Latest TCS Aptitude Question SOLUTION: A number in integer base n when written in base n1 becomes 1030n1 .The same number when written in base n+1 becomes 444n+1.The original number written in base 10 is

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(#M40154711) TCS Numerical Ability question Number system Keep an EYE

A number in integer base n when written in base n1
becomes 1030n1
.The same number when
written in base n+1 becomes 444n+1.The original number written in base 10 is

Asked In TCS ABHISHEK KUMAR (8 years ago)

Unsolved Read Solution (2)

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sundar  (8 years ago)

1 * (n - 1)^3 + 3 * (n - 1) = 4 * (n + 1)^2 + 4 * (n + 1) + 4
(n - 1) * ((n - 1)^2 + 3) = 4 * ((n + 1)^2 + (n + 1) + 1)
(n - 1) * (n^2 - 2n + 1 + 3) = 4 * (n^2 + 2n + 1 + n + 2)
(n - 1) * (n^2 - 2n + 4) = 4 * (n^2 + 3n + 3)
n^3 - 2n^2 + 4n - n^2 + 2n - 4 = 4n^2 + 12n + 12
n^3 - 2n^2 - n^2 - 4n^2 + 4n + 2n - 12n - 4 - 12 = 0
n^3 - 7n^2 - 6n - 16 =0
n = 8

1 * (8 - 1)^3 + 3 * (8 - 1) =>
1 * 7^3 + 3 * 7 =>
343 + 21 =>
364

Is this solution Helpfull? Yes (8) | No (1)  

Ajit Kumar Pandey  (8 years ago)

let n=8;
when base is n+1
364%9 =4
40%9=4
4%9=4
444 base n+1

when base is n-1
364%7=0
52%7=3
7%7=0
1%7=1
1030 base n-1

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