×

# 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
4 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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

Last digit are.....560

- 4 years ago

560

- 4 years ago

- 4 years ago

560

- 4 years ago

2013+42+1337=3392

- 4 years ago

3392

- 4 years ago

560

- 4 years 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,

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

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

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

- 4 years ago

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

- 4 years ago

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

- 4 years ago

That is correct! Solution?

- 4 years ago

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

- 4 years ago

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

- 4 years ago

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

- 4 years ago