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11 Aug 2007, 08:19
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I need help on how to solve this problem. The question is attached as a jpg image.
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11 Aug 2007, 08:55
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The product = 1*2*3*4*5*6*7*8
Since a and n are greater than 1, the only possibilities are 2^6 and 3^2.
Stmt 1: a^n = 64 = 2^6 = 4^3 = 8^2. insufficient.
Stmt 2: n=6. Only one possibility. 2^6. sufficient.
Hence, answer should be B.
Since a and n are greater than 1, the only possibilities are 2^6 and 3^2.
Stmt 1: a^n = 64 = 2^6 = 4^3 = 8^2. insufficient.
Stmt 2: n=6. Only one possibility. 2^6. sufficient.
Hence, answer should be B.
11 Aug 2007, 10:23
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sumande replay is perfect !!
8! = 1*2*3*4*5*6*7*8
break this number into primes:
2*3*(2*2)*5*(2*3)*7*(2*2*2)
statement 1
since a^n = 64
64 = 2*2*2*2*2*2 or 4*4*4
2*3*(2*2)*5*(2*3)*7*(2*2*2)/(2*2*2*2*2*2) = 2*3*3*5*7
2*3*(2*2)*5*(2*3)*7*(2*2*2)/(4*4*4) = 2*3*3*5*7
insufficient
statement 2
8! = 2*3*(2*2)*5*(2*3)*7*(2*2*2) = 2^7*3^2*5*7
the only way for n to be 6 is when a = 2
sufficient
8! = 1*2*3*4*5*6*7*8
break this number into primes:
2*3*(2*2)*5*(2*3)*7*(2*2*2)
statement 1
since a^n = 64
64 = 2*2*2*2*2*2 or 4*4*4
2*3*(2*2)*5*(2*3)*7*(2*2*2)/(2*2*2*2*2*2) = 2*3*3*5*7
2*3*(2*2)*5*(2*3)*7*(2*2*2)/(4*4*4) = 2*3*3*5*7
insufficient
statement 2
8! = 2*3*(2*2)*5*(2*3)*7*(2*2*2) = 2^7*3^2*5*7
the only way for n to be 6 is when a = 2
sufficient
