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# 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
3 years, 12 months ago

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## Comments

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560

- 3 years, 12 months ago

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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,

$$4e_4 = e_3 p_1 - e_2 p_2 + e _1 p_3 - p_4$$

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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That is correct! Solution?

- 3 years, 12 months ago

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

Last digit are.....560

- 3 years, 11 months ago

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560

- 3 years, 12 months ago

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- 3 years, 12 months ago

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560

- 3 years, 12 months ago

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2013+42+1337=3392

- 3 years, 12 months ago

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3392

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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

- 3 years, 12 months ago

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