# CMC - Problem 1

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 $a^4+b^4+c^4+d^4+4abcd$.

Note by Cody Johnson
6 years, 6 months ago

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560

- 6 years, 6 months ago

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

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

- 6 years, 6 months ago

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{aligned} 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{aligned}

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

- 6 years, 6 months ago

Same solution; nice problem! Very similar to one of my Brilliant problems. ;)

- 6 years, 6 months ago

I think the answer is $\boxed{560}$

- 6 years, 6 months ago

That is correct! Solution?

- 6 years, 6 months ago

560

- 6 years, 6 months ago

- 6 years, 6 months ago

560

- 6 years, 6 months ago

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

Last digit are.....560

- 6 years, 5 months ago

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

- 6 years, 6 months ago

Post your answer only, then post a solution. Refer to the rules here.

- 6 years, 6 months ago

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

- 6 years, 6 months ago

3392

- 6 years, 6 months ago

2013+42+1337=3392

- 6 years, 6 months ago