Number System - Questions & Discussions

@mayurdhingra thanks mayur but it'd really help if you could give me the complete solution.
@Pulkit1990 that's the thing, I neither have the answer nor the solution..I hope ur answer is right...it'd help if u posted the solution...

@yasaswimr i just applied first sum of a finite GP formula and then applied euler theorem for finding the remainder....hope it will work for you...
@Pulkit1990 answer cannot be 12. It cannot be greater than 9 since you are dividing by 9.
@yasaswimr said:
Throw some light on this plzQ.) What is the remainder when 1+7+7^2+7^3+....+7^2009 is divided by 9?
[1+7+4]%9=3
so 9 terms will provide with the remainder 0
there are 2010 terms
2010 is of the form 9k+3
and we know that 3 terms is giving a remainder of 3 so
3 is the answer
@yasaswimr there are 2010 terms in total which is a multiple of 3.
First three terms give a total of 1+7+4 = 12 as remainder i.e. 3 as remainder.
Similary next 3, also give 3 as remainder.\

Total remainder = 3+3+.... (670 times) = 2010/9 = 3. Hence the answer is 3.
@mayurdhingra & @REVEALED thanks a lot guys...really appreciate it.
@Pulkit1990 u r on the right path...u just missed the last step..thanx a lot

@Reetu107
is 47 the OA???

Please help me with this -

Next number of this series?
1,4,108,294,?



Thank you
Need help here:

Q) When the smallest three-digit number which when divided by 7, 8 and 11 leaves respective remainders of 4, 3 and 7, is divided by 9, the remainder is

a) 5
b) None of these
c) 8
d) 7


pls elaborate the process involved...
thanks in advance !
@Sandeeppappu said:
Need help here:Q) When the smallest three-digit number which when divided by 7, 8 and 11 leaves respective remainders of 4, 3 and 7, is divided by 9, the remainder isa) 5 b) None of these c) 8 d) 7pls elaborate the process involved...thanks in advance !
None of This.....remainder is 3
@rituraj123 hey Dude....
its 4
@Sandeeppappu said:
Need help here:Q) When the smallest three-digit number which when divided by 7, 8 and 11 leaves respective remainders of 4, 3 and 7, is divided by 9, the remainder isa) 5 b) None of these c) 8 d) 7pls elaborate the process involved...thanks in advance !
7a+4=8b+3
56c+11=11d+7

616n+403
403 mod 9=7
determine the number of zeros at the end of (2^123-2^122-2^121)X(3^234-3^233-3^232)
ans options
A)0
b)1
c)121
D)none of these

m confused as i feel its option A
but the ans is given as B
plzzz help
@niranjan031286
@niranjan031286 it is 1
as it can be simplified to
2^121*3^232*5
hence one zero
@messicat thanks
koi plz bata do Number System ki theory k liye kya prefer kiya jaye . I want to strong my Number System . thanks in advance .......
@Neel453 number system arun sharma ki quant book mein kaafi sahi samjhaya hai. I would strongly recommend that.
@mayurdhingra thnx a lot brother ...:)

Hi Guys i have a doubt in Number system, Kindly give your inputs :

(Non MCQ type question- CL package)

The product of three positive integers is 36. if one knows the sum of 3 nos., but is unable to uniquely identify the nos., what is the sum of three nos. ?

to uniquely identify the numbers , they have to be distinct.


so if u look for triplets which give such a product
we find only 1 triplet which is (2,3,6)
whose sum is 11