How Can You Find Seven Unit Fractions That Sum to One?

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Homework Help Overview

The original poster seeks to find seven different unit fractions that sum to one, presenting a challenge in the area of number theory. Additionally, a second question involves determining the number of different six-digit numbers that can be formed using the digits 1, 2, 5, 6, 7, and 9, and whether these numbers can be divisible by 6.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the possibility of using a guess and check method for finding the unit fractions. Questions arise regarding the conditions for the six-digit numbers, including whether digits can be repeated and how to determine divisibility by 6.

Discussion Status

Some participants have offered insights into the second question, suggesting that all six-digit numbers formed from the given digits are divisible by 3. However, there is ongoing uncertainty about how to specifically count those divisible by 6. The first question remains largely focused on the guess and check approach without further methods being proposed.

Contextual Notes

Participants note that the second question does not explicitly state whether digits can be reused, leading to different interpretations of the problem. There is also a mention of simplifying fractions in relation to the first question, though the relevance of this to the original poster's inquiry is unclear.

courtrigrad
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1. Find seven different unit fractions whose sum is 1. So

[tex]\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e} + \frac{1}{f} + \frac{1}{g} = 1[/tex] Would this just be purely guess and check?

2. How many different 6 digit numbers can you make using [tex]1,2,5,6,7,9[/tex]. Would it just be [tex]6![/tex]? Also how would you find how many of these numbers are divisble by 6?

Thanks! :smile:
 
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i think i can answer your 2nd question

yes, the combination is 6!

all 6 digits numbers form by those numbers are divisible by 3 (because 1,2,5,6,7,9 add up to 30, which is divisible by 3) therefore, all 6 digits numbers which ended with 2 and 6 are divisible by 6
 
but then for #2 how do you find how many are divisble by 6

also for #1 I don't see another way than gues and check.

thanks
 
courtrigrad said:
but then for #2 how do you find how many are divisble by 6

also for #1 I don't see another way than gues and check.

thanks

For the second part, what are the number of possibilities with a terminal digit of 2 ? With a terminal digit of 6 ? Add these up.

It should be 2*(5!)
 
Thanks a lot

So I guess for the first one we have to guess and check. Just take fractions see if they add to 1, and then take the reciprocal of the reciprocal?
 
courtrigrad said:
2. How many different 6 digit numbers can you make using [tex]1,2,5,6,7,9[/tex]. Would it just be [tex]6![/tex]?
The question doesn't state explicitly that each number may only be used once.
So then 111111 and 965759 etc. are also solutions and the answer would be [tex]6^6[/tex].
 
thanks a lot. i guess the second question is impossible then?
 
I think you have to simplify all of he fractions for #2. IS this right?

Thanks :smile:
 

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