Q1. A = 333....333(51 digits), B = 666...666(51 digits) Find 52nd digit of (AXB) from the right. (a)7, (b)9, (c)1, (d)2.
Q2. 10000! = (100!)^k - P . P is prime number & P,k both are integers. (a)105, (b)106, (c)104, (d)102, (e)None of these.
Those who can solve the qs plz give explanation of these two stuffs.
Regards, Dipanjan. __________________
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Sorry 4 the mistake in QS NO. 2. Tht qs is:
Q2. 10000! = (100!)^k - P . P is prime number & P,k both are integers.Find the max value of k.
(a)105,
(b)106,
(c)104,
(d)102,
(e)None of these.
Thank You in Advance,
Dipanjan
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trail n error method
considrr...33*66
3rd digit 4m right =7
sim..333*666
4th digi 4m right=7
By mere observation we see that
3x6=18 ... 1 digit number and second digit from right in ans is 1
33x66=2178 ... 2 digit number and third digit from right in ans is 1
333x666=221778 ... 3 digit number and fourth digit from right in ans is 1
....
Thus since in q, its a 51 digit no, 52nd digit from right is 1
Also we can note that, from 52nd digit onwards till the leftmost digit all digits are 2, 51st digit is 1, 2nd to 50th digits are all 7 and units place is 8..
oops..i took it 4m left
when we observe the pattern, for 33X66=2178 ( 3rd digit from right is 1)
333X666= 221778 ( 4th from right is 1)
3333X6666 = 22217778 ( 5th digit from right is 1)........
So the answer will be c) 1
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