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olga55 [171]
3 years ago
15

If we list all the natural numbers below 20 that are multiples of 7 or 11, we get 7, 11 and 14. The sum of these multiples is 32

. What is the sum of all the multiples of 7 or 11 up to and including 1337?
Mathematics
1 answer:
velikii [3]3 years ago
8 0
Number of multiples of 7 up to 1337: \left\lfloor\dfrac{1337}7\right\rfloor=191
Number of multiples of 11 up to 1337: \left\lfloor\dfrac{1337}{11}\right\rfloor=121
Number of multiples of 77 up to 1337: \left\lfloor\dfrac{1337}{77}\right\rfloor=17

This means there are 191+121-17=295 distinct multiples of 7 *or* 11 up to 1337.

The sum of these multiples is

\displaystyle\sum_{k=1}^{191}7k+\sum_{k=1}^{121}11k-\sum_{k=1}^{17}77k

which can be computed using the well-known formula,

\displaystyle\sum_{k=1}^nk=\dfrac{n(n+1)}2

So you have

\displaystyle\sum_{k=1}^{191}7k+\sum_{k=1}^{121}11k-\sum_{k=1}^{17}77k=7\dfrac{191\times192}2+11\dfrac{121\times122}2-77\dfrac{17\times18}2=197762
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