1. ## Fractions

Hi

I spent a lot of time trying to sovle this so I would really appreciate it if someone could just give me some hints what I am doing wrong.

Vicky.

There are positive intergers k, n, m such that

19/20 < 1/k + 1/n + 1/m < 1

What is the smallest possible value of k+n+m?

19 < 20 ( 1/k + 1/n + 1/m ) < 20

19 < 20 ( nm + km + kn/ knm ) < 20

nm + km + kn = try to make the value about 19.01

knm = 20

I did a lot of guessing and checking but now I am starting to see my solution doesn't make any sense..

Vicky.

2. Originally Posted by Vicky1997
Hi

I spent a lot of time trying to sovle this so I would really appreciate it if someone could just give me some hints what I am doing wrong.

Vicky.

There are positive intergers k, n, m such that

19/20 < 1/k + 1/n + 1/m < 1

What is the smallest possible value of k+n+m?

just a try

$\displaystyle \frac{19}{20} < \frac{1}{k}+\frac{1}{n}+\frac{1}{m} < 1$

$\displaystyle 19 < \frac{20}{k} + \frac{20}{n} + \frac{20}{m} < 20$

take the numbers 2,3,6

$\displaystyle \frac{20}{2} + \frac{20}{3} + \frac{20}{6} = 10 + 6.6666 + 3.3333$

2+3+6 = 11 I think it is the smallest number

3. Originally Posted by Amer
$\displaystyle \frac{20}{2} + \frac{20}{3} + \frac{20}{6} = 10 + 6.6666 + 3.3333$

2+3+6 = 11 I think it is the smallest number
Not quite. 10 + 6.6666... + 3.3333... = 20 (we have repeating decimals), and the sum of the fractions has to be less than 20.

01

4. Originally Posted by yeongil
Not quite. 10 + 6.6666... + 3.3333... = 20 (we have repeating decimals), and the sum of the fractions has to be less than 20.

01
the sum is less than 20

10+3.3333333333... + 6.66666666... = 19.999999999999... < 20 right

5. Originally Posted by Amer
the sum is less than 20

10+3.3333333333... + 6.66666666... = 19.999999999999... < 20 right
No. Go back to the fraction form:
$\displaystyle \frac{20}{2} + \frac{20}{3} + \frac{20}{6}$
$\displaystyle = \frac{60}{6} + \frac{40}{6} + \frac{20}{6}$
$\displaystyle = \frac{120}{6}$
$\displaystyle = 20$

Also, I believe it's been proven that 0.999.... (repeating forever) = 1.

01

6. Originally Posted by yeongil
No. Go back to the fraction form:
$\displaystyle \frac{20}{2} + \frac{20}{3} + \frac{20}{6}$
$\displaystyle = \frac{60}{6} + \frac{40}{6} + \frac{20}{6}$
$\displaystyle = \frac{120}{6}$
$\displaystyle = 20$

Also, I believe it's been proven that 0.999.... (repeating forever) = 1.

01

I think we should take 2,3

ok take 2,3,7

7. Originally Posted by Amer
I think we should take 2,3

ok take 2,3,7
Again no, you get a fraction less than 19/20 (about 0.80952380952381).

I'm starting to think that there is no solution.

01

8. Originally Posted by yeongil
Again no, you get a fraction less than 19/20 (about 0.80952380952381).

I'm starting to think that there is no solution.

01

9. I take that back. You're right -- 2, 3, and 7 does work. I was using a spreadsheet when figuring out 1/k + 1/n + 1/m and I must have inputted the formula wrong. My bad.

01

EDIT: I found out what I did. I plugged in 2, 6 & 7 by mistake. Oops!

10. Originally Posted by yeongil
I take that back. You're right -- 2, 3, and 7 does work. I was using a spreadsheet when figuring out 1/k + 1/n + 1/m and I must have inputted the formula wrong. My bad.

01

EDIT: I found out what I did. I plugged in 2, 6 & 7 by mistake. Oops!

no problem we all make mistakes thanks

11. Thanks everyone for helping me out.

Vicky

12. Originally Posted by Amer
just a try

$\displaystyle \frac{19}{20} < \frac{1}{k}+\frac{1}{n}+\frac{1}{m} < 1$

$\displaystyle 19 < \frac{20}{k} + \frac{20}{n} + \frac{20}{m} < 20$

take the numbers 2,3,6

$\displaystyle \frac{20}{2} + \frac{20}{3} + \frac{20}{6} = 10 + 6.6666 + 3.3333$

2+3+6 = 11 I think it is the smallest number
Is this the equation I should have set up?
And start plugging and checking the values?

Vicky.

13. Originally Posted by Vicky1997
Is this the equation I should have set up?
And start plugging and checking the values?

Vicky.
you can leave it but I prefer my equation

$\displaystyle 19<\frac{20}{k}+\frac{20}{m}+\frac{20}{n}<20$

since the both right side and the left side are integers then I just start to guess numbers first take 2,2,2 that dose not work cuz the sum is 30

it is clear that if we choose two numbers (2) the sum of them will be more than 20 so try

2,3,3 dose not work too >20

try
2,3,4 dose not work too >20

2,4,4 =20 the same

3,3,3 < 19 so hold 2

2,3,5 >20

2,3,6 = 20 as you see

r
so
2,3,7 work

you can see that if we take bigger number the sum $\displaystyle 20/k + 20/n + 20/m$ become smalle

14. Thank you.

Vicky.