Question

4. Find the sum of all the numbers from 100 to 199 that are not divisible by 13.

Answer 1

There are 100 numbers in the range 100-199. Find the smallest and largest multiple of 13 in this range. We have

104 = 8•13

195 = 15•13

so there are 15 - 8 + 1 = 8 multiples of 13 between 100 and 199.

Then the sum we want is

$\displaystyle \sum_{k=100}^{199} k - \sum_{\ell=8}^{15}13\ell$

or equivalently,

(100 + 101 + 102 + … + 199) - 13 (8 + 9 + … + 15)

To compute these sums, recall the following formula:

$\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

Then

$\displaystyle \sum_{k=100}^{199} k = \sum_{k=1}^{199} k - \sum_{k=1}^{99} k = \frac{199\cdot200}2 - \frac{99\cdot100}2 = 14,950$

and

$\displaystyle \sum_{\ell=8}^{15} 13\ell = 13 \left(\sum_{\ell=1}^{15} \ell - \sum_{\ell=1}^7 \ell\right) = 13 \left(\frac{15\cdot16}2 - \frac{7\cdot8}2\right) = 1,196$

which means the sum we want is 14,950 - 1,196 = 13,754.