math teaser [solved]

[dolor]

The sum of two numbers is 28, and their product is 7. Find the sum of the reciprocals of the numbers.

[NotLKH]

(a+b)/ab

[Acidic]

Well since 7+1 < 28 one of the numbers has to be larger than 7, and the other has to be a fraction. But to get them to add up to 28 is not easy, still working on it.

but if one is a fraction, and they should add up to 28, the non-fraction should be bigger than 27.

but then 27 is a prime number which is not nice. It doesn't multiply with a fraction to give nice numbers.

So (if I'm right)
lets call the two numbers x and y.
x+y=28
x*y=7
27<x<28
0<y<1

but there are not many decimals that multiply with other decimals to give integers (if any).

I think this is insolvable.



hold on, new method
substitute so you get:
x+(7/x)=28
x²-28x+7=0 (multiply by x)
x=27.74772708
or x=0.2522729151

therefore, lets call:
x=27.74772708
and
y=0.2522729151
x^-1+y^-1=4.000000001 (unless I typed it in on the calc. wrong.)
Anyway, I think the answer is 4.

[NotLKH]

a + b = 28
ab = 7

What is (1/a) + (1/b) = ???

(1/a) + (1/b) = (b/b)*(1/a) + (a/a)*(1/b)
==> (b/ab) + (a/ab)
==> (a+b)/(ab)

Nuff said

[TheManWhoCan]

Did that by any chance originate from a product/sum of quadratic roots problem?

[dolor]

Originally posted by TheManWhoCan
Did that by any chance originate from a product/sum of quadratic roots problem?

nope

[Something Else]

so its not (a+b)/ab?

[dolor]

Originally posted by Something Else
so its not (a+b)/ab?

no, i think acidic is on the right track

[Something Else]

So, where is my math wrong?

You said a + b = 28
and a*b = 7

and you want (1/a) + (1/b)

Now I reckon if you multiply 1/a BY 1 YOU STILL HAVE 1/A.
sAME THING WITH 1/B.

So, (b/b)*(1/a) = b/ab
and (a/a)*(1/b) = a/ab
so that means (1/a) + (1/b) = (b/ab) + (a/ab)
and this becomes (a+b)/ab.

so how can (a+b)/ab not be the answer?

[kedaman]

(a+b)/ab=28/7=4

[dolor]

sorry, i misread the reply (several times i guess). for some reason, i thought you meant a = 7 and b = 28 then you plug them into (a + b)/ab ...sorry bout that. good show.

[NotLKH]

-Lou

[dolor]

Originally posted by NotLKH

-Lou

ive found this problem im having lately...i can't seem to read english

[Acidic]

so everyone was right, how nice

[NotLKH]

You're right!


Dolor,
What solution did you have in mind? If its different, it'd be interesting to see.

-Lou

[dolor]

Originally posted by NotLKH
Dolor,
What solution did you have in mind? If its different, it'd be interesting to see.

-Lou

Actually, I was doing it the way Acidic went about it and got 4, but like I said, I can't read English (it happens every other day) and so I misunderstood what you were saying. When I plugged in the nums the way I thought you said to do it, I got 5/28 So that's why I thought you were wrong. Anywho, it's all settled now. Good show.

[sql_lall]

Ahh...Symmetric Polynomials are wonderous things...

anyway, NotLKH is completely right in their first post, using 'poets' way, i.e. short and beautiful.

However, people seem to want to find the numbers, let's bring symmetric polynomials into play.
As you may or may not know, you are looking for two numbers, and have been given their two symmtric polynomials (a+b, and ab)

so, we do a little thing like:

(x-a)(x-b) = x^2 - (a+b)x + (ab)
replacing with what we know, a and b are the roots of
x^2 - 28x + 7
use the quadratic formula, find the two roots, whoila!

(i believe the general is 1/2 *( S +/- sqrt(S^2 - 4P) ) where S = sum, P = product)

[da_silvy]

but then 27 is a prime number which is not nice.

3 * 9 = ?


***...

[Acidic]

*Whacks myself over head repeatedly*

[da_silvy]

Hehehe, i've done something like that in an exam before :/

[alkatran]

a+b = 28
a*b = 7

b=7/a
b=28-a
28-a = 7/a
a(28-a) = 7

-a^2 + 28a - 7 = 0

[-28 +- sqr(28^2 - 4*-1*-7)] / (2*-1) = 14 +- sqr(756)/-2
sqr(756)/-2 ~= -13.7477
a = 14 +- 13.7477 = .2523 or 27.7477


The two numbers are .2523 and 27.7477
1/.2523 + 1/27.7477 = 3.9635 + .03603 = 3.9995
(may be 4, some precision was lost with calculations, and all I have is the computer's calculator).



*was done before reading replies* aw, that was sortof fun, I figured it would be some sort of trick...