find the sum of all 3-digit numbers which are not multiples of 6

I couldn't read through all the posts carefully.
Here is what I would do.
Multiple of 6: 102, 108, 114, 120, ..., 996
102+108+114+120+...+996 = 6(17+18+19+20+...+166)= 6[(1+2+3+...+166)-(1+2+...+16)]=6[166*167/2 - 16*17/2}= 6[13861-136] = 6[13725] = 82350

Sum of 100+101+102+...+ 999 = (1+2+3 +...+999) - (1+2+...+99) =999*1000/2 - 99*100/2 = 499500-4950 = 494550
So the answer is 494550-82350 = 412200

I would not use a's and d's. But then again that is my style so maybe this is not for you.
 
You must log in or register to reply here.
Top