And what is the quickest way to solve this with a calculator?
$\endgroup$3 Answers
$\begingroup$Between 100 to 1000 are 900 numbers. 900/14 =~ 64 so there will be 64 multiples of 14. The go from 112 = 8*14 to 71*14 = 994. The sum of these numbers are 14(8 + ..... + 71). The sum 8 + .... + 71 = 79*64/2 = 2528. So the sum of all 3 digit multiples of 14 is 14*2528 = 35392.
Now we must subtract those that are multiples of 21. Well we only have even multiples of 21 so these are the multiples of 42. 900/42 =~ 21 of theses. From 42*3 = 126 to 432*23 = 966. The sum of these are 42(3 + .... + 23) = 42(26*21/2) = 11466.
So the sum is 35,392 - 11,466 = 23,926.
$\endgroup$ 1 $\begingroup$Zero.
If you have all the 3-digit positive integers with such conditions and all the 3-digit negative integers with such conditions and then add them all up, you will get zero.
$\endgroup$ 6 $\begingroup$Presumably "number" here is interpreted as "nonnegative integer".
Hint: take the sum of those numbers divisible by $14$ and subtract the sum of those numbers divisible by both $14$ and $21$ (and thus by the least common multiple of $14$ and $21$). Use the fact that
$$ \sum_{i=1}^b i = \dfrac{b(b+1)}{2}$$
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