Sum of Reciprocals

First of all, the sum of the reciprocals of the 4 unknown integers is $1-\frac 1 3 - \frac 1 7- \frac 1 9- \frac 1 {11}- \frac 1 {33}= \frac{202}{693}$ . We let $\frac1{5a}+\frac1{5b}+\frac1{5c}+\frac1{5d}=\frac{202}{693}$ , where $a >b>c>d$ are odd integers. Then $\frac1a+\frac1b+\frac1c+\frac1d=\frac{1010}{693}$

If $d \neq 1$ , the left hand side would be not greater than $\frac13+\frac15+\frac17+\frac19=\frac{248}{315}$ . But $\frac{248}{315} < 1 < \frac{1010}{693}$ , which is a contradiction. Hence, $d=1$ and $\frac1a+\frac1b+\frac1c = \frac{317}{693}$ .

Similarly, if $c \neq 3$ , the left hand side would be not greater than $\frac15+\frac17+\frac19 = \frac{71}{315}$ . But $\frac {71}{315} <\frac{317}{693}$ , which is a contradiction. Hence, $c=3$ and $\frac 1 a+\frac 1 b =\frac {86} {693}$ .

This gives $693(a+b) =86ab \Rightarrow 693\cdot 86(a+b) =86^2 ab \Rightarrow (86a-693)(86b-693) = 693^2=3^4\cdot7^2\cdot11^2$ . Since $86a-693 > 86b-693$ $\Rightarrow 86b-693\leq 693$ .Thus $0<b<17$ . A quick test shows that only when $b=9$ gives that $86b-693=81$ is a factor of $693^2$ . Thus $b=9, a=77$ . As a result, we have $a=77, b=9, c=3, d=1$ . Thus the nine integers are $3, 5, 7, 9, 11, 15, 33, 45, 385$ , and $3+5+7+9+11+15+33+45+385=\boxed{513}$ .

After removing the first five fractions, you are left with 202/693. If you skip 5 and use 15,25,35, and 45 (and their corresponding fractions) you don't get high enough. So 5 (and its fraction 1/5) must be included. Similarly, by taking out 1/5, you find that 15 (and its fraction 1/15) must be included as well. You are left with 86/3465. From there you see that 1/45 is the next highest fraction that is possible. Solving the equation gives 385 for the final number. My solution is admittedly inelegant...I would love to see an elegant one!

There are five (5) given and four (4) unknown integers. It is said that the sum of their reciprocals is 1. This means that if you have to add these fractions, the least common multiple of the denominators should be equal to its resulting numerator.

The LCM of the 5 given integers is 693. Since the 4 unknown integers all have 5 as its units digit, then these integers are sure to be multiples of 5. This follows that the working LCM to be used is 693*5 which is equal to 3465.

$\frac {1}{3} + \frac {1}{7} + \frac {1}{9} + \frac {1}{11} + \frac {1}{33} + \frac {1}{5a} + \frac {1}{5b} + \frac {1}{5c} + \frac {1}{5d} = \frac {3465}{3465}$ where a, b, c, d are distinct integers.

By solving, $1155 + 495 + 385 + 315 + 105 + \frac {693}{a} + \frac {693}{b} + \frac {693}{c} + \frac {693}{d} = 3465$

$\frac {693}{a} + \frac {693}{b} + \frac {693}{c} + \frac {693}{d} = 1010$

Given that $693 = 3^2 * 7 * 11$ , the possible values of a, b, c, d are ${1, 3, 7, 9, 11, 21, 33, 63, 77, 99, 231, \ldots and all possible multiplication pairings of the factors}$ to make each of the remaining fractions integer. Since the expected sum of these four fractions (see the simplified equation above) is 1010 then the values of a, b, c, d to be considered are ${1, 3, 9, 77}$ in any assignments. Note: Since 1010 is the aimed sum, then most of the four needed values (a, b, c, d) must be minimal to reach this sum.

Therefore, $3 + 7 + 9 + 11 + 33 + 5a + 5b + 5c + 5d = 63 + 5(1) + 5(3) + 5(9) + 5(77) = **513**$ .

