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Originally Posted by samarth007 @ Sat Oct 02, 2004 4:40 pm Q.1 find the lowest of three nos. as described: if the cube of the first no exceeds the product by 2, the cube of 2nd no. is smaller than therir product by 3, and the cube of the third no. exceeds their product by 3.
A) 3^1/3 * *b)9^1/3
c)2 * * * * * *d)any of these |
Go by the options....
we can see product of 3^1/3, 9^1/3 and 2 is 6 and each one of them satisfies 1 given condition. So we just have to find the least of these 3 no.s which is 3^1/3 or option A.
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Originally Posted by samarth007 @ Sat Oct 02, 2004 4:40 pm
Q.2 *find the Gcd of (2^100 -1, 2^120 -1)
a) 2^20 -1 * * *b)2^40 - 1
C)2^60 -1 * * d)2^10 *-1 |
GCD of 100 and 120 is 20 thus option A should be the answer 2^20 -1
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Originally Posted by samarth007 @ Sat Oct 02, 2004 4:40 pm
Q.3 * find the gcd of (111111....1111 hundred ones ; 1111....111 sixty ones)
A) 1111....forty ones * * * * * * * *b)11111.....twenty five ones
C) 11111.... twenty ones * * * * * d) none of these |
Same way option C should be the answer