If n = a^4 × b^3 × c^7, where the three distinct prime numbers a, b, c > 2; what is the number of perfect squares which are factors of n?
A. 48
B. 24
C. 18
D. 12
E. 9
OA B
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a, b, c > 2
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sanju09
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Last edited by sanju09 on Thu Apr 23, 2009 2:30 am, edited 1 time in total.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
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Lucknow-226001
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Let the perfect square be denoted: S=a^x+b^y+c^z. x y and z must all be even for S to be a perfect square.
Possible values of x are 0, 2, 4 (Total 3)
Possible values of y are 0, 2 (Total 2)
Possible values of z are 0, 2, 4, 6 (total 4)
So total possible factors = 3*2*4 = 24
I actually disagree with this answer. What is there to say that a, b and c are different? If a, b and c all equaled 3. Then n = 3^14 and would be divisible by 8 perfect squares (powers of 3: 0, 2, 4, 6, 8, 10, 12, 14).
Possible values of x are 0, 2, 4 (Total 3)
Possible values of y are 0, 2 (Total 2)
Possible values of z are 0, 2, 4, 6 (total 4)
So total possible factors = 3*2*4 = 24
I actually disagree with this answer. What is there to say that a, b and c are different? If a, b and c all equaled 3. Then n = 3^14 and would be divisible by 8 perfect squares (powers of 3: 0, 2, 4, 6, 8, 10, 12, 14).
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sanju09
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Good!!
Only one problem: Why take the perfect square as a^x + b^y + c^z?
Your objection is honoured and the question is given the necessary repair.
Only one problem: Why take the perfect square as a^x + b^y + c^z?
Your objection is honoured and the question is given the necessary repair.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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Uri
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i have arrived at a solution, perhaps without thinking too much mathematically and surprisingly the answer is correct.
this problem is similar to "how many factors does a number have?" but the twist is that we have to consider taking only upto the greatest possible even degree for any prime factor and then consider that a pair of any prime factor makes 1 unit.
from a^4 we get 2 units, from b^3 we get 1 unit and from c^7 we get 3 units.
So, the solution becomes (2+1)*(1+1)*(3+1)=24
this problem is similar to "how many factors does a number have?" but the twist is that we have to consider taking only upto the greatest possible even degree for any prime factor and then consider that a pair of any prime factor makes 1 unit.
from a^4 we get 2 units, from b^3 we get 1 unit and from c^7 we get 3 units.
So, the solution becomes (2+1)*(1+1)*(3+1)=24
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Vemuri
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The question is asking what are the number of perfect squares which are factors of n. The perfect squares are a^2, a^2, b^2, c^2, c^2, c^2. But, the list also should have combinations of these factors, such as a^2*b^2, a^2*c^2, ....etc
I am not sure I understand the question properly. Is it asking for number of distinct perfect square factors?
I am not sure I understand the question properly. Is it asking for number of distinct perfect square factors?
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doclkk
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Does this mean that if the question readVemuri wrote:The question is asking what are the number of perfect squares which are factors of n. The perfect squares are a^2, a^2, b^2, c^2, c^2, c^2. But, the list also should have combinations of these factors, such as a^2*b^2, a^2*c^2, ....etc
I am not sure I understand the question properly. Is it asking for number of distinct perfect square factors?
If n = a^8 × b^6 × c^14, where the three distinct prime numbers a, b, c > 2; what is the number of perfect squares which are factors of n?
5X4X8 = 160
A - 4 units
B - 3 Units
C - 7 Units
(A+1 )* (B+1) * (C+1) = 160?
If yes, thanks so much - you're great. If no, please explain your units theory - its simple and sizzling and sexy for time constraints.
