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Master | Next Rank: 500 Posts
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by dferm » Wed Nov 28, 2007 8:37 am
Hey can anybody help !!!!

IF n and y are positive integers and 450y=n^3, which of the following must be an integer?

I. y/3x2^2x5

II. y/3^2x2x5

III. y/3x2x5^2

Reminder x is multiplication...

A. NONE
B. I only
C. II only
D. III only
E. I, II, and III
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by sankruth » Wed Nov 28, 2007 8:49 am
Solve using prime factors

450y = 3 x 3 x 5 x 5 x 2 x (y) = n^3

Therefore y must contain atleast (5 x 3 x 2 x 2)

I .y/3x2^2x5 - YES (Denominator has 5 x 3 x 2 x 2)

II. y/3^2x2x5 - NO (Denominator only has one 2, we need 2 x 2)

III. y/3x2x5^2 - NO (Same as II)

Option B
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by dferm » Wed Nov 28, 2007 9:13 am
How are you getting 2 x 2 when I used the prime factorization I am only getting 3 x 3 x 5 x5 x 2
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by dferm » Wed Nov 28, 2007 9:13 am
How are you getting 2 x 2 when I used the prime factorization I am only getting 3 x 3 x 5 x5 x 2
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by sankruth » Wed Nov 28, 2007 9:25 am
General rule: x^3 has three x's

Similarly 450 has 3x3, 5x5 and 2. So, if 450y = n^3 it needs one 3, one 5 and two 2's to (i.e. 3x5x2x2) which should come from Y

Another perspective...
If y = 3 x 5 x 2 x 2 then,

450y= 3 x 3 x 5 x 5 x 2 x 3 x 5 x 2 x 2 (each number is present 3 times) and will be a perfect cube.

Hope that explains!
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