A Special Sequence - For my Friend Sharky Kesa!

Algebra Level 4

A sequence of numbers x 1 , x 2 , … , x 100 x_{1},x_{2},\ldots ,x_{100} has the property that for every integer k k between 1 and 100 inclusive the number x k x_{k} is k k less than the sum of other 99 numbers. Given that x 50 = m n x_{50}=\dfrac mn , where m m and n n are relatively prime positive integer, find m + n m+n .


The answer is 173.

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Department 8 by Department 8 , 27, Israel

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2 solutions

Aareyan Manzoor
Dec 15, 2015

we found that 2 x n = S − n 2x_n=S-n where is the sum of all the variables. so: 22 x 1 = S − 1 2 x 2 = S − 2 … … 2 x 99 = S − 99 2 x 100 = S − 100 \begin{array}{c}22x_1&=&S&-&1\\ 2x_2&=&S&-&2\\ \dots\dots\\ 2x_{99}&=&S&-&99\\ 2x_{100}&=&S&-&100\end{array} we add them to find 2 ( x 1 + x 2 + . . . + x 99 + x 100 ) = 101 S − ( 1 + 2 + . . + 99 + 100 ) 2(x_1+x_2+...+x_{99}+x_{100})=101S-(1+2+..+99+100) 2 S = 100 S − 5050 2S=100S-5050 S = 5050 98 = 2525 49 S=\frac{5050}{98}=\frac{2525}{49} we generalize for x n x_n : x n = S − n 2 = 2525 49 − n 2 = 2525 − 49 n 98 x_n=\frac{S-n}{2}=\frac{\frac{2525}{49}-n}{2}=\frac{2525-49n}{98} put n=50 x 50 = 2525 − 49 ∗ 50 98 = 75 98 x_{50}=\frac{2525-49*50}{98}=\frac{75}{98} and 75 + 98 = 173 75+98=\boxed{173}

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Simple Explanation ! +1

Venkata Karthik Bandaru - 5 years, 5 months ago
Julian Poon
Dec 15, 2015

It is easy to see that 2 x k + k = S 2x_k+k=S , where S = ∑ n = 1 100 x n \displaystyle S=\sum _{n=1}^{100}x_n .

So S = 2 x k + k = 2 x k + 1 + k + 1 S=2x_k+k=2x_{k+1}+k+1

x k + 1 = 1 2 ( 2 x k − 1 ) x_{k+1}=\frac{1}{2}\left(2x_k-1\right)

This gives x k = x 1 − k − 1 2 x_k=x_1-\frac{k-1}{2}

Now,

2 x 1 + 1 = S = ∑ n = 1 100 ( x 1 − n − 1 2 ) = 100 x 1 − 2475 2x_1+1=S=\sum _{n=1}^{100}\left(x_1-\frac{n-1}{2}\right)=100x_1-2475 x 1 = 2476 98 x_1=\frac{2476}{98}

x 50 = x 1 − 49 2 = 75 98 x_{50}=x_1-\frac{49}{2}=\boxed{\frac{75}{98}}

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Moderator note:

Good observation with comparing the terms.

I did the same

Shreyash Rai - 5 years, 6 months ago

How is 2 x k + k = S 2x_k + k = S ?

neelesh vij - 5 years, 6 months ago

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x k = x 1 + x 2 + . . . + x k − 1 + x k + 1 + . . + x 100 − k x_k=x_1+x_2+...+x_{k-1}+x_{k+1}+..+x_{100}-k add x k + k x_k+k to both sides: 2 x k + k = x 1 + x 2 + . . . . + x 99 + x 100 2x_k+k=x_1+x_2+....+x_{99}+x_{100} 2 x k + k = S 2x_k+k=S in the case this helps, feel free to upvote my solution.

Aareyan Manzoor - 5 years, 6 months ago

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Got it thanks!!

neelesh vij - 5 years, 6 months ago

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