Display Number as a sum so that sum of reciprocals of the summands is 1

Question

Good Evening,

Right now I am solving some Mathematic Puzzles and want to create a formula for the following:

Display the Number 2008 as a sum of natural numbers so that the sum of the reciprocals of the summands is 1?

You can use as many summands as you want!

My thougt:

2008 is divisable by 8 so the sum of the summands only has to be 251 and the sum of the reciprocals of the summands 0,125

The solution is that all summands must be multiples of the prime factors, so in this case 2 and 251

Possible solutions:

8+3*16+11*32+21*64+2*128=2008

and 1/8+3/16+11/32+21/64+2/128=1

3*8+2*16+15*32+64+128+256+2 *512 = 2008

and 3*8^-1+2*16^-1+15*32^-1+64^-1+128^- 1+256^-1+2*512^-1 = 1

8+16+24*32+3*64+4*256

and 1*8^-1+1*16^-1+24*32^-1+3*64^-1+4*256^-1 = 1

Can anyone help me to create a formula to solve this problem?

right now I am stuck at this Sum formulas, I do not know how to go on...

Answer 1

Divisibility is not so important feature here (as Ross Millikan noted properly above).
The Problem is rich on different solutions.
Optimized brute force may be helpful here (?).

A few examples:

\begin{array}{rl} \dfrac{1}{2}+ \dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{55}+\dfrac{1}{420}+\dfrac{1}{429}+\dfrac{1}{1092}=1; & \qquad 2008 = 2+3+7+55+420+429+1092; \\ \dfrac{1}{2}+\dfrac{1}{3}+2\cdot\dfrac{1}{16}+3\cdot\dfrac{1}{73}+\dfrac{1}{1752}=1; & \qquad 2008 = 2 + 3+ 2\cdot 16 + 3 \cdot 73 + 1752; \\ 2\cdot\dfrac{1}{4}+8\cdot\dfrac{1}{25}+18\cdot\dfrac{1}{100}=1; & \qquad 2008=2\cdot 4 +8\cdot 25 +18\cdot 100; \\ 2\cdot\dfrac{1}{6}+5\cdot\dfrac{1}{12}+22\cdot\dfrac{1}{88}=1; & \qquad 2008 = 2\cdot 6 +5\cdot 12 +22\cdot 88; \\ 3\cdot\dfrac{1}{8}+6\cdot\dfrac{1}{32}+28\cdot\dfrac{1}{64}=1; & \qquad 2008 = 3\cdot 8 + 6\cdot 32 + 28 \cdot 64; \\ \cdots & \qquad \cdots \end{array}

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If we search $n$ terms $d_1,d_2,\ldots,d_n$ ($d_1\le d_2 \le \ldots \le d_n$), such that:

$$ \left\{ \begin{array}{c} d_1+d_2+\ldots+d_n=2008; \\ \frac{1}{d_1}+\frac{1}{d_2}+\ldots+\frac{1}{d_n}=1; \end{array} \right. $$

then for small $n$ algorithm has following bounds for $d_1,d_2,...,d_n$:

$$ 2\le d_1 \le \min\{n,2008\}; $$ (otherwise $d_1$ isn't the smaller term);

$$ \left\{ \begin{array}{rcl} d_1 \le & d_2 & \le 2008-d_1; \\ \dfrac{1}{1-\frac{1}{d_1}} \le & d_2 & \le \dfrac{n-1}{1-\frac{1}{d_1}}; \end{array} \right. $$

if $\dfrac{1}{d_1}+\dfrac{1}{d_2}\ne 1$, then $$ \left\{ \begin{array}{rcl} d_2 \le & d_3 & \le 2008-d_1-d_2; \\ \dfrac{1}{1-\frac{1}{d_1}-\frac{1}{d_2}} \le & d_3 & \le \dfrac{n-2}{1-\frac{1}{d_1}-\frac{1}{d_2}}; \end{array} \right. $$

and so on.

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As I see, minimal amount of terms is $7$.

Here is list of such $7$-tuples:

 2  3  7  55  420  429  1092;
 2  3  9  20  336  630  1008;
 2  4  5  24  168  665  1140;
 2  4  5  35   58  280  1624;
 2  4  6  20   32  864  1080;
 2  5  5  14   38  684  1260.

List of such $8$-tuples is much wider: here is begin of list:

 2  3  7  66  385  418  462  665;
 2  3  7  70  243  486  567  630;
 2  3  7  70  252  429  585  660;
 2  3  7  70  266  380  608  672;
 2  3  7  70  273  380  532  741;
 2  3  7  70  288  336  630  672;
 2  3  7  70  315  319  522  770;
 2  3  7  70  330  360  396  840;
 ... ... ...  ...  ...  ...  ...