Latest TCS Aptitude Question SOLUTION: 12341234..........400 digits when divided by 909 leaves reminder?

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(#M40025978) TCS question Keep an EYE

12341234..........400 digits when divided by 909 leaves reminder?

Asked In TCS (10 years ago)

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manoj chauhan  (10 years ago)

1234124341234.......upto 400 digits = 1234(10^0 + 10^4 + 10^8 + ... 10^396)
1234(10^0 + 10^4 + 10^8 + ... 10^396) mod 909
= 123400 mod 909 = 685 (As 10^4k mod 909 is always 1)

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prakash gupta  (10 years ago)

Ans 325.
last 4 digit is 1234 which when divide by 909 leave remainder 325......correct me if i wrong

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Abdil khader  (10 years ago)

N=12341234…upto 400digits =1234[10^396+10^392+10^298+...+10^0] =1234[(10^4)^99+(10^4)^98+(10^4)^97+....+(10^4)^0] =A*B(let) So we can write N%909=(A*B)%909

=1234%909*[1+1+...100 terms]%909[As10^4=1(mod909)]

=32500%909=685[Here % implies Mod which determines the remainder]

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Chandra Bhatt  (10 years ago)

(1234)group of 4 digits whose sum of digits is 10, when repeated 100 times makes overall 400 digits like 12341234.....1234.
check rem. for 9 which is = 1
rem. for 101 =91
ans.92

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Guru Pranav  (10 years ago)

685 will the remainder !!!

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