**Problem 1 (5 points).** Let \(a,b,c,d\) be complex numbers satisfying

\[a+b+c+d=42\text{,}\] \[ab+ac+ad+bc+bd+cd=2013\text{, and}\] \[a^3+b^3+c^3+d^3+abc+abd+acd+bcd=1337\]

Find the last three digits of \(a^4+b^4+c^4+d^4+4abcd\).

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## Comments

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TopNewest560

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Use Newton's Sum

We have \(e_1 = p_1 = 42 \)

\(2e_2 = e_1 p_1 - p_2 \Rightarrow 2(2013) = 42(42) - p_2 \Rightarrow p_2 = -2262 \)

\(3e_3 = e_2 p_1 - e_1 p_2 + p_3 \Rightarrow 3e_3 = (2013)(42)-(42)(-2262) + p_3\)

\(\Rightarrow 3e_3 - p_3 = 179550 \)

Given \(e_3 + p_3 = 1337 \), solve them simultaneously gives \(e_3 = 45221.75, p_3 = -43884.75 \)

Lastly,

\(4e_4 = e_3 p_1 - e_2 p_2 + e_1 p_3 - p_4 \)

\( 4e_4 + p_4 = 45221.75(42) - 2013(-2262) + 42(-43884.75) = 4609560 \)

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Aaaaaand the 5 points goes to Pi Han Goh!

Official solution:

Let \(a,b,c,d\) be roots of a polynomial

\[f(x)=x^4-42x^3+2013x^2-(1337-(a^3+b^3+c^3+d^3))x+abcd)\]

Add \(f(a)+f(b)+f(c)+f(d)=0\) to get

\[\begin{align*} 0&=a^4+b^4+c^4+d^4+4abcd-42(a^3+b^3+c^3+d^3)\\&+2013(42^2-2(2013))-1337(42)+42(a^3+b^3+c^3+d^3)\\&\equiv (a^4+b^4+c^4+d^4+4abcd)-560\pmod{1000} \end{align*}\]

so that \(a^4+b^4+c^4+d^4+4abcd\equiv\boxed{560}\pmod{1000}\)

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I think the answer is

\(\boxed{560}\)Log in to reply

That is correct! Solution?

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I got....... a^{4} + b^{4} + c^{4} + d^{4} + 4abcd =4609560

Last digit are.....560

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560

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Problem 2.

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560

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2013+42+1337=3392

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3392

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a^{2} + b^{2} + c^{2} + d^{2} = -2262

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Post your

answer only, then post a solution. Refer to the rules here.Log in to reply

4(abc + abd + acd + bcd) = 180887

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