CMC - Problem 1

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

a+b+c+d=42,a+b+c+d=42\text{,} ab+ac+ad+bc+bd+cd=2013, andab+ac+ad+bc+bd+cd=2013\text{, and} a3+b3+c3+d3+abc+abd+acd+bcd=1337a^3+b^3+c^3+d^3+abc+abd+acd+bcd=1337

Find the last three digits of a4+b4+c4+d4+4abcda^4+b^4+c^4+d^4+4abcd.

Note by Cody Johnson
5 years, 10 months ago

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17 votes

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560

Pi Han Goh - 5 years, 10 months ago

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Use Newton's Sum

We have e1=p1=42e_1 = p_1 = 42

2e2=e1p1p22(2013)=42(42)p2p2=22622e_2 = e_1 p_1 - p_2 \Rightarrow 2(2013) = 42(42) - p_2 \Rightarrow p_2 = -2262

3e3=e2p1e1p2+p33e3=(2013)(42)(42)(2262)+p33e_3 = e_2 p_1 - e_1 p_2 + p_3 \Rightarrow 3e_3 = (2013)(42)-(42)(-2262) + p_3

3e3p3=179550\Rightarrow 3e_3 - p_3 = 179550

Given e3+p3=1337e_3 + p_3 = 1337 , solve them simultaneously gives e3=45221.75,p3=43884.75e_3 = 45221.75, p_3 = -43884.75

Lastly,

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

4e4+p4=45221.75(42)2013(2262)+42(43884.75)=4609560 4e_4 + p_4 = 45221.75(42) - 2013(-2262) + 42(-43884.75) = 4609560

Pi Han Goh - 5 years, 10 months ago

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Aaaaaand the 5 points goes to Pi Han Goh!

Official solution:

Let a,b,c,da,b,c,d be roots of a polynomial

f(x)=x442x3+2013x2(1337(a3+b3+c3+d3))x+abcd)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)=0f(a)+f(b)+f(c)+f(d)=0 to get

0=a4+b4+c4+d4+4abcd42(a3+b3+c3+d3)+2013(4222(2013))1337(42)+42(a3+b3+c3+d3)(a4+b4+c4+d4+4abcd)560(mod1000)\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 a4+b4+c4+d4+4abcd560(mod1000)a^4+b^4+c^4+d^4+4abcd\equiv\boxed{560}\pmod{1000}

Cody Johnson - 5 years, 10 months ago

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

Ahaan Rungta - 5 years, 10 months ago

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I think the answer is 560\boxed{560}

A Former Brilliant Member - 5 years, 10 months ago

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That is correct! Solution?

Cody Johnson - 5 years, 10 months ago

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560

Rajas Salpekar - 5 years, 10 months ago

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Problem 2.

Cody Johnson - 5 years, 10 months ago

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560

Marviliour Wikki - 5 years, 10 months ago

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

Last digit are.....560

Abhishek Pal - 5 years, 10 months ago

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

Akbarali Surani - 5 years, 10 months ago

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

Cody Johnson - 5 years, 10 months ago

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

Akbarali Surani - 5 years, 10 months ago

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3392

sunitha bhadragiri - 5 years, 10 months ago

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2013+42+1337=3392

sunitha bhadragiri - 5 years, 10 months ago

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