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prateek.agarwal

3. The remainder when

2² + 22² + 222² + 2222² + ?.. ( 2222?.2)²

45 times

is divided by 9

Answer: 3

nick

2^2

22^2 can be written as (18+4)^2 this term when expanded will leave remainder as 4^2.........RIGHT.....

222^2 CAN BE WRITTEN AS (216 +6)^2 leaving remainder 6^2
as 9 can be taken out common
and the whole lot can be written as
9(m) +2^2+4^2......................................90^2

so on and so forth,the ramainders left till 45 222222222222....

will be

2^2+4^2+6^2...................................90^2

SEE DOWN FOR SOLUTION

nick

2^2+4^2......................................90^2

CAN be written as 4(1 +2^2 +3^2..................................45^2)..........(1)

RIGHT??!!!(see above)

now using summation formula for expression inside the bracket

which is n(n+1)(2n+1)/6

here n = 45

therefore 1 becomes 4*45*46*91/6

i.e.125580

this can be written as 9(13953) +3

hence the REMAINDER IS 3.

PLSS rate my efforts peopleeeeeee

kislay

HEY NICK HOW THE REMAINDERS LEFT WILL BE 18^2,,36^2. ........90^2
PLS EXPLAIN

kislay

2^2

22^2 can be written as (18+4)^2 this term when expanded will leave remainder as 4^2.

222^2 CAN BE WRITTEN AS (216 +6)^2 leaving remainder 6^2
as 9 can be taken out common
and the whole lot can be written as
9(m) +2^2+4^2......................................90^2.......................(1)

BUT THE EQN 1 HAS ALSO THE TERMS
18^2,36^2,54^2,72^2,90^2, INCLUDED IN THE REMAINDER LIST.

HENCE THE REMAINDER IS
2^2+4^2......................................90^2 - (18^2+36^2+54^2+72^2+90^2)

=>4(1 +2^2 +3^2..................................45^2) -18^2(1 +2^2 +3^2.......5^2)

=>125580(USING FORMULA) -18^2*55

=>143400

=9*15933 + 3

hence the REMAINDER IS 3

THIS IS THE CORRECT SOLN NICK.
HEY USERS PLS RATE ME .

nick

NO I AM NOT

SEE
find the remainder when
1+2+3+4 is divided by 3

according to me i add up all ,which gives 10 leaving ramainder 1

which is correct

but according to u it should be
1+2+4-3 which gives 4 and remainder 1

SO??? we both ARE CRT ONLY I HAPPEN TO USE a direct formula
and go easy

shine

hey nick has done a wonderful job
bt heres another way
see
2^2 +22^2+ 222^2.........................
=2^2(1^2 + 11^2 +111^2...................)
=2^2( 1+ (9+2)^2 +(99+ 9+3 )^2.+ (999+ 99 +9 +4)^2 +..................)

now on dividing whole by 9 or say 3^2 we get
=4((1/3)^2 + ((9+2)/3)^2....................)
we get the series of remainders as
4(1 + 2^2 + 3^2 + 4^2................+ 45^2)
=4.45.46.91/6
=125580 when divided by 9 gives 3 as remainder

shine

nd kislay ur solution nd nicks solution both r correct....................when u r saying th eq
18^2,36^2,54^2,72^2,90^2, INCLUDED IN THE REMAINDER LIST.
u may see tht they already r multiples of 9 as they r foremed by ano multiple of 9
so no need to apply a lenthy approach
ne ways take the pts

nick

good work shine!!

three solutions!!!! to a problem

BRILLIANT.....work by everybody.......!!!!

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