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What is the sum of all the natural numbers between $500$ and $1000$ (extremes included) that are multiples of $2$ but not of $7$?

JKnecht
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RafaSashi
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    Have you tried anything yet? – fosho Feb 11 '16 at 12:47
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    Working on it using http://math.stackexchange.com/questions/513846/find-the-sum-of-all-the-integers-between-1-and-1000-which-are-divisible-by-7 and http://math.stackexchange.com/questions/9259/find-the-sum-of-all-the-multiples-of-3-or-5-below-1000 am I on the right path? – RafaSashi Feb 11 '16 at 12:50
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    Looks like a good one to emulate, yes. How about: Find the sum of all even numbers in the given range. Find the sum of all multiples of 14 in the given range. Subtract. – Harald Hanche-Olsen Feb 11 '16 at 12:53

4 Answers4

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In general, the sum of an arithmetic progression is $$S_n = \sum_{i=1}^{n} a_i=\frac{n}{2}(a_1+a_n)$$

So, the sum of all even numbers in your interval

$$ = \frac{251}{2}(500+1000)$$

And the sum of all multiples of $14$ in your interval $$ = \frac{36}{2}(504+994)$$

Subtracting these two answers will give you the result you are looking for.

fosho
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\begin{align*} \sum_{i=0}^{250}(500+2i)-\sum_{i=0}^{35}(504+14i) &=251\cdot500+2\sum_{i=0}^{250}i-36\cdot504-14\sum_{i=0}^{35}i\\ &=251\cdot500+2\biggl[\frac{251(0+250)}2\biggr]-36\cdot504-14\biggl[\frac{36(0+35)}2\biggr]\\ &=161286 \end{align*} using the arithmetic progression formula (see here).

Cm7F7Bb
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hint1: for divisibility by 2 find the sum of arithmetic numbers of distance2

hint2: Numbers divisible of 7 are either multiples of 14 or divisible by 14 when added 7, so try to subtract sequence of difference 14 from your sum

Abr001am
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Heres a code solution using a C# console application:

// hold all numbers to be summed 

List<int> multiplesOfTwoCollection = new List<int>();<br>

// count numbers 

for (int naturalNumber = 500; naturalNumber <= 1000; naturalNumber++)

{ if (naturalNumber % 2 == 0) { multiplesOfTwoCollection.Add(naturalNumber); }

// log contents

 Console.WriteLine("natural number:{0}", naturalNumber); }

// multiples of 7

int sumOfInts = 0;
foreach (var item in multiplesOfTwoCollection)
 if (item % 7 == 0) { Console.WriteLine("multiple of 7:{0}", item);
     } else { sumOfInts += item; }

Console.WriteLine("sum of numbers between 500 and 1000 that are multiples of 2 not 7:{0}", sumOfInts);

Returns

161286

RafaSashi
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