Latest Elitmus Aptitude Question SOLUTION: Q. What is the remainder when expression 2^2+22^2+222^2+2222^2+....+22222...48 times^2 divided by 9.

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Q. What is the remainder when expression
2^2+22^2+222^2+2222^2+....+22222...48 times^2 divided by 9.

Asked In Elitmus (10 years ago)

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RAKESH  (10 years ago)

[2^2+22^2+222^2+2222^2+....+22222...48 times^2]/9

=[2^2(1^2+11^2+111^2+1111^2+....+11111...48 times^2 )]/9

=4[1+(9*1+2)^2+(9*12+3)^2+(9*123+4)^2+(9*1234+5)^2+....]/9

=Rem[4(1+2^2+3^2+4^2+...+48^2)]/9

=Rem[(4*48*49*97/6)/9]

=Rem[(32*49*97)/9]

=Rem[(5*4*7)/9]

=Rem(5)

remainder = 5

Is this solution Helpfull? Yes (89) | No (12)  

AVIJIT  (10 years ago)

divide each term by 9
then the remainder will be 4,7,0, 4,7,0 ....so on
and these will be reapetd after 3 term
total of 1st 3 term is 11.
there have 48 term. so it will be reapeted 16 times.
so total of all term is 16*11=176
176/9 then qutient is 18 and remainder is 5.

Is this solution Helpfull? Yes (62) | No (10)  

Mani kandan  (10 years ago)

hai rakesh how become this Rem[(4*48*49*97/6)/9]
explain me

Is this solution Helpfull? Yes (4) | No (1)  

abhishek maini  (9 years ago)

@manikandhan it is formula of 1^2+2^2+3^3.....n^2=n*(n+1)*(2n+1)/6

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khushboo sharma  (9 years ago)

sum of n^2 term is n(n+1)(2n+1)/6

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Sanjeev Raj  (8 years ago)

sorry! A little correction w.r.t previous answer. instead of "42 times" it's "21 times".

rem[ (2^2+22^2+222^2+2222^2+....+22222...48 times^2) /9]
rem[ (2^2+4^2+6^2+8^2+1^2 + 3^2+5^2+3^2+5^2+....(3^2+5^2)...21 times+ 3^2) /9]
rem[ (4+16+36+64+1+ 21*(9+25) + 9 ) /9]
rem[ (121+ 504 + 9) /9]
rem[ (634) /9] = 4

So Answer should be 4.

Is this solution Helpfull? Yes (2) | No (1)  

SOURABH KUMAR  (8 years ago)

ans is :- 6

Is this solution Helpfull? Yes (1) | No   

aarizoo dosh  (10 years ago)

REM[4(1+2^2+3^2.........+48^2)-(9^2+18^2+27^2+36^2+45^2)]/9....
4[38024-3455]/9
0

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anujnandwana  (10 years ago)

ans is :- 5

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thakur  (9 years ago)

ITS NOT 24 ITS INFINITY

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Sanjeev Raj  (8 years ago)

rem[ (2^2+22^2+222^2+2222^2+....+22222...48 times^2) /9]
rem[ (2^2+4^2+6^2+8^2+1^2 + 3^2+5^2+3^2+5^2+....+3^2+5^2...42 times+ 3^2) /9]
rem[ (4+16+36+64+1+ 42*(9+25) + 9 ) /9]
rem[ (121+ 1428 + 9) /9]
rem[ (1558) /9] = 1

Ans: 1

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