Since we have the number ending with 5 Sum of 1/3 + 1/7 + 1/9 + 1/11+ 1/33 = 491/693 Which is approx .7 We are left with .3 Without 1/5 and 1/15 we cannot reach .3 as 1/25 is just .04 and then the values keep diminishing Therefore 1/5 + 1/15 + 491/693 = 3379/3465 so we are left with 86/3465 The only way we can split 3465/5 = 693 to get 86 is 77 and 9 Thus the last 2 numbers are 77*5 = 385 and 9 * 5 = 45

Therefore overall total = 3 + 7 + 9 + 11 + 33 + 5 + 15 + 45 + 385 = 513

Firstly, $\frac{1}{3}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{33}=\frac{491}{693}$ . Therefore, the reciprocals of the other 4 integers add up to $\frac{203}{693}$ . Because we know that the units digit of all 4 of these numbers is 5, we can express the sum of their reciprocals as \frac{1}{10a+5} + \frac{1}{10b+5} + \frac{1}{10c+5} + \frac{1]{10d+5} = \frac{202}{693} . Multiplying by a $5$ on both sides gives us \frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}+\frac{1]{2d+1} = \frac{1010}{693} . Next, factor $693$ into $3^{2} \times 7 \times 11$ . We want to find 4 numbers $n_{1}, n_{2}, n_{3}, n_{4}$ that multiply to $693k$ and that satisfy the identity $n_{1} \times n_{2} \times n_{3} + n_{2} \times n_{3} \times n_{4} + n_{1} \times n_{2} \times n_{4} + n_{1} \times n_{3} \times n_{4} = 1010k (expression1)$ , where k is an even integer. While $k=1$ , The possible choices for the four numbers are $(3,3,7,11),(1,3,3,77),(1,3,11,21),(1,3,7,33)$ etc... The more stratified the four numbers are, the larger expression will be because the larger numbers will be contained in $3/4$ of the terms. ${1,3,11,21}$ yields a value of 1020, while ${1,3,7,33}$ and ${1,9,7,11}$ , the closest to ${1,3,11,21}$ in terms of "stratification," give us values of $1044$ and $932$ respectively. This means that for $k=1$ , there are no sets of $n_{1}, n_{2}, n_{3}, n_{4}$ that give a value of 1010 for expression1. Now lets examine $k=3$ . Then the four numbers have to multiply to $2079$ and expression1= $3030$ . Some experimenting gives us the four numbers $1,3,9,77$ , which satisfy expression1. Therefore, the four remaining numbers are $5, 15, 45,$ and $385$ . This gives us a total sum of $513$ .

Motivation : Because all the numerators of the fractions are $1$ , Egyptian fraction is a valid approach.

Let the other 4 integers be $5a, 5b, 5c, 5d$ for positive integers $a<b<c<d$ , then

$\begin{aligned} 1 - \left ( \frac {1}{3} + \frac {1}{7} + \frac{1}{9} + \frac {1}{11} + \frac {1}{33} \right ) & = & \frac {1}{5a} - \frac {1}{5b} - \frac{1}{5c} - \frac {1}{5d} \\ \frac {202}{693} & = & \frac {1}{5a} + \frac {1}{5b} + \frac{1}{5c} + \frac {1}{5d} \\ \frac {1010}{693} & = & \frac {1}{a} + \frac {1}{b} + \frac{1}{c} + \frac {1}{d} \\ 1 + \frac {317}{693} & = & \frac {1}{a} + \frac {1}{b} + \frac{1}{c} + \frac {1}{d} \Rightarrow a = 1 \\ \frac {317}{693} & = & \frac {1}{b} + \frac{1}{c} + \frac {1}{d} \\ \frac {1}{ \lceil{\frac {693}{317} } \rceil} + \frac {- 693 \bmod{317} }{693 \lceil{\frac {693}{317} } \rceil } & = & \frac {1}{b} + \frac{1}{c} + \frac {1}{d} \\ \frac {1}{ 3} + \frac {258 }{2079 } & = & \frac {1}{b} + \frac{1}{c} + \frac {1}{d} \Rightarrow b = 3 \\ \frac {258 }{2079 } & = & \frac{1}{c} + \frac {1}{d} \\ \frac {1}{ 9 } + \frac {1 }{77} & = & \frac{1}{c} + \frac {1}{d} \Rightarrow c = 9,d = 77 \\ \end{aligned}$

Hence, the answer is $3 + 7 + 9 + 11 + 33 + 5(1) + 5(3) + 5(9) + 5(77) = \boxed{513}$

Let's start with summing the reciprocals of 3, 7, 9, 11 and 33, and subtract it from 1.

$1 - (\frac {1}{3} + \frac {1}{7} + \frac {1}{9} + \frac {1}{11} + \frac {1}{33}) = \frac {202}{693}$

The remaining 4 numbers are multiples of 5, let's call them 5a, 5b, 5c and 5d (with a, b, c and d odd numbers).

