Another Divisible

Algebra Level pending

3 2 n − 8 n − 1 i s d i v i s i b l e b y s o m e i n t e g e r X s i m p l y f i n d t h e g r e a t e s t X { 3 }^{ 2n }-8n-1 \quad is\quad divisible\quad by\quad some\quad integer\quad X\\ simply\quad find\quad the\quad greatest\quad X


The answer is 64.

Amr Saber by Amr Saber , 24, Egypt

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1 solution

Amr Saber
Apr 26, 2014

3 2 n − 8 n − 1 = ( 3 2 ) n − 8 n − 1 = 9 n − 8 n − 1 = ( 1 + 8 ) n − 8 n − 1 = ( 1 + n C 1 8 + n C 2 8 2 + n C 3 8 3 + . . . + n C n 8 n ) − 8 n − 1 b u t n C 1 = n = 1 + n 8 + n C 2 8 2 + n C 3 8 3 + . . . + n C n 8 n − 8 n − 1 = n C 2 8 2 + n C 3 8 3 + . . . + n C n 8 n = 8 2 ( n C 2 + n C 3 8 + n C 4 8 2 + . . . + n C n 8 n − 2 ) = 8 2 ( s o m e i n t e g e r s ) w h i c h i s d i v i s i b l e b y 8 2 = 64 { 3 }^{ 2n }-8n-1\quad =\quad ({ 3 }^{ 2 })^{ n }-8n-1\\ =\quad { 9 }^{ n }-8n-1\quad =\quad (1+8)^{ n }-8n-1\\ =\quad (1+^{ n }{ C }_{ 1 }8+^{ n }{ C }_{ 2 }{ 8 }^{ 2 }+^{ n }{ C }_{ 3 }{ 8 }^{ 3 }+\quad ...\quad +^{ n }{ C }_{ n }{ 8 }^{ n })-8n-1\quad but\quad ^{ n }{ C }_{ 1 }=n\\ =\quad 1+n8+^{ n }{ C }_{ 2 }{ 8 }^{ 2 }+^{ n }{ C }_{ 3 }{ 8 }^{ 3 }+\quad ...\quad +^{ n }{ C }_{ n }{ 8 }^{ n }-8n-1\\ =\quad ^{ n }{ C }_{ 2 }{ 8 }^{ 2 }+^{ n }{ C }_{ 3 }{ 8 }^{ 3 }+\quad ...\quad +^{ n }{ C }_{ n }{ 8 }^{ n }\\ =\quad { 8 }^{ 2 }(^{ n }{ C }_{ 2 }+^{ n }{ C }_{ 3 }{ 8 }+^{ n }{ C }_{ 4 }{ 8^{ 2 }+ }\quad ...\quad +^{ n }{ C }_{ n }{ 8 }^{ n-2 })\\ =\quad { 8 }^{ 2 }\quad (some\quad integers)\quad which\quad is\quad divisible\quad by\quad { 8 }^{ 2 }\quad =\quad 64

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