Denary is not ordinary

Algebra Level pending

f ( x ) = ( a − 3 ) x 5 + ( b − 3 a ) x 4 + ( c − 3 b ) x 3 + ( d − 3 c ) x 2 + ( e − 3 d ) x + ( 1 − 3 e ) \displaystyle f(x)= (a - 3) x^5 + (b - 3a) x^4 + (c - 3b) x^3 + (d - 3c) x^2 + (e - 3d) x + (1 - 3e) We have a function f(x) with a, b, c, d, and e as single-digit natural numbers. If f(10) = 0, then find 1 a b c d e ‾ \overline{1abcde} .


The answer is 142857.

Rajen Kapur by Rajen Kapur , 72, India

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1 solution

Rajen Kapur
Jul 10, 2015

Putting x = 10 in f(x) = 0 and transposing 3 ( 1 0 5 + a . 1 0 4 + b . 1 0 3 + c . 1 0 2 + d . 10 + e ) = a . 1 0 5 + b . 1 0 4 + c . 1 0 3 + d . 1 0 2 + e . 10 + 1 3(10^5 + a.10^4 + b.10^3 + c.10^2 + d.10 + e) = a.10^5 + b.10^4 + c.10^3 + d.10^2 + e.10 + 1 → 3 ∗ ( 1 a b c d e ‾ ) = a b c d e 1 ‾ \rightarrow 3*(\overline{1abcde}) = \overline{abcde1} This problem is now manipulated by a well-known method of back calculation to give 3 * (142857) = 428571.

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