a math puzzle

[Albert]

$a,b,c,d,e,f,g \in N$

$a<b<c<d<e<f<g$

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}+\dfrac{1}{f}+\dfrac{1}{g}=1$

please find one possible solution of a,b,c,d,e,f,g

(you should find it using mathematical analysis,and show your logic,don't use any

program)

 

[kaliprasad]

$a,b,c,d,e,f,g \in N$

$a<b<c<d<e<f<g$

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}+\dfrac{1}{f}+\dfrac{1}{g}=1$

please find one possible solution of a,b,c,d,e,f,g

(you should find it using mathematical analysis,and show your logic,don't use any

program)

we have
1 = 1/2 + 1/2
= 1/2 + 1/4 + 1/4
= 1/2 + 1/4 + 1/8 + 1/8
= 1/2 + 1/4 + 1/8 + 1/16 + 1/16
= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/32
= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 3/96
= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + (2+1)/96
= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/48 + 1/96
so a =2 , b= 4, c = 8, d= 16,e=32, f= 48,g =96

this follows from 1/a = 1/(2a) + 1/(2a) and
1/(2a) = 1/ (3a) + 1/(6a)

 

[Opalg]

$a,b,c,d,e,f,g \in N$

$a<b<c<d<e<f<g$

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}+\dfrac{1}{f}+\dfrac{1}{g}=1$

please find one possible solution of a,b,c,d,e,f,g

Just to liven the problem up a bit, what are the smallest and largest possible values of $g$?

The smallest I have found so far is $g=30$, from the sum $$\frac13 + \frac15 + \frac16 + \frac19 + \frac1{10} + \frac1{18} + \frac1{30} = \frac{30 + 18 + 15 + 10+9+5 + 3}{90} = \frac{90}{90} = 1.$$ (I found that example by taking a number with many divisors, in this case $90$, and looking for seven of its divisors that added up to $90$.)

The largest value I can find for $g$ is $10\,650\,056\,950\,806$. That came from repeatedly using the relation $\dfrac1n = \dfrac1{n+1} + \dfrac1{n(n+1)}$, as follows: $$1 = \frac12 + \frac13 + \frac16,$$ $$1 = \frac12 + \frac13 + \frac17 + \frac1{42},$$ $$1 = \frac12 + \frac13 + \frac17 + \frac1{43} + \frac1{1806},$$ $$1 = \frac12 + \frac13 + \frac17 + \frac1{43} + \frac1{1807} + \frac1{3\,263\,442},$$ $$1 = \frac12 + \frac13 + \frac17 + \frac1{43} + \frac1{1807} + \frac1{3\,263\,443} + \frac1{10\,650\,056\,950\,806}. $$ Any advance on ten trillion?

 

[Albert]

$(\dfrac {1}{1} - \dfrac {1}{2}) +(\dfrac {1}{2} - \dfrac {1}{3}) +(\dfrac {1}{3}- \dfrac {1}{4}) + (\dfrac {1}{4} - \dfrac {1}{5}+) (\dfrac {1}{5} - \dfrac {1}{6})+(\dfrac {1}{6} - \dfrac {1}{7})$
$=\dfrac {1}{2} + \dfrac {1}{6}+\dfrac {1}{12} + \dfrac {1}{20} + \dfrac {1}{30}+ \dfrac {1}{42}
=\dfrac {6}{7}=1-\dfrac {1}{7}$
$\therefore \,\, \dfrac {1}{2} + \dfrac {1}{6}+\dfrac {1}{7} + \dfrac {1}{12}+\dfrac {1}{20}+\dfrac {1}{30}+\dfrac {1}{42}=1$