Factorization

C

castelobrz

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Being [imath]\sqrt{a^{2}+1}+\sqrt{b^{2}+4}+\sqrt{c^{2}+9}=10[/imath], find the value of 9abc.

a) 81
b) 64
c) 72
d) 128
e) 156
 
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To get help, you need to be actively involved in getting that help. You need to tell us where you are stuck and show us the work you have done so far, even if know it is wrong, so we can show you where you are making any mistakes. All this was stated in the posting guidelines. Did you read them?
 
topsquark said:
But 9abc here isn't one of the options we can choose.

-Dan
Click to expand...
Yes, I'm aware. My point is that the question written as-is is faulty. By presenting the answers as multiple choice, it's giving off the impression that the solution is unique, which it isn't. One shouldn't be expected to work backwards from the answer.
 
Let c = b = 2.60610545856893950539
a = sqrt( (10 - sqrt(b^2+4) - sqrt(c^2+9) )^2 - 1 ) ≈ 2.55210252473053590658
Then a*b*c*9 ≈ 156

Let c = b = 2.06427941885509084676
a = sqrt( (10 - sqrt(b^2+4) - sqrt(c^2+9) )^2 - 1 ) ≈ 3.33757086001313540265
Then a*b*c*9 ≈ 128

But I wonder if it's possible to specify some exact values that will yield one of the supplied options?
 
Cubist said:
Let c = b = 2.60610545856893950539
a = sqrt( (10 - sqrt(b^2+4) - sqrt(c^2+9) )^2 - 1 ) ≈ 2.55210252473053590658
Then a*b*c*9 ≈ 156

Let c = b = 2.06427941885509084676
a = sqrt( (10 - sqrt(b^2+4) - sqrt(c^2+9) )^2 - 1 ) ≈ 3.33757086001313540265
Then a*b*c*9 ≈ 128

But I wonder if it's possible to specify some exact values that will yield one of the supplied options?
Click to expand...
If we additionally assume that the numbers (a,b,c) are non-negative integers - then 9 * a * b * c can be either 81 or 72. Thus a * b * c = 9 or 8

I did not do it - but may be we can apply brute-force now.
 
Cubist said:
But I wonder if it's possible to specify some exact values that will yield one of the supplied options?
Click to expand...

Yes it's possible.

I found an answer via brute force as @Subhotosh Khan suggested. I'm not sure if I should just provide the answer on here. For now here are some hints...

There are (different) rational values for a and b, and both have the same denominator. They are both vulgar. And c has an integer value

9 * a * b = <+ve integer, with no denominator>
 
Cubist said:
Yes it's possible.

I found an answer via brute force as @Subhotosh Khan suggested. I'm not sure if I should just provide the answer on here. For now here are some hints...

There are (different) rational values for a and b, and both have the same denominator. They are both vulgar. And c has an integer value

9 * a * b = <+ve integer, with no denominator>
Click to expand...

With your hints, the route I took got me to values where the result 9abc is not
one of the answer choices.
a = 5/3
b = 8/3
c = 4

9abc = 9(5/3)(8/3)(4) = 160
 
Subhotosh Khan said:
If we additionally assume that the numbers (a,b,c) are non-negative integers - then 9 * a * b * c can be either 81 or 72. Thus a * b * c = 9 or 8

I did not do it - but may be we can apply brute-force now.
Click to expand...
I got as far as this.
 
lookagain said:
With your hints, the route I took got me to values where the result 9abc is not
one of the answer choices.
a = 5/3
b = 8/3
c = 4

9abc = 9(5/3)(8/3)(4) = 160
Click to expand...
Darn, I tried letting c=4 and could not find b and c. I spent way too much time on this problem (about 45 minutes).
 
Cubist said:
I'm not sure if I should just provide the answer on here.
Click to expand...
I think it's fine for you to post the solution. You're solving a completely different problem than the OP. You're solving for a, b, c given 9abc=81or 72. Not the other way around.
 
BigBeachBanana said:
I think it's fine for you to post the solution. You're solving a completely different problem than the OP. You're solving for a, b, c given 9abc=81or 72. Not the other way around.
Click to expand...
I know that there's infinite solutions to the OP. The original question is probably incomplete. Out of my own interest I've been looking for sets of a,b,c in the form of rational numbers that produce integer results for 9abc (and now I'm looking for abc in the form of a surd times a rational number). This is a very good guess at what the original problem might be. I do hope that the OP will respond with clarification, but somehow I doubt it.

lookagain said:
With your hints, the route I took got me to values where the result 9abc is not
one of the answer choices.
Click to expand...
a = 4/3
b = 8/3
c = 4
these satisfy both of...
9abc = 128
sqrt(a^2+1) + sqrt(b^2+4) + sqrt(c^2+9) = 10

Also
a=0
b=sqrt(14)*4/5
c=sqrt(14)*6/5
give the integer result 9abc = 0 but obviously this isn't one of the given options
 
Steven G said:
...What is never done on this forum is giving answers to students....
Click to expand...

1 hour 20 later...

Steven G said:
...Can you please tell me one solution?
Click to expand...

:ROFLMAO:

It does seem a good strategy for a student to just disappear and then wait for the helpers to take an interest in the question. Sometimes helpers discuss a question and then a potential answer gets revealed with no work at all shown by the OP. I'm obviously guilty of this, occasionally. It might be nice if there was a way for known helpers to discuss a thread with each other (within the particular thread) in such a way that students can't immediately see the "between helper" posts.

Anyway, back to this thread, there could be another set of a,b,c (a set that can be written in some algebraic form) that yields one of the other options.
 
Cubist said:
a = 4/3
b = 8/3
c = 4
these satisfy both of...
9abc = 128
sqrt(a^2+1) + sqrt(b^2+4) + sqrt(c^2+9) = 10
Click to expand...

I understand the correction you made. I did use the value of a initially as you meant,
but ...
after I substituted it into \(\displaystyle \ \sqrt{ a^2 + 1\ } , \ \) I got 5/3 and wrongly substituted that as the
a-value into 9abc.
 
Guys, I solved this problem but the question needs to be more rigorous as to the condition of existence.

I took this question of a exercise list to "Colégio Naval".

1649619190945.png
 
Sorry for my late, the time zone in Brasil is different. When I posted the question, I was going to sleep
 
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