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I got some fractions such as $1/30$, $7/30$,$11/30$,$13/30$,$17/30$,$19/30$,$23/30$, $29/30$. Rest of them can be created by adding 1 or 2 or 3...or 9 to all the aforementioned terms. How to find the sum of all such numbers

This is one answer. How do we get to this?

Note that there are also 8 terms between 1 and 2 which we can obtain by adding 1 to each of our first 8 terms. For example, $1+\frac{19}{30}=\frac{49}{30}.$ Following this pattern, our answer is $4(10)+8(1+2+3+\cdots+9)=\boxed{400}.$

Sherlock Watson
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    I think I got something. If I add one to the first fraction, it will be $1+1/30$, i.e. $31/30$. $31/30$ added to the last fraction will give me 2 as a total. – Sherlock Watson Nov 24 '18 at 07:55

3 Answers3

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Use the inclusion-exclusion!

Add all fractions: $$\frac{1}{30}+\frac{2}{30}+\cdots +\frac{298}{30}+\frac{299}{30}=\frac{300\cdot 299}{2\cdot 30}=299\cdot 5=1495.$$

Subtract the fractions with the numerator of multiple of $2,3,5$: $$\frac2{30}+\frac4{30}+\cdots +\frac{298}{30}=\frac{300\cdot 149}{2\cdot 30}=745;\\ \frac3{30}+\frac6{30}+\cdots +\frac{297}{30}=\frac{300\cdot 99}{2\cdot 30}=495;\\ \frac5{30}+\frac{10}{30}+\cdots +\frac{295}{30}=\frac{300\cdot 59}{2\cdot 30}=295;\\$$ Add the fractions with the numerator of multiple of $6,10,15$: $$\frac6{30}+\frac{12}{30}+\cdots +\frac{294}{30}=\frac{300\cdot 49}{2\cdot 30}=245;\\ \frac{10}{30}+\frac{20}{30}+\cdots +\frac{290}{30}=\frac{300\cdot 29}{2\cdot 30}=145;\\ \frac{15}{30}+\frac{30}{30}+\cdots +\frac{285}{30}=\frac{300\cdot 19}{2\cdot 30}=95;\\$$ Subtract the fractions with numerator of multiple of $30$: $$\frac{30}{30}+\frac{60}{30}+\cdots+\frac{270}{30}=\frac{300\cdot 9}{2\cdot 30}=45.$$ Hence: $$1495-(745+495+295)+(245+145+95)-(45)=400.$$

farruhota
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Hint: Notice the same-colored pairs. $$\color{blue}{\frac{1}{30}}+\color{purple}{\frac{7}{30}}+\color{green}{\frac{11}{30}}+\color{red}{\frac{13}{30}}+\color{red}{\frac{17}{30}}+\color{green}{\frac{19}{30}}+\color{purple}{\frac{23}{30}}+\color{blue}{\frac{29}{30}}$$

There is a very clear pattern in their sums.

By adding $1$ to all the fractions to get those between $1$ and $2$, the expression can be rewritten as shown below.

$$\color{blue}{\frac{31}{30}}+\color{purple}{\frac{37}{30}}+\color{green}{\frac{41}{30}}+\color{red}{\frac{43}{30}}+\color{red}{\frac{47}{30}}+\color{green}{\frac{49}{30}}+\color{purple}{\frac{53}{30}}+\color{blue}{\frac{59}{30}} = 8+\bigg(\color{blue}{\frac{1}{30}}+\color{purple}{\frac{7}{30}}+\color{green}{\frac{11}{30}}+\color{red}{\frac{13}{30}}+\color{red}{\frac{17}{30}}+\color{green}{\frac{19}{30}}+\color{purple}{\frac{23}{30}}+\color{blue}{\frac{29}{30}} \bigg)$$

This pattern repeats.

KM101
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  • Wonderful, got it. – Sherlock Watson Nov 24 '18 at 08:55
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    As a small addition, you get $4+12+20+…+76$, which is an arithmetic sequence. There are ten terms, so you can use $$S_n = \frac{n(u_1+u_n)}{2}$$ which yields $$S_{10} = \frac{10(4+76)}{2} = 5(80) = 400$$ – KM101 Nov 24 '18 at 09:10
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Notice that all such fractions must be in the form $\frac{\gcd(a,300)=1}{30}, a < 300.$ Since the sum of all numbers less than and relatively prime to a number $n$ is $\frac{n \cdot \phi(n)}{2},$ if $n=300,$ we have:

$$\frac{300 \cdot \phi(300)}{2}= \frac{300 \cdot 80}{2}=12000.$$

Now, we can't forget that we have a denominator of $30$ so we must divide by $30,$ leaving us with $\frac{12000}{30}=400.$ And thus $400$ is our final answer.

Note: $\phi(n)$ is the totient of $n,$ it counts the number of integers less than $n$ that are relatively prime to $n.$

Note: The formula $\frac{n \cdot \phi(n)}{2}$ comes from the fact that for every integer $a$ that is relatively prime to $n,$ $\gcd(a,n)=1,$ and therefore $\gcd(n-a,n)=1$ by the euclidean algorithm. Thus every number relatively prime to $n$ will have a "partner" with which when added to, produces $n.$

Thus we count the number of numbers relatively prime to $n$ and divide it by $2$ to find the number of "partners." Since we know that every pair of "partners" will add up to $n,$ we multiply the expression by $n.$ Hence:

$$\frac{n \cdot \phi(n)}{2}.$$

Hope this helps! (Pls upvote!)