Finding a Solution to an Inequality in Natural Numbers

[Albert1]

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

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

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

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?

 

[Albert1]

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

 

[CellCoree]

As a scientist, my approach to finding a solution to this inequality would involve using mathematical analysis and logic to determine the values of a, b, c, d, e, f, and g that satisfy the given conditions.

First, let's consider the condition that a<b<c<d<e<f<g. This means that the numbers must be consecutive and increasing. One possible set of numbers that satisfies this condition is 1, 2, 3, 4, 5, 6, and 7.

Next, we need to find values for a, b, c, d, e, f, and g that satisfy the equation $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}+\dfrac{1}{f}+\dfrac{1}{g}=1$. One way to do this is to start with the largest number, g, and work backwards. We know that g must be greater than 1, since it is the largest number. So, let's start with g=7.

Plugging in g=7 into the equation, we get:

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}+\dfrac{1}{f}+\dfrac{1}{7}=1$

Next, we can simplify this equation by multiplying both sides by 7, giving us:

$7(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}+\dfrac{1}{e}+\dfrac{1}{f}+\dfrac{1}{7})=7$

Simplifying further, we get:

$\dfrac{7}{a}+\dfrac{7}{b}+\dfrac{7}{c}+\dfrac{7}{d}+\dfrac{7}{e}+\dfrac{7}{f}+1=7$

Now, we can see that in order for this equation to be satisfied, the sum of the fractions must be equal to 6 (since 7+1=6). Since we know that the numbers must be consecutive and increasing, we can assign values to