Bunuel wrote:
Math Revolution and GMAT Club Contest Starts!
QUESTION #10:If a and b are positive integers, let \(n = a^3*b^4\), how many different factors n has?
(1) a and b are prime numbers
(2) n has only prime factors 5 and 7
Check conditions below:
Math Revolution and GMAT Club ContestThe Contest Starts November 28th in Quant Forum
We are happy to announce a Math Revolution and GMAT Club Contest
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PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize:
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MATH REVOLUTION OFFICIAL SOLUTION:Since we have 3 variables (n, a, b) and 1 equation (\(n=a^3b^4\)) in the original condition, we need 2 equations to match the number of variables and the number of equations. Since we need both 1) and 2), the correct answer is likely C.
Using con 1) & 2), we get \(n=5^37^4\) or \(n=7^35^4\). The number of different factors is (3+1)(4+1)=20. This is unique and sufficient. Therefore,
the correct answer is C.
However, since this is an “integer” question, which is one of the key questions, we should apply Common Mistake Type 4(A).
In case of con 1), if a=b, \(n=a^7\) → (7+1)=8. However, if a≠b, \(n=a^3b^4\) → (3+1)(4+1)=20. So this is not unique and sufficient.
In case of con 2), \(n=5^37^4\) → (3+1)(4+1)=20. However, \(n=1^335^4=5^47^4\) → (4+1)(4+1)=25. This is not unique and sufficient. Therefore the correct answer is C.
Note : For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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