There are different ways to solve this

One way

6*1 = 6

6*166 = 996

There are 166 multiples of 6 greater than 0 and less than 1000 (6,12,18...996)

It can be treated as an ARITHMETIC PROGRESSION

Sn = n/2(2a1+(n-1)d
Sn->sum of n terms
n->number of terms (166 in this case)
a1->first term in sequence (a1=6)
d->common difference(d=6)

WE get 83*1002 as the sum. This divided by number of terms 166 is

83*1002/166 = 501

C)

I need to memorize this! thanks! thanksgivn.

6 + 12 + 18 + ..... + 696

6 ( 1+2+3+...166)

1+2+...+n = nx(n+1)/2

SO the MEAN of 166 numbers is :

= 6x166x167/2x166

= 167x3

Hence, C

i see

i was thinking in terms or evenly spaced consecutive numbers, where the mean would be equal to the median. i thought since 0 and 1000 and not included, there are 998 values. A and E sorta fit the bill on that idea.

could anyone advise how one would solve this question using that idea?

could anyone advise how one would solve this question using that idea?

Yes it can be done like that also(mean=median )

It will 166/2 = 83


a83+a84/2

a83 = 6*83
a84 would be 6*84

6*83+6*84/2

6(83+84)/2 = 3 (167) = 501

And I am sure you have already seen that the median and the mean of consecutive integers are the same:

6x1 = 6 and 6x166 = 996

So the answer is : 996+6 = 1002/2 = 501

supersonic !