1ok ani...........for this ques its x100..........i was talkin of a generalk method for all ques of this type.........
wel ive cum across many questions of this type..........BUT......... wat is d exact procedure doin such questions???????????
-What is the number of positive integers which divide (210)2006 but not (210)2005........
one more........................
-Wats d last digit of 17+27+......(999)7....... please tel d right method not d answer......
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341For the second one, group the terms as
(17+9997)+(27+9987)+....+(4997+5017)+5007
Since xn+yn is divisible by x+y when n is odd, each of the bracketed terms will be divisible by 1000.
5007 is of course divisible by 1000
So, not only the last digit, but the last three digits are zero's
9k im doing ist q
210 = 7X5X3X2
acc to q we need to find nos were atleast one of powers in factor is 2006
hence no of ways = (2007)4-(2006)4
62i will give an example
a≡b (mod c)
means that b-a is divisible by c
that is remainder when a is divided by c is same as when b is divided by c :)
the following statements are true
4≡1 (mod 3)
14≡2 (mod 3)
1≡1 (mod 3)
The following are false
4≡2 (mod 3)
14≡1 (mod 3)
The properties of congruence are these
a≡b (mod c)
implies all of these
ka≡bk (mod c)
an≡bn (mod c)
These are all simple things and have simple derivations...
all we do is to formally (in symbols write the long statements that we would otherwise have written!!!)
1i solved 2nd by repititions but how can one solve for a power like 1232
62yeah u do akand..
and sometimes these are the formal methods... :)
They check ur presence of mind ....
had the problem been 51232 on division by 32 gives?
One method is congruence (probably the easiest)
Another is by taking binomial theorem
for this kind of problem though there is one way generally.. (to find repetions!)
1OPTIONS- a)10,00,000 b)more than 10,00,000
c)less than 10,00,000 d)none of these
OPTIONS- a)0 b)1 c)2 d)none of these
11No need for last digit of nonsense. You have an x100 so answer will end with 0 anyway!
1ok.............so basically we need 2 check d last digit of each term then check for any repetiton in other terms...........then find d rest.........dats d method????? K got it.............but dont v hav a more formal method???
621st one is a brillaint problem .. i remember a similar problem sometime back that we discussed... I would like to see someone from the users solve that one :)
Aragaon ur second method is perfect :)
It is the correct answer :)
By the way for those who want to read more into this subject of remainders... do read something called congruence (I wud nto suggest this at this time for the class xii or dropping students.. but definitly for someone who is in class xi!) It will take max 1 day to understand it! and it will make mockery of many of the remainder problems ;)
1wel may be........u and ani are gettin d same...............d method is imp....gimme that.......
11For the second one, the last digits sequence from 17+27...+97 repeats itself for, I think, a 100 times. So the sum will be 100x(nonsense), which always ends with 0. This, I hope, is the correct answer :)
P.S: I haven't bothered to see olympiad problems before; so this one's the first time for me. So clarify with someone else for the method!