#2: Check my answer

Find the smallest integer value of n for which 120n is a perfect square.

My answer is n = 7.5

Is it correct?

If not, could you please show me your workings? I want to learn, thanks. :)

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Originally Posted by Yunie

Find the smallest integer value of n for which 120n is a perfect square.

My answer is n = 7.5

Is it correct?

If not, could you please show me your workings? I want to learn, thanks. :)

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I'm not sure about the strategy for this one, probably someone else will be helping you. In the meantime, notice you are looking for an integer and 7.5 isn't one.

You have:

120 * n = b*b ' with b*b showing the result is a perfect square!

The left side could be rearranged to say

(120/n *n) *n = b * b
Looks stupid at the first look, but think about it, the multiplication shall be a perfect square (n*n) is one step toward that, no we have the 120/n part to take care off. This one has to be a perfect sqare as well, start the search with the smallest result(1, where n is 120), this will give you a solution, but not the smallest. The next possible result would be 4 (since 2 and 3 are not perfect sqares). To get 4 you need a n of 30 and that is your answer.

There might be better ways to get this one, but..........

Quote:

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Originally Posted by opus

You have:

120 * n = b*b ' with b*b showing the result is a perfect square!

The left side could be rearranged to say

(120/n *n) *n = b * b
Looks stupid at the first look, but think about it, the multiplication shall be a perfect square (n*n) is one step toward that, no we have the 120/n part to take care off. This one has to be a perfect sqare as well, start the search with the smallest result(1, where n is 120), this will give you a solution, but not the smallest. The next possible result would be 4 (since 2 and 3 are not perfect sqares). To get 4 you need a n of 30 and that is your answer.

There might be better ways to get this one, but..........

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That's really clever, I like it! :thumb:

I think the intended method is probably to look at the prime factors of 120:

120n = x²

-> (12 x 10)n = x²
-> (2 x 2 x 3 x 2 x 5)n = x²

-> to make the LHS a square you need at least another 2, 3 and 5.

-> 2x3x5 = 30



Also, I notice Yunie has a new strategy; provide a wild hazarded guess in answer to the question, and then ask for workings if it is wrong. I suggest that this is countered by asking why before providing an answer, which will be the most helpful to Yunie if she really does want to learn.

Quote:

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Originally Posted by zaza

I think the intended method is probably to look at the prime factors of 120:

120n = x²

-> (12 x 10)n = x²
-> (2 x 2 x 3 x 2 x 5)n = x²

-> to make the LHS a square you need at least another 2, 3 and 5.

-> 2x3x5 = 30



Also, I notice Yunie has a new strategy; provide a wild hazarded guess in answer to the question, and then ask for workings if it is wrong. I suggest that this is countered by asking why before providing an answer, which will be the most helpful to Yunie if she really does want to learn.

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I really like your solution, and I think you're right, let's try to do it the way you suggest.

Quote:

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Originally Posted by zaza

I suggest that this is countered by asking why before providing an answer, which will be the most helpful to Yunie if she really does want to learn.

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Sounds like a good plan.thumb:

Quote:

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Originally Posted by opus

Sounds like a good plan.thumb:

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:thumb: !!!
(missing one dot in the colon)... Just wanted to correct your thumb! :D