[vishrock]

Originally Posted by swargupta

find the remainder when (2)^2 + (22)^2 + (222)^2 + .... + (222...49 times) is divided by 9.

My take :- 6

[simrat_sandy]

Originally Posted by swargupta

find the remainder when (2)^2 + (22)^2 + (222)^2 + .... + (222...49 times) is divided by 9.

is the remainder 1?
given 2^2 +(22)^2.. +(22...49times)^2

(2)^2 +(2*11)^2 ...(2* 111...49times)^2

4/9 + 4*(11)^2/9 +4*(111)^2/9+ ....

4*6{ 1+ (2^2) +(3^2)+ ...(8^2)} +1

4*6{n(n+1)(2n+1)}/9*6

=24*8*9*17/9 +1/9
=27*17/9 +1/9
=0*17/9 +1/9
=1
it seems correct to me now, i resquest ppl to check this

[Ankurb]

Originally Posted by simrat_sandy

given 2^2 +(22)^2.. +(22...49times)^2

(2)^2 +(2*11)^2 ...(2* 111...49times)^2

4/9 + 4*(11)^2/9 +4*(111)^2/9+ ....

4{ 1+ (2^2) +(3^2)+ ....(49^2)}

4{n(n+1)(2n+1)}/9*6

(i can't cancel numerator and denominator right?)

so i get a answer like this 36/54
remainder= 36

dude, you solved it correctly
4{n(n+1)(2n+1)}/9*6, put n = 49 and solve, you'll get remainder as 6
I don't see any problem there.

[ABhishek009]

Originally Posted by Ankurb

my take:-
6 will be the remainder

WIl u kindloy explain ur method??

I took 2^2 and lost midway.......

[uphorik_85]

Originally Posted by prateek7563

the remainder when 10^10+10^100+......+10^10000000000 is divided by 7 :
1. 0
2. 1
3. 2
4. 5
5. 4

also explain the method.

Hi prateek,

if u had known fermat's little theorem..dis problem wud hav been a piece of cake...
Eulers theorem states that:
if 2 nos m& n r coprime to each other i,e h.c.f(m,n)=1..and n=a^x.b^y..then remainder of m^phi(n)%n=1,where phi(n)=n*(1-1/a)*(1-1/b)..phi(n) is known as euler's totient function..

now fermat's little theorem states that..
if n in the above euler's theorem is prime..then
phi(n)=n*(1-1/n)
since n has only one factr..
so remaider of m^n-1 /n=1

so in the above problem.
z=10^10+10^100+10^1000......
here let m=10 and n=7..now 7 is prime..
so according to fermats little theorem..
10^6 %7=1
now
remainder when 10^10%7=remainder of(10^6*10^4)%7
10^6 %7=1
so,remainder of (10^4)%7 =rem of (3^4)%7 =4
10^100=10^96*10^4
so rem of 10^100%7=rem of ((10^6)^16*10^4)%7=4.
so all give a remainder of 4.
hence total remainder =4*10=40
remainder of 40 with 7 is 5
so the ans is 5

hope you got my point..

[simrat_sandy]

Originally Posted by Ankurb

dude, you solved it correctly
4{n(n+1)(2n+1)}/9*6, put n = 49 and solve, you'll get remainder as 6
I don't see any problem there.

you wont get 49... i made a mistake sorry i corrected my ans
after one division i will only get 1^2 ,2^2.... uptill 8 6 times and 1 for the 49th term

[simrat_sandy]

i find it difficult to understand questions like these , pls help
->the sum of two numbers is equal to 15 and their arithmetic mean is 25 percent greater than its geometric mean. find the number.

(no options available, so i want to learn the method)

[swargupta]

Originally Posted by Ankurb

my take:-
6 will be the remainder

This is the correct answer dude. How did u get it. I tried getting the answer, couldnt get it though.

[kamra.abhinav]

Originally Posted by simrat_sandy

i find it difficult to understand questions like these , pls help
->the sum of two numbers is equal to 15 and their arithmetic mean is 25 percent greater than its geometric mean. find the number.

(no options available, so i want to learn the method)

let the 2 nos be a and b.....
the question says sum ie a+b=15.....(1)
so A.M is a+b/2 that comes out 2 be 15/2
now v r given A.M of no .is 25% > its G.M
so we can simply write A.M=1.25 G.M
so we get G.M =6
and ab=(G.M)^2 which vl give us ab=36.....(2)
from (1) & (2) we get the no.s as 3 and 12

[ragsagar.yad]

yes dude , the answer is 4