Symmetric System's Cyclic Sum

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Symmetric System's Cyclic Sum

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worthawholebean

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worthawholebean

#1 h Feb 11, 2008, 4:15 PM • 3 Y

Y by Khalid999, Adventure10, Mango247

If $a+1=b+2=c+3=d+4=a+b+c+d+5$ , then $a+b+c+d$ is

$\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$

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xpmath

#2 h Feb 11, 2008, 10:13 PM • 2 Y

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kentok

#3 h Jul 3, 2022, 12:55 PM

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Note that $a=d+3, b=d+2,$ and $c=d+1$ , so we have
$d+4=a+b+c+d+5\Rightarrow d+4=(d+3)+(d+2)+(d+1)+5\Rightarrow d=-\frac{7}{3}.$ So, $a+b+c+d=4d+6=-\frac{10}{3}$ .

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#4 h Jul 3, 2022, 1:35 PM

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$a+1=b+2=c+3=d+4=a+b+c+d+5=k$ result $a=k-1,b=k-2,c=k-3,d=k-4$ so from $a+b+c+d+5=k\Leftrightarrow 4k-5=k\Rightarrow k=\frac{5}{3}\Rightarrow a++b+c+d=-\frac{10}{3}$

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s12d34

#5 h Jul 3, 2022, 2:54 PM

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This post has been edited 3 times. Last edited by s12d34, Jul 3, 2022, 2:58 PM
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#6 h Jul 4, 2022, 2:00 AM

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$\begin{align*} a + 1 &= a + b + c + d + 5 \\ b + 2 &= a + b + c + d + 5 \\ c + 3 &= a + b + c + d + 5 \\ d + 4 &= a + b + c + d + 5 \\ \line(1,0){40} & \line(1,0){100} \ + \\ a + b + c + d + 10 &= 4(a + b + c + d) + 20 \\ 3(a + b + c + d) &= - 10 \\ \therefore \ \ \ \ \ a + b + c + d &= - \frac{10}{3} \end{align*}$

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#7 h May 21, 2025, 4:09 PM

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Solution 1: Subtracting each of the smaller equations from the large equation, we have $1 = b+c+d+5\\ 2=a+c+d+5\\ 3=a+b+d+5\\ 4=a+b+c+5.$ Adding these together, we have $10=3a+3b+3c+3d+20 \implies 3(a+b+c+d)=-10\implies a+b+c+d=\boxed{\dfrac{-10}{3}}$ .

Solution 2: Expressing $a$ , $b$ , $c$ , and $d$ in terms of $a$ , we have $a=a\\ b=a-1\\ c=a-2\\ d=a-3.$ Therefore, we have $a+b+c+d+5=4a-1$ . We know that $a+b+c+d+5=a+1$ , so $4a-1=a+1 \implies a=\dfrac{2}{3}$ , so $a+b+c+d=4a-6=\boxed{\dfrac{-10}{3}}$ .

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