Online Course Discussion Forum

MCIII Number Theory 2.28

 
 
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MCIII Number Theory 2.28
by Tina Jin - Wednesday, 27 December 2023, 6:12 PM
 

Hello,


I found that it must be 3 since the sum that is subtracted from 4 cannot be greater than 1. I also found 1/2+2/3+8/9+17/18=3, but how do I find all of them?


Thanks,

Tina Jin

 
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Re: MCIII Number Theory 2.28
by Dr. Kevin Wang - Friday, 29 December 2023, 10:28 PM
 

$n=3$ is correct, and the example you found is good.

To find all the sets, let $a < b < c < d$ and $$\frac{a-1}{a} + \frac{b-1}{b} + \frac{c-1}{c} + \frac{d-1}{d}=3.$$ So it turns out that $$\frac{1}{a}+\frac{1}{b} +\frac{1}{c} + \frac{1}{d}=1.$$

The choices for $a$ are very limited. Similar for $b$.  Once you fix $a$ and $b$, you can solve for $c$ and $d$.

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Re: MCIII Number Theory 2.28
by Tina Jin - Friday, 5 January 2024, 11:41 AM
 

I still have a question on this problem:

Here's what I did:

Note a=2 and only 2 because if a≥3, then 1/b+1/c+1/d≥2/3 but max(1/b+1/c+1/d)<2/3.

I fixed a=2:

Fix b=3:

1/c+1/d=1/6

Expand and factor: (c-6)(d-6)=36, then get corresponding solutions for corresponding factor pairs.

And the same for b=4, 6 (b=5 doesn't give integers c and d). But some solutions overlap. 

How do I know if a value of b will give an integer, and how do I know if a solution will overlap? Do I have to test everything out until the biggest number in the variable set (42 I think, since 1/2+2/3+6/7+41/42=3 and 41/42 is the closest solution I got to 1)

Thanks,

Tina Jin

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Re: MCIII Number Theory 2.28
by Dr. Kevin Wang - Saturday, 6 January 2024, 1:02 AM
 

Each solution is an ordered quadruple $(a,b,c,d)$, so there is no such thing as "overlap" of solutions.

There is only one case for $a$: $a=2$.  There are only two cases for $b$: $b=3$ and $b=4$.  Then for each case you solve for $c$ and $d$, and only a few choices in each case.  You are already on the right track of finding them, so why not just carry it out?

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Re: MCIII Number Theory 2.28
by Tina Jin - Sunday, 14 January 2024, 10:07 AM
 

I am still a little bit confused. How do you know there can only be two cases, b=3,4? 

By Simon's Favorite Factoring Trick,

If b=5: 1/c+1/d=3/10

10c+10d=3cd so 3cd-10c-10d=0, so (c-10/3)(d-10/3)=10/9. Now multiply both sides by 9 to get (3c-10)(3d-10)=10 c=4 and d=5 work, but now d=b so there are no solutions.

But of course it is not possible to do this for every integer b>5.

How do you prove it must be b≤4?

Thanks,

Tina Jin