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"Let s be an integer greater than 6. A solid cube of side s has a square hole of side x < 6 drilled directly through from one face to the opposite face (so the drill removes a cuboid). The volume of the remaining solid is numerically equal to the total surface area of the remaining solid. Determine all possible integer values of x."
We also note that s > x.
Equating the surface area and the volume gives;
So
But s > x, so
which simplifies to
But x < 6, so
which simplifies to
This has critical values at
But s > 6 and s is an integer, and must also satisfy the inequality, so the possible values of s are:
s = 7, 8, 9, 10, 11, 12
I substituted these values for s into one of my equations for x (the original equation I set up, for surface area = volume, which was an equation in s and x), and I only got one solution which was an integer solution smaller than 6, which was:
s = 10, x = 5
Can someone confirm if this is correct, or if there are any other solutions? Or if there's a better way of doing it?
(I wish I could check the mark scheme, but they don't publish them online.)
hi zetafunc
Nice solution!
And I agree with every stage.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi,
Thanks... I just wish there was a way to check my answers! This question was from the British Mathematical Olympiad (Round 1), but I can't seem to find any solutions anywhere. They seem to publish them in video format for a short time after the paper is sat, then remove them. It makes it a little harder to prepare for...
Which year is it from?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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It says "Round 1: Thursday, 2 December 2010" and "2010/11 British Mathematical Olympiad" at the top.
And I do not agree with zetafunc.'s solution.
The description "solid cube of side s has a square hole of side x" gives me an image as in the attached picture.
From it, the surface of the remaining solid is: 4*s^2 + 2*(s^2-x^2) + 4*s*x. Here:
4*s^2 - for four untouched sides of the original cube
2*(s^2-x^2) - for face and back of the original cube
4*s*x - for the inner surface of the hole
The volume of the remaining solid is: s^3-s*x^2.
Therefore we have an equation:
From this we can see that if s=6, then x=0. But if s=7, then x=7/5. And with increasing of the s, x also increases. (in fact, for s=10, x=5).
Therefore, the final answer: x in {1, 2, 3, 4, 5}.
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hi White_Owl
That's the same interpretation as zetafunc.
And the same equation.
He jumped a few steps but gets to the same solution.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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But I don't understand how zetafunc. got the starting equation.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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It's a re-arrangement of White-Owl's equation
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Oh, yes. I see it now. I am just wondering if he started his work directly from that equation or if he started with the same equation as White_Owl did.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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He may have working on paper that he assumed we could do without.
But I wouldn't be surprised if he said he did the re-arrangement in his head.
He is a very good mathematician.
Made any progress with his factorials question?
Oh and by the way, if you want to try some of these questions yourself, they can be found at
http://www.bmoc.maths.org/home/bmo.shtml
Just don't expect me to know how to do them.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob
I looked at the problems from Round 1 Year 2011/12. Those are too easy. I managed to solve 4/6 for now.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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That's the same interpretation as zetafunc.
And the same equation.
He jumped a few steps but gets to the same solution.
ummmm.... Ok... I see now, were we went into different directions.
Although I still think my solution is more clear.
Well it is more clear for me
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