An algebra problem by K. J. W.

Algebra Level 4

Let X= ${ \left( -1+\frac { 2 }{ -2+\frac { 2 }{ -2+\frac { 2 }{ -2+\frac { 2 }{ -2+... } } } } \right) }^{ 1000 }$ Reverse the last two integral digits of X to form a two-digit number Y.

Find the value of $\left\lfloor \frac { Y }{ 3 } \right\rfloor$ .

P.S. -1 is not an error

The answer is 3.

1 solution

K. J. W.
Jul 13, 2014

$Let\quad Z=-2+\frac { 2 }{ -2+\frac { 2 }{ -2+... } } \\ Then\quad X={ \left( Z+1 \right) }^{ 1000 }\\ Also,\quad Z=-2+\frac { 2 }{ Z } \\ i.e.\quad { Z }^{ 2 }=-2Z+2\\ Thus\quad { Z }^{ 2 }+2Z-2=0\\ Z=-1\pm \sqrt { 3 } \\ Therefore\quad Z+1=\pm \sqrt { 3 } \\ \pm { \sqrt { 3 } }^{ 1000 }\\ ={ 9 }^{ 250 }\\ Also\quad note\quad that\\ 81(20a+1(mod\quad 100))\\ \equiv 1620a+81(mod\quad 100)\\ \equiv 1600a+20a+81(mod\quad 100)\\ \equiv 20a-19(mod\quad 100)\\ \equiv 20a-20+1(mod\quad 100)\\ \equiv 20(a-1)+1(mod\quad 100)\\ Therefore,\quad since\quad the\quad last\quad two\quad digits\quad of\quad { 9 }^{ 2 }\quad are\quad 81,\quad the\quad last\quad two\quad digits\quad of\\ { 9 }^{ 4 },{ 9 }^{ 6 }...\quad are\quad 61,41,21...\\ Following\quad by\quad this,\quad the\quad last\quad two\quad digits\quad of\quad { 9 }^{ 250 }\quad are\quad 01.\\ Reverse\quad these\quad to\quad get\quad Y=10.\\ Then\left\lfloor \frac { Y }{ 3 } \right\rfloor =3.$

As that complex expression is going upto infinity.. So let that expression be z.. And we can take any no of branches bcz its going upto infinity... So by equating any 2 branches we get a quadratic equation in z and z =0.. Therefore X = 1^1000 = 01

Y = 10

[Y/3] = 3

Kïñshük Sïñgh - 6 years, 11 months ago

An algebra problem by K. J. W.

The answer is 3.

1 solution

Y = 10

[Y/3] = 3

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