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20 Nov 2024, 07:21
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Difficulty:
55%
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Question Stats:
63% (01:40) correct
37%
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wrong
based on 287
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If N is a positive integer and \(N^2\) is a multiple of 150, which of the following is the minimum number of positive factors that N has?
(A) 3
(B) 8
(C) 10
(D) 12
(E) 15
(A) 3
(B) 8
(C) 10
(D) 12
(E) 15
miag
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Expert reply
27 Nov 2024, 00:56
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i) We are given that N^2 = 150 (c) where c>=1
ii) Prime factorize 150, 150 = 5^2 * 3 * 2
iii) Subbing back in, N^2 = 5^2 * 3 * 2 * c. We know that, for LHS to be a perfect square, powers on the RHS need to be even. So c needs to contain atleast one 2 & one 3. This means N^2 should at the minimum be 5^2 * 3^2 * 2^2
iv) This gives us that N is atleast 5*3*2. Number of factors of n = (1+1)* (1+1) * (1+1) = atleast 8
>> Hence option b)
ii) Prime factorize 150, 150 = 5^2 * 3 * 2
iii) Subbing back in, N^2 = 5^2 * 3 * 2 * c. We know that, for LHS to be a perfect square, powers on the RHS need to be even. So c needs to contain atleast one 2 & one 3. This means N^2 should at the minimum be 5^2 * 3^2 * 2^2
iv) This gives us that N is atleast 5*3*2. Number of factors of n = (1+1)* (1+1) * (1+1) = atleast 8
>> Hence option b)
General Discussion
21 Nov 2024, 05:25
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siddhantvarma
In this question, it is given that: N^2=150I.
⇒ N\(^2\)/(2×3×5^2)=I.
As, N is positive integer,
⇒ 2×3×\(5^2\) in the denominator needs to be cancelled with the numerator, hence the minimum possible value of N is 30.
i.e., 30=\(2^1\)×\(3^1\)×\(5^1\).
As we know that, any prime number raise to the power n , collectively will have n+1 positive factors, hence applying the same rule we get;
\(2^1\) will have 2 positive factors, similarly, \(3^1\) will have 2 positive factors, and \(5^1\) will also have 2 positive factors.
Therefore, Number of positive factors of N =2×2×2=8 .
