Problem 1 (5 points). Let \(a,b,c,d\) be complex numbers satisfying
a+b+c+d=42,a+b+c+d=42\text{,}a+b+c+d=42, ab+ac+ad+bc+bd+cd=2013, andab+ac+ad+bc+bd+cd=2013\text{, and}ab+ac+ad+bc+bd+cd=2013, and a3+b3+c3+d3+abc+abd+acd+bcd=1337a^3+b^3+c^3+d^3+abc+abd+acd+bcd=1337a3+b3+c3+d3+abc+abd+acd+bcd=1337
Find the last three digits of a4+b4+c4+d4+4abcda^4+b^4+c^4+d^4+4abcda4+b4+c4+d4+4abcd.
Note by Cody Johnson 7 years, 6 months ago
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Use Newton's Sum
We have e1=p1=42e_1 = p_1 = 42 e1=p1=42
2e2=e1p1−p2⇒2(2013)=42(42)−p2⇒p2=−22622e_2 = e_1 p_1 - p_2 \Rightarrow 2(2013) = 42(42) - p_2 \Rightarrow p_2 = -2262 2e2=e1p1−p2⇒2(2013)=42(42)−p2⇒p2=−2262
3e3=e2p1−e1p2+p3⇒3e3=(2013)(42)−(42)(−2262)+p33e_3 = e_2 p_1 - e_1 p_2 + p_3 \Rightarrow 3e_3 = (2013)(42)-(42)(-2262) + p_33e3=e2p1−e1p2+p3⇒3e3=(2013)(42)−(42)(−2262)+p3
⇒3e3−p3=179550\Rightarrow 3e_3 - p_3 = 179550 ⇒3e3−p3=179550
Given e3+p3=1337e_3 + p_3 = 1337 e3+p3=1337, solve them simultaneously gives e3=45221.75,p3=−43884.75e_3 = 45221.75, p_3 = -43884.75 e3=45221.75,p3=−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 4e4+p4=45221.75(42)−2013(−2262)+42(−43884.75)=4609560
Aaaaaand the 5 points goes to Pi Han Goh!
Official solution:
Let a,b,c,da,b,c,da,b,c,d be roots of a polynomial
f(x)=x4−42x3+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)f(x)=x4−42x3+2013x2−(1337−(a3+b3+c3+d3))x+abcd)
Add f(a)+f(b)+f(c)+f(d)=0f(a)+f(b)+f(c)+f(d)=0f(a)+f(b)+f(c)+f(d)=0 to get
0=a4+b4+c4+d4+4abcd−42(a3+b3+c3+d3)+2013(422−2(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}0=a4+b4+c4+d4+4abcd−42(a3+b3+c3+d3)+2013(422−2(2013))−1337(42)+42(a3+b3+c3+d3)≡(a4+b4+c4+d4+4abcd)−560(mod1000)
so that a4+b4+c4+d4+4abcd≡560(mod1000)a^4+b^4+c^4+d^4+4abcd\equiv\boxed{560}\pmod{1000}a4+b4+c4+d4+4abcd≡560(mod1000)
@Cody Johnson – Same solution; nice problem! Very similar to one of my Brilliant problems. ;)
I think the answer is 560\boxed{560}560
That is correct! Solution?
Problem 2.
I got....... a^{4} + b^{4} + c^{4} + d^{4} + 4abcd =4609560
Last digit are.....560
a^{2} + b^{2} + c^{2} + d^{2} = -2262
Post your answer only, then post a solution. Refer to the rules here.
4(abc + abd + acd + bcd) = 180887
3392
2013+42+1337=3392
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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560
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Use Newton's Sum
We have e1=p1=42
2e2=e1p1−p2⇒2(2013)=42(42)−p2⇒p2=−2262
3e3=e2p1−e1p2+p3⇒3e3=(2013)(42)−(42)(−2262)+p3
⇒3e3−p3=179550
Given e3+p3=1337, solve them simultaneously gives e3=45221.75,p3=−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
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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)=x4−42x3+2013x2−(1337−(a3+b3+c3+d3))x+abcd)
Add f(a)+f(b)+f(c)+f(d)=0 to get
0=a4+b4+c4+d4+4abcd−42(a3+b3+c3+d3)+2013(422−2(2013))−1337(42)+42(a3+b3+c3+d3)≡(a4+b4+c4+d4+4abcd)−560(mod1000)
so that a4+b4+c4+d4+4abcd≡560(mod1000)
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I think the answer is 560
Log in to reply
That is correct! Solution?
560
Problem 2.
560
I got....... a^{4} + b^{4} + c^{4} + d^{4} + 4abcd =4609560
Last digit are.....560
a^{2} + b^{2} + c^{2} + d^{2} = -2262
Log in to reply
Post your answer only, then post a solution. Refer to the rules here.
4(abc + abd + acd + bcd) = 180887
3392
2013+42+1337=3392