Question 1164975: the multiples of 3 and 7 are removed from the series 1+2+3...+150. the resulting sum is Found 2 solutions by Boreal, solver91311:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The sum without removal is (1/2)(150*151)=11325
multiples of 3 are 50,
3+6+9+...+150
that sum Is (n/2)(a1+an)=25*(3+150)=3825
multiples of 7 are 21;
7+14+21+28+...+147
that sum is (21/2)(7+147)=1617
the sum of both is 5442
double counted are the product of both or multiples of 21
21+42+63+84+105+126+147=(n/2)(168)=588
so the amount to be subtracted is 5442-588=4854
the answer is 11325-4854=6471
You can put this solution on YOUR website!
Add all the numbers from 1 to 150. From that subtract the sum of all the numbers from 1 to 150 that are divisible by 3. Then subtract the sum of all the numbers from 1 to 147 (the last number before 150 that is divisible by 7) that are divisible by 7. Then calculate the sum of the numbers from 1 to 147 that are divisible by 21 and add them back because you have subtracted them twice -- once when you did the 3s and once when you did the 7s.
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
