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If positive integer m is a multiple of 8 and of 9, is the positive square root of m between two consecutive integers?
(1) m is not a multiple of 48. (2) m is not a multiple of 27.
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A is the answer.
m = k*(2^3)*(3^2)
1) 48 = (2^4)*3
since m is not a multiple of 48, then k is an odd number -> sqrt(m) is not an integer -> it's between two integers -> sufficient 2) 27 = 3^3 -> k is not a multiple of 3 -> insufficient
k = 2 -> sqrt(m) = 4*3 = 12 -> not between integers k = 5 -> sqrt(m) = 6*sqrt(5*2) = 6*sqrt(10) -> between integers
m is a multiple of 8 and 9. this means m = 2*2*2*3*3 or it's multiples Stmt 1: m is not a multiple of 48. It means that there will never be four 2s among the factors of m. But for m to be a perfect square, there should be even number of 2s and 3s. This means m can never be a perfect square. So the sqaure root of m will always lie between two consecutive integers. Hence suff.
Stmt 2: m is not a multiple of 27. There i no restriction on 2s. We can have even/odd number of 2s. This leads m to be a perfect square in some cases and not a square in other cases. Hence Insuff
answer is 'A'
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