"L'alphabet du millénaire"

Algebra Level 2

A = 2000 B = A − 999 C = A + B − 998 D = A + B + C − 997 ⋮ Z = A + B + C + . . . + Y − 975 \begin{aligned} A&=&2000\\ \\ B&=&A-999\\ \\ C&=&A+B-998\\ \\ D&=&A+B+C-997\\ & \vdots & \\ Z&=&A+B+C+...+Y-975 \end{aligned}

How much is Z + 1 2 25 ? \frac { Z+1 }{ { 2 }^{ 25 } } ?


The answer is 501.

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Antony Diaz by Antony Diaz , 25, Canada

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3 solutions

Antony Diaz
Apr 24, 2014

F i r s t , I g o t t h a t B = 1001. A n d w e c a n s e e t h a t C = 2 B + 1 ⇒ 2 B + ( 2 − 1 ) D = 2 C + 1 = 4 B + 3 ⇒ 2 2 B + ( 2 2 − 1 ) E = 2 D + 1 = 8 B + 7 ⇒ 2 3 B + ( 2 3 − 1 ) F = 2 E + 1 = 16 B + 15 ⇒ 2 4 B + ( 2 4 − 1 ) . . . Z = 2 Y + 1 ⇒ 2 24 B + ( 2 24 − 1 ) N o w , Z + 1 2 25 = [ 2 24 ( B + 1 ) − 1 ] + 1 2 25 = B + 1 2 = 501. First,\quad I\quad got\quad that\quad B=1001.\quad And\quad we\quad can\quad see\quad that\quad \\ \\ C=2B+1\Rightarrow 2B+(2-1)\\ \\ D=2C+1=4B+3\Rightarrow { 2 }^{ 2 }B+({ 2 }^{ 2 }-1)\\ \\ E=2D+1=8B+7\Rightarrow { 2 }^{ 3 }B+({ 2 }^{ 3 }-1)\\ \\ F=2E+1=16B+15{ \Rightarrow 2 }^{ 4 }B+({ 2 }^{ 4 }-1)\\ .\\ .\\ .\\ Z=2Y+1\quad \Rightarrow \quad { 2 }^{ 24 }B+({ 2 }^{ 24 }-1)\\ \\ Now,\quad \frac { Z+1 }{ { 2 }^{ 25 } } =\frac { \left[ { 2 }^{ 24 }(B+1)-1 \right] +1 }{ { 2 }^{ 25 } } =\frac { B+1 }{ 2 } =501.

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Note, use \ ( \ ) only for math equations / statement. For 'normal' words, there is no need to use them. I've edited your question for your reference.

Calvin Lin Staff - 7 years, 1 month ago

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Thanks!

Antony Diaz - 7 years, 1 month ago
Otto Bretscher
Apr 30, 2015

I'm setting B = x 0 = 1001 , C = x 1 = 2003 , . . . , Z = x 24 B=x_0=1001, C=x_1=2003,... ,Z=x_{24} . We have the recursive formula x n = 2 x n − 1 + 1 x_{n}=2x_{n-1}+1 , or, x n + 1 = 2 ( x n − 1 + 1 ) x_{n}+1=2(x_{n-1}+1) , so that x n + 1 = 2 n ( x 0 + 1 ) = 1002 ( 2 n ) x_{n}+1=2^n(x_0+1)=1002(2^n) . Thus z + 1 2 25 = x 24 + 1 2 25 = 1002 ( 2 24 ) 2 25 = 501 \frac{z+1}{2^{25}}=\frac{x_{24}+1}{2^{25}}=\frac{1002(2^{24})}{2^{25}}=\boxed{501}

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How you know that 2(xsubn-1+1) is equal to 2n(xsub0+1)?

Sergio Alejandro Acelas Avila - 6 years, 1 month ago

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I'm applying the formula x n + 1 = 2 ( x n − 1 + 1 ) x_{n}+1=2(x_{n-1}+1) over and over again: x n + 1 = 2 ( x n − 1 + 1 ) = 4 ( x n − 2 + 1 ) + . . . x_{n}+1=2(x_{n-1}+1)=4(x_{n-2}+1)+... More formally, I'm using induction.

Otto Bretscher - 6 years, 1 month ago
Pop Wong
May 4, 2020

Noted Z + 1 2 25 \frac{Z+1}{2^{25}} = 2 Y + 1 + 1 2 25 \frac{{2Y+1}+1}{2^{25}} = Y + 1 2 24 \frac{Y+1}{2^{24}}

so Z + 1 2 25 \frac{Z+1}{2^{25}} = Y + 1 2 24 \frac{Y+1}{2^{24}} = X + 1 2 23 \frac{X+1}{2^{23}} = .... = C + 1 2 2 \frac{C+1}{2^{2}} = B + 1 2 1 \frac{B+1}{2^{1}} = A − 999 + 1 2 \frac{A-999+1}{2} = 501

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