP=average problems? also, does this formula apply for all sum of series? thanks.

godemol wrote:sum of n terms in AP is n/2 * (1st term + last term)

thanks for ur quick response...one more thing, AP=average problems? also, does this formula apply for all sum of series? thanks.

sum of n terms is n/2 * (2a +(n-1)d)
= n/2 * (a + a+ (n-1)d)
a is the first terms
a+(n-1)d is nothing but the last term

and of course it is applicable to all series in arithmetic progression

This might not be the proper way, but this is how I reached the answer.
Lowest number which is mulitple of 6 = 6
Higheset number which is multiple of 6 = 996

Taking average of both (6+996)/2 = 501

This is the solution of this problem:
Formula for sum for arithmetic sequence:

S= N/2(A1+An)

N- number of terms in the sequence.
A1- 1st term
An- last term.
So problem asks a multiple of six greater than 0 and smaller than 1000.

Such will be 6(1st term),12,18...966(last term)
Number of terms- 166
Now just apply the formula for sequence and you have:
S= 166/2(6+966)= 83*1002= 13 778
Now, you need to find the mean. Since the sum= 13778 and number of terms= 166; Mean=13778/166=501
Make sure that you memorize this formula. It is being tested quite frequently on the actual test.

the only best short cut i caould think of is this:

Sum= 166/2(6+966)/166(number of terms in order to find the mean)
this leaves us with 1002/2= 501

eustudent wrote:the only best short cut i caould think of is this:

Sum= 166/2(6+966)/166(number of terms in order to find the mean)
this leaves us with 1002/2= 501

Well, actually, in the options, the only one divisible by 6 is 501. But HellandBack method applies to any bizarre case in GMAT exam, thanks Hell.

used the following method:

No. of multiples of 6 (< 1000) = 1000/6 = 166 (ignore the decimal)
Avg = (83rd multiple of 6 + 84th multiple of 6)/2
= 498 + 504/2
= 501

Hope this helps