$\frac {1}{5a} + \frac {1}{5b} + \frac {1}{5c} + \frac {1}{5d} = \frac {202}{693} \rightarrow \frac {1}{a} + \frac {1}{b} + \frac {1}{c} + \frac {1}{d} = \frac {1010}{693}$

If we factorize 693, we get 693 = 1 $*$ 3 $*$ 3 $*$ 7 $*$ 11.

Using a < b < c < d, the first trials are a=1and b=3:

$\frac {1}{c} + \frac {1}{d} = \frac {1010}{693} - \frac{1}{1} - \frac {1}{3} = \frac {1010}{693} - \frac{693}{693} - \frac {231}{693} = \frac {1010-924}{693} =\frac{86}{693} < \frac {1}{7}$

So the next value in line is c=3 $*$ 3=9, which makes the remaining equation:

$\frac {1}{d} = \frac{86}{693} - \frac {1}{9} = \frac {86}{693} - \frac{77}{693} = \frac {9}{693} = \frac {1}{77}$

We now have the last four numbers: 5a, 5b, 5c and 5d $\rightarrow$ 5, 15, 45 and 385.

If we sum all 9 numbers, the result is:

$3+7+9+11+33+5+15+45+385 = \fbox {513}$

I don't know how to solve this problem algebraically, but I did solve it using some logic and trial-and-error. First, it is given that \begin{align} \frac{1}{3} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{33} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 1 \end{align} where a, b, c, and d are integers with 5 in the units place.

We first note that \begin{align} a = 5 \end{align} because if it did not equal 5, it would be impossible for the terms to sum to 1. The greatest sum without a = 5 is \begin{align} \frac{1}{3} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{33} + \frac{1}{15} + \frac{1}{25} + \frac{1}{35} + \frac{1}{45} = \frac{1667}{1925} < 1 \end{align}

We can continue by using the "greedy" algorithm, where we add the largest possible unit fraction that we can which keeps the sum below 1.

Note that \begin{align} \frac{1}{3} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{33} + \frac{1}{5} + \frac{1}{15} = \frac{3379}{3465} < 1 \end{align}

so we can set \begin{align} b = 15 \end{align}

Continuing, the next largest unit fraction that we can add is \begin{align} \frac{1}{45} \end{align}

so \begin{align} c = 45 \end{align}

Finally, we see that \begin{align} \frac{1}{3} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{33} + \frac{1}{5} + \frac{1}{15} + \frac{1}{45} = \frac{384}{385} \end{align}

so \begin{align} d = 385 \end{align}

Thus, \begin{align} 3 + 7 + 9 + 11 + 33 + a + b + c + d =\\ 3 + 7 + 9 + 11 + 33 + 5 + 15 + 45 + 385 = \boxed{513} \end{align}

1/3 +1/7 +1/9 +1/11 +1/33 =491/693 So,sum of other reciprocals will be 202/693 Now,all other 4 integers have 5 as unit digit. So, I multiplied both numerator and denominator by 5. So, I got 10140/3465 which is more than largest possible reciprocal 1/5. I took 5 as one of the 9 numbers. In this way I took 15 and 45 into my account. Atlast I got 9/3465 which is equal to 1/385. So, I got 385 the last number. Added all 9 numbers and got the result.

Let the numbers be $x_1, x_2, x_3, x_4, 3, 7, 9, 11, 33$ . Then we have

$\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}=\frac{202}{693}$

We are given that $x_1 \equiv x_2 \equiv x_3 \equiv x_4 \pmod 5$ , so let $x_i=5y_i$ for $i=1,2,3,4$ . Then we get

$\frac{1}{y_1}+\frac{1}{y_2}+\frac{1}{y_3}+\frac{1}{y_4}=\frac{1010}{693}$

and $y_1, y_2, y_3, y_4$ are odd. WLOG $y_1<y_2<y_3<y_4$ . Note that

$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}<\frac{1010}{693}$

and so $y_1=1$ . By a similar bounding argument, $y_2=3$ . Thus

$\frac{86}{693}=\frac{1}{y_3}+\frac{1}{y_4}$

$\implies 86y_3y_4=693(y_3+y_4)$

Note that $693=3^2 \cdot 7 \cdot 11$ . Since $693 \mid y_3y_4$ and $86 \mid y_3+y_4$ , it is easy to find $y_3=9$ and $y_4=77$ . Adding the numbers gives $513$ .