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Algebra Level 3

Given that x 1 , x 2 , x 3 , ⋯ x 7 x_1, x_2, x_3, \cdots x_7 satisfy the system of equations below:

{ x 1 + 4 x 2 + 9 x 3 + 16 x 4 + 25 x 5 + 36 x 6 + 49 x 7 = 1 4 x 1 + 9 x 2 + 16 x 3 + 25 x 4 + 36 x 5 + 49 x 6 + 64 x 7 = 12 9 x 1 + 16 x 2 + 25 x 3 + 36 x 4 + 49 x 5 + 64 x 6 + 81 x 7 = 123 \begin{cases} \begin{aligned} x_1 + 4 x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 & = 1 \\ 4 x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 & = 12 \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 & = 123 \end{aligned} \end{cases}

Find the value of 16 x 1 + 25 x 2 + 36 x 3 + 49 x 4 + 64 x 5 + 81 x 6 + 100 x 7 16x_1 + 25x_2 + 36x_3 + 49x_4 + 64x_5 + 81x_6 + 100x_7 .

Source: Problem 8 AIMEP 1989


The answer is 334.

Yugendra Uppalapati by Yugendra Uppalapati , 16, USA

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2 solutions

Chew-Seong Cheong
Aug 29, 2020

Given that { x 1 + 4 x 2 + 9 x 3 + 16 x 4 + 25 x 5 + 36 x 6 + 49 x 7 = 1 . . . ( 1 ) 4 x 1 + 9 x 2 + 16 x 3 + 25 x 4 + 36 x 5 + 49 x 6 + 64 x 7 = 12 . . . ( 2 ) 9 x 1 + 16 x 2 + 25 x 3 + 36 x 4 + 49 x 5 + 64 x 6 + 81 x 7 = 123 . . . ( 3 ) \begin{cases} x_1 + 4 x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 = 1 & ...(1) \\ 4 x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 = 12 & ...(2) \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 = 123 & ...(3) \end{cases}

{ ( 2 ) − ( 1 ) : 3 x 1 + 5 x 2 + 7 x 3 + 9 x 4 + 11 x 5 + 13 x 6 + 15 x 7 = 11 . . . ( 4 ) ( 3 ) − ( 2 ) : 5 x 1 + 7 x 2 + 9 x 3 + 11 x 4 + 15 x 5 + 15 x 6 + 17 x 7 = 111 . . . ( 5 ) ( 5 ) − ( 4 ) : 2 x 1 + 2 x 2 + 2 x 3 + 2 x 4 + 2 x 5 + 2 x 6 + 2 x 7 = 100 . . . ( 6 ) \begin{cases} (2)-(1): & 3 x_1 + 5 x_2 + 7 x_3 + 9 x_4 + 11 x_5 + 13 x_6 + 15 x_7 = 11 & ...(4) \\ (3)-(2): & 5 x_1 + 7 x_2 + 9 x_3 + 11 x_4 + 15 x_5 + 15 x_6 + 17 x_7 = 111 & ...(5) \\ (5)-(4): & 2 x_1 + 2 x_2 + 2 x_3 + 2 x_4 + 2 x_5 + 2 x_6 + 2 x_7 = 100 & ...(6) \end{cases}

Then ( 3 ) + ( 5 ) + ( 6 ) : 16 x 1 + 25 x 2 + 36 x 3 + 49 x 4 + 64 x 5 + 81 x 6 + 100 x 7 = 123 + 111 + 100 = 334 (3)+(5)+(6): \ \ 16x_1 + 25x_2 + 36x_3 + 49x_4 + 64x_5 + 81x_6 + 100x_7 = 123+111+100 = \boxed{334} .

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@Yugendra Uppalapati , I have amended the problem statement for you. There may not be real numbers x 1 , x 2 , x 3 , ⋯ x 7 x_1, x_2, x_3, \cdots x_7 satisfy the system of equations.

Chew-Seong Cheong - 9 months, 2 weeks ago

I solved this problem the same way. Also I forgot to mention this in the body and the title, but x 1, x 2, x 3,...,x 7 are meant to be real numbers.

Yugendra Uppalapati - 9 months, 2 weeks ago

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I am a moderator and I can change your problem statement. As I have mentioned I have removed the real numbers from your problem statement. They may be wrong because we may not be able to find all real numbers of x 1 , x 2 , x 3 , ⋯ x 7 x_1, x_2, x_3, \cdots x_7 that satisfy the system of equation. But it is definitely true that if they are complex numbers. Real numbers are complex numbers but complex numbers may not necessary be real numbers.

Chew-Seong Cheong - 9 months, 2 weeks ago

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Yes I understand that all real numbers are complex and I consent with the amendment. I should mention that I am not the original author of this problem and I will be linking the source of the problem: https://artofproblemsolving.com/wiki/index.php/1989AIMEProblems/Problem_8

Yugendra Uppalapati - 9 months, 2 weeks ago

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@Yugendra Uppalapati – I have added the source in the problem statement.

Chew-Seong Cheong - 9 months, 2 weeks ago
Pop Wong
Aug 29, 2020

Let R i R_i be i t h i^{th} row

( 1 4 9 16 25 36 49 4 9 16 25 36 49 64 9 16 25 36 49 64 81 ) ⋅ ( x 1 x 2 x 3 x 4 x 5 x 6 x 7 ) = ( 1 12 123 ) \left( \begin{array}{c}&1 &4 &9 &16 &25 & 36 &49 \\ &4 &9 &16 &25 & 36 &49 &64\\ &9 &16 &25 & 36 &49 &64 &81 \\ \end{array} \right) \cdot \left( \begin{array}{c}&x_1 \\ &x_2 \\ &x_3 \\ &x_4 \\ &x_5 \\ &x_6 \\ &x_7 \\ \end{array} \right) = \left( \begin{array}{c}&1 \\ &12 \\ &123 \end{array} \right)

R 3 − 2 R 2 + R 1 = 2 ( x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 ) = 100 [ ∗ ] R_3 - 2R_2 + R_1 = 2 (x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 ) = 100 \hspace{10mm} \text{ [}* \text{]} \\

∴ 16 x 1 + 25 x 2 + 36 x 3 + 49 x 4 + 64 x 5 + 81 x 6 + 100 x 7 = 2 ⋅ R 3 − R 2 + [ ∗ ] = 2 ( 123 ) − 12 + 100 = 334 \begin{aligned} \therefore 16x_1 + 25x_2 + 36x_3 + 49x_4 +64 x_5 + 81x_6 + 100x_7 &= 2\cdot R_3 - R_2 + [*] \\ &= 2(123) - 12 + 100 \\ &= \boxed{334} \end{aligned}

3 ( n + 2 ) 2 − 3 ( n + 1 ) 2 + n 2 = ( 3 n 2 + 12 n + 12 ) − ( 3 n 2 + 6 n + 3 ) + n 2 = n 3 + 6 n + 9 = ( n + 3 ) 2 \begin{aligned} 3(n+2)^2 - 3(n+1)^2 + n^2 &= (3n^2+12n+12) - (3n^2+6n + 3 ) + n^2 \\ &= n^3 + 6n + 9 \\ &= (n+3)^2 \end{aligned}

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