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450y - how to approach

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by netigen » Mon Jun 09, 2008 7:17 pm
When you see a question like this, think factors

find factors of 450 = 2 x 3 x 3 x 5 x 5

We know that 450Y = n^3

which means Y should atleast have 2 x 2 x 3 x 5 X N where N is of the form A^3 where A is an integer

Y = 2^2 x 3 x 5 x A^3

Now lets look at the option

I. we are left with A^3 which is an int
II We are left with (2A^3)/3 which may or may not be an int
III We are left with (2A^3)/5 which may or may not be an int

so only I is an integer for sure, ans is B
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by VerbalAttack » Tue Jun 10, 2008 12:29 am
awesome netigen...
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by leovonp » Tue Jun 10, 2008 2:29 am
When you see a question like this, think factors

find factors of 450 = 2 x 3 x 3 x 5 x 5

We know that 450Y = n^3

which means Y should at least have 2 x 2 x 3 x 5 X N where N is of the form A^3 where A is an integer
Sorry netigen, why do you say that Y should at least have 2 x 2 x 3 x 5 x N?

How do you get from the factors of 450 to these factors?

Thx
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by Neo2000 » Tue Jun 10, 2008 5:00 am
leovonp wrote:
When you see a question like this, think factors

find factors of 450 = 2 x 3 x 3 x 5 x 5

We know that 450Y = n^3

which means Y should at least have 2 x 2 x 3 x 5 X N where N is of the form A^3 where A is an integer
Sorry netigen, why do you say that Y should at least have 2 x 2 x 3 x 5 x N?
How do you get from the factors of 450 to these factors?
Thx
For the left Hand side to equal the Right Hand side ( and given that both n and y are positive integers, it implies that the Left Hand side has to be the cube of some number
For e.g. we know that X^3 = 8 then X = 2 because 8 = 2^3 therefore X^3 = 2^3
You can therefore say that if X^3 = 16Y then at the very least Y = 4 and X = 4 since 16 = 4^2 and you need another 4 to satisfy the given equation.
Last edited by Neo2000 on Tue Jun 10, 2008 5:55 am, edited 1 time in total.
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by Ian Stewart » Tue Jun 10, 2008 5:49 am
Neo2000 wrote: You can therefore say that if X^3 = 16Y then Y = 4 and X = 4 since 16 = 4^2 and you need another 4 to satisfy the given equation.
You can't be sure that Y is 4 and that X is 4, which is what the above seems to imply. They certainly *could* have those values, but there are infinitely many other possibilities. For example, X could be 12 = (2^2)*3, and Y could be (2^2)*(3^3) = 108.

What you can be certain of, if X and Y are integers, is that Y is divisible by 4, and the same must be true of X.

Neo2000 wrote: You can therefore say that if X^3 = 16Y then Y = 4 and X = 4 since 16 = 4^2 and you need another 4 to satisfy the given equation.
The logic is also problematic- you really do need to get down to prime factors here. If instead you have the equation:

X^3 = 64Y

by the logic quoted above, you might then say "since 64 = 8^2, you need another 8 to satisfy the equation", and conclude that Y must be divisible by 8. This would not be correct, of course. Y could be 1, and X could be 4, to give one example. You must break down to primes, and only then look at the exponents.
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by Neo2000 » Tue Jun 10, 2008 5:53 am
Umm, the previous poster had underline the word "atleast" i was aiming to explain that. Guess I wasn't clear enough :)
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by aatech » Tue Jun 10, 2008 12:16 pm
Good one netigen.. OA is B
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by egybs » Tue Jun 10, 2008 3:18 pm
yeah - that was sweet!
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Still not clear - Any help ?

by mbaobsessed » Fri Nov 28, 2008 11:35 am
Hello,

I still did not understand this.


450 is factored to 2x3^2x5^2

How or why should Y be atleast 2 x 2 x 3 x 5 X N ? :?:


- a dumbo :(
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by coffee5251 » Fri Jan 16, 2009 5:22 am
netigen wrote:When you see a question like this, think factors

find factors of 450 = 2 x 3 x 3 x 5 x 5

We know that 450Y = n^3

which means Y should atleast have 2 x 2 x 3 x 5 X N where N is of the form A^3 where A is an integer

Y = 2^2 x 3 x 5 x A^3
Why is Y=2^2 x 3 x 5 x A^3 if the prime factorization of 450 only has one 2?
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by Neo2000 » Fri Jan 16, 2009 8:41 am
(450) x (Y) = Cube of a number

If the prime factorization of 450 has only a single 2, the n Y should contain 2^2 so that the condition of Cube is satisfied
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different approach

by yvichman » Fri Jan 16, 2009 9:10 am
i did this in a completely different approach and maybe just got lucky with the correct answer. I always approach math problems with Plugging in if at all possible. I look at 450y =N^3 and start thinking how I can possibly plug in. Well if the question was 450y=n^2 ... i would assume y= 2 and n = 30 ... so I do some quick math and figure out that 30^3 = 27000 and divide that by 450 = 60
so assuming Y= 60 and N= 30 i plug 60 into the answers into the choices given.
and only choice 1 gives me an integer.

Will my approach work everytime or did I just get lucky?
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