Let f ( x ) be a quartic polynomial in the form a x 4 + b x 3 + c x 2 + d x + e where
f ( − 7 ) = 4 6 3 6 , f ( − 6 ) = 2 5 0 4 , f ( − 5 ) = 1 2 1 2 .
Given that a , b , c , d and e are all positive integers less than 5. Evaluate the addition of the two 3-digit integers, a b c + c d e .
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Great solution using the idea of negative integer bases.
Lagrange interpolation can be used as well. We must have (the last three terms define the unique quadratic polynomial taking the desired values at − 7 , − 6 , − 5 , and the first term makes f ( x ) as general a quartic as possible): f ( x ) = ( α x + β ) ( x + 5 ) ( x + 6 ) ( x + 7 ) + 2 4 6 3 6 ( x + 5 ) ( x + 6 ) − 2 5 0 4 ( x + 5 ) ( x + 7 ) + 2 1 2 1 2 ( x + 6 ) ( x + 7 ) for some values of α and β . The coefficients of x 4 , x 3 and 1 in f ( x ) are α , 1 8 α + β and 7 3 5 2 + 2 1 0 β respectively. Since each of these coefficients must be integers between 1 and 4 , inclusive, it follows that α and β must both be integers, and indeed that α = 2 and β = − 3 5 . Thus f ( x ) = 2 x 4 + x 3 + 4 x 2 + 3 x + 2 and so the answer is 2 1 4 + 4 3 2 = 6 4 6 .
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This was my inspiration: Negative Integer Number Base
We can use negative integers bases of -5 and below since we are told that a , b , c . d . e < 5 and are all positive.
Hence we can take f ( − 5 ) = 1 2 1 2 :
1 2 1 2 in base − 5 can be calculated as:
1 2 1 2 ÷ − 5 = − 2 4 2 , remainder 2
− 2 4 2 ÷ − 5 = 4 9 , remainder 3
4 9 ÷ − 5 = − 9 , remainder 4
− 9 ÷ − 5 = 2 , remainder 1
2 ÷ − 5 = 0 , remainder 2
Reading the remainders from bottom to top we get 1 2 1 2 = 2 1 4 3 2 − 5
2 ⋅ ( − 5 ) 4 + 1 ⋅ ( − 5 ) 3 + 4 ⋅ ( − 5 ) 2 + 3 ⋅ ( − 5 ) 4 + 2 ⋅ ( − 5 ) 0 = 1 2 1 2
Therefore the polynomial is 2 x 4 + 1 x 3 + 4 x 2 + 3 x + 2
a b c + c d e = 2 1 4 + 4 3 2 = 6 4 6