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How to solve this Question?
:
1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1
1/z means 1 over z a fraction.
The letters can be any number.
How to find seven numbers that fit into the equation?
(the numerator is always 1)
What if its 8 sets of fraction? is there any technique?
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Are you allowed to have them all as 1/7?
Why did the vector cross the road?
It wanted to be normal.
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How to solve this Question?
:1/z+ 1/y+ 1/x + 1/w + 1/v + 1/t + 1/s = 1
1/z means 1 over z a fraction.
The letters can be any number.
How to find seven numbers that fit into the equation?(the numerator is always 1)
What if its 8 sets of fraction? is there any technique?
Well, if you're allowed to use mathsyperson's technique, then 1/number of fractions that sum up to 1 would be your technique. So for 8 sets of fractions, it's just 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 = 1
Of course the reason this works is simply that if seven fractions sum up to 1, then the denominator is 1/7 of 1.
Don't quote me on that.
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I thought of a "better" technique to use if that first one is cheating. Kinda nifty too, because it can be modified to fit any number of fractions (nearly).
First, note that 1/3 + 1/6 = 1/2.
Therefore, 1/2 + 1/3 + 1/6 = 1, and so there's your answer for three fractions.
However, you can half each of those fractions and then add half to get 1/2(1/2+1/3+1/6) + 1/2
= 1/4 + 1/6 + 1/12 + 1/2 = 1, which is your answer for four fractions.
Repeat this method to get the answer for as many fractions as you want. So, in your initial question, the answer would be 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/48 + 1/96.
In general, to get n fractions which add up to 1, the first (n-2) fractions should start with 1/2 and continue with each fraction being half of the previous one.
Then the last two would be 1/3 and 1/6, multiplied by the fourth-to-last fraction. I apologise if that made no sense.
So anyway, that method gets you a solution (although there are almost certainly others) for any amount of fractions more than 2.
If you only want one fraction, then you have the trivial answer of 1/1 = 1.
And I'm pretty sure that there isn't an answer for two different fractions.
Why did the vector cross the road?
It wanted to be normal.
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mathsy strikes again! That's beautiful!
````````````
ignore the ````` marks; it's for spacing since the spacebar is too thin.
```````````````````````````1/2`+1/3`+1/6``=1
``````````````````````1/2`+1/4`+1/6`+1/12`=1
`````````````````1/2`+1/4`+1/8`+1/12+1/24`=1
````````````1/2`+1/4`+1/8`+1/16+1/24+1/48`=1
```````1/2`+1/4`+1/8`+1/16+1/32+1/48+1/96`=1
``1/2`+1/4`+1/8`+1/16+1/32+1/64+1/96+1/192=1
So you go by 50% mores and then do 66.66% more and 50% more to finish.
igloo myrtilles fourmis
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I began with the same as John E. Franklin.
1/3 + 1/6 = 1/2
That means only 1/2 remains.
So, I added the two largest fractions whose sum will not surpass 1/2:
1/4 + 1/5 = 9/20
9/20 is only 1/20 less the 1/2 that remained.
So, I added two numbers together that would get me really close to 1/20, but leave a little for the final fraction.
I tried 1/40 + 1/30 = 7/120. That didn't work because 1/20 = 6/120.
I changed 1/30 to 1/60, so I could still have 120 as my denominator.
1/40 + 1/60 = 5/120, which is only 1/120 away from the 1/20 that remained.
Therefore
1/3 + 1/6 + 1/4 + 1/5 + 1/40 + 1/60 + 1/120 = 1
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May I get another chance to prove myself a genius?
'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.
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