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| Re: ques -
26-04-2003, 12:16 AM
Quote: |
Originally Posted by morning_star Hi
Here's a question..I guess its quite simple...but then my maths is pathetic so never mind 
Find the remainder if 30^40 is divided by 17.
Luv n Luk | 30^40 /17
= 900^20/17
=(901-1)^20/17
=(53*17 - 1)^20/17
hence applying binomial theorem........ we get the remainder as 1 
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26-04-2003, 02:45 PM
hey just wanna ask 1 question regarding this...
did these numbers strike you or u know tables pretty welll that u cud see thru the solution.
let me know...
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26-04-2003, 03:09 PM
well its just a method to solve such questions nothing to do with remembering tables
take the powers of 30 untill you get something like 17*m +/- 1
if it wud not have been 900 .....i wud have tried representing 2700 like 17*m +/- 1
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26-04-2003, 06:35 PM
why is it -1?
I heard M$ is coming up with a vacuum cleaner. I guess now we will finally have an M$ product that wont suck. | | | | | The Following 2 Users Say NO Thank You to Silken_Touch For This Un-useful Post: | | | | | |
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26-04-2003, 07:11 PM
dont u think it shud be +1..??
(901 + (-1))^20/17
as (-1)^20 =1.so the remainder left is one....!!
what do ya say guys...??
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26-04-2003, 07:55 PM
yeah u r correct ........silken pointed this out to me
i was assuming it for odd powers
just forgot to edit it again
thanks
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27-04-2003, 08:19 PM
Quote:
2. If (AB) x (AB) = XYA, then X stands for :
(1) 3 (2) 1 (3) 6 (4) 9
| Hey guys,
How to go abt this?
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27-04-2003, 09:21 PM
Quote:
. If (AB) x (AB) = XYA, then X stands for :
(1) 3 (2) 1 (3) 6 (4) 9
Hey guys,
How to go abt this?
| lemme try this..
a 2-digit no. havin a 3-digit no as its square..so the 2digit no is<32.. in other words A is 1,2 or 3.
but as B^2 has its last digit as A ..therefore A cant be 2 or 3.Hence A=1..and this leads us to B=9..so 19^2=361...
And X=3..
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| Que on Numbers.... -
29-04-2003, 05:52 AM
An interesting Question on Numbers funda..
Let D be a decimal of the form, D = 0.0a1a2a1a2a1a2……….., where digits a1 & a2 lie between 0 and 9. Then which of the following numbers necessarily produces an integer, when multiplied by D?
(1) 18 (2) 108
(3) 1980 (4) 208
Will come up with answers if there is no answer from Junta....
BFN...
shyam
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29-04-2003, 05:55 AM
one more sitter on Numbers
Let S be the set of prime numbers greater than or equal to two and less than 100.
All such existing terms are then multiplied, then how many consecutive zeros would the final digit end with.
(1) 1 (2) 4
(3) 5 (4) 10
Lemme know if posting questions seperately will help the cause. Otherwise i wud compile them together and post them
cheers
Shyam
"Reason is man's only means of perceiving reality, his only source of knowledge, his only guide to action, and his basic means of survival. " - Ayn Rand
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