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Math PS

Expert replies
by ukr.net » Thu Nov 29, 2007 6:34 am
If K is the square of an integer N, which of the choice below could be the number of factors of K?

I. Odd Number
II. Even Number
III. Prime Number
I only
II only
III only
I and II
I and III
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Source: — Problem Solving |

by ri2007 » Thu Nov 29, 2007 7:42 am
I would go with 1 & 3 (last choice)

If n = 1, K = 1, so no of factors is 1, which is odd
if n = 2, K = 4, so no of factors = 3, which is both odd and a prime number.

However if K is a square of n, it cannot have even numbers. Here is my thinking. say k = x^2y^2

no of factors = (2+1) (2+1)

both are odd, and o * o = o

Pls confirm
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by sankruth » Thu Nov 29, 2007 9:22 am
K = N^2

So, 1, N and K must be factors of K. i.e. Odd(I) and Prime (III)

If N = 1 then there is only one factor (1), so odd (I)

IMO A (Is it the correct way to solve this??)
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DS

by ukr.net » Sat Dec 01, 2007 3:43 am
A is the correct one
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by swle24 » Sat Dec 01, 2007 4:22 am
the question asks which of the following COULD be the number of factors of K so the answer should be E as ri2007 explain and i agree with this explanation. If it asked which of the following MUST be the number of factors of K then the answer would be A because if N=1 then the number of factors K is 1 which is odd but if N=2 then K=4 and then k has 3 factors which is odd and prime. so in either case the number of factors must be odd not odd+prime.
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by sankruth » Sun Dec 02, 2007 10:16 am
swle24 wrote:the question asks which of the following COULD be the number of factors of K so the answer should be E as ri2007 explain and i agree with this explanation. If it asked which of the following MUST be the number of factors of K then the answer would be A because if N=1 then the number of factors K is 1 which is odd but if N=2 then K=4 and then k has 3 factors which is odd and prime. so in either case the number of factors must be odd not odd+prime.
I beg to differ on this point. Consider 36, one could argue its factors are just 9 and 4, i.e. 2 factors (EVEN). The question does not specify prime factors.

I guess there is a flaw in the way the question is framed.
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