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

AI Thread Summary
To find seven different unit fractions that sum to one, the discussion suggests that guess and check may be the only viable method. For the second question about creating six-digit numbers using the digits 1, 2, 5, 6, 7, and 9, the initial calculation of 6! is correct, but the possibility of repeated digits leads to a total of 6^6 combinations. It is noted that all six-digit numbers formed are divisible by 3, and only those ending in 2 or 6 are divisible by 6. The conversation highlights the complexity of determining the exact count of numbers divisible by 6 and the need to clarify the use of digits. Overall, the discussion emphasizes the challenges in both mathematical problems presented.
courtrigrad
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1. Find seven different unit fractions whose sum is 1. So

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

2. How many different 6 digit numbers can you make using 1,2,5,6,7,9. Would it just be 6!? 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 1,2,5,6,7,9. Would it just be 6!?
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 6^6.
 
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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