Dare you to find the actual number.

Find the sum of the last three digits of 9 105 9^{105} .


The answer is 13.

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2 solutions

Alex Wang
Dec 6, 2014

We write 9 105 = ( 10 1 ) 105 9^{105} = (10-1)^{105} .

We then apply the binomial theorem. Terms that have powers of 10 that are greater than 1 0 2 10^2 do not contribute to the last three digits.

However, ( 105 2 ) 1 0 2 = 546000 \binom{105}{2} 10^2 =546000 and doesn't contribute to the last 3 digits either.

Therefore, the only terms that contribute to the last 3 digits is 105 10 1 = 1049 105 \cdot 10-1=1049 .

Therefore, the last 3 digits are 049, and our answer is 0 + 4 + 9 = 13 0+4+9=\boxed{13} .

Same method ;-)

Parth Bhardwaj - 6 years, 3 months ago
William Isoroku
Jan 20, 2015

Do 9 105 m o d 1000 { 9 }^{ 105 }\quad mod\quad 1000

We know that 9 5 49 ( m o d 1000 ) { 9 }^{ 5 }\equiv 49\quad (mod\quad 1000)

So 9 105 = ( 9 5 ) 21 49 21 ( m o d 1000 ) { 9 }^{ 105 }={ ({ 9 }^{ 5 } })^{ 21 }\equiv { 49 }^{ 21 }\quad (mod\quad 1000)

We also know that 49 3 m o d 1000 = 649 { 49 }^{ 3 }\quad mod\quad 1000=649

So 49 21 = ( 49 3 ) 7 649 7 ( m o d 1000 ) { 49 }^{ 21 }={ ({ 49 }^{ 3 }) }^{ 7 }\equiv { 649 }^{ 7 }\quad (mod\quad 1000)

Another thing we know: 649 2 m o d 1000 = 201 { 649 }^{ 2 }\quad mod\quad 1000=201

So 649 7 = ( 649 2 ) 3 649 201 3 649 ( m o d 1000 ) { 649 }^{ 7 }={ ({ 649 }^{ 2 }) }^{ 3 }\cdot 649\equiv { 201 }^{ 3 }\cdot 649\quad (mod\quad 1000)

And so 649 7 601 649 ( m o d 1000 ) { 649 }^{ 7 }\equiv 601\cdot 649\quad (mod\quad 1000)

390049 ( m o d 1000 ) \equiv 390049\quad (mod\quad 1000)

49 ( m o d 1000 ) \equiv 49\quad (mod\quad 1000) by the law that A B ( m o d n ) A\equiv B\quad (mod\quad n) is equal to A m o d n = B m o d n A\quad mod\quad n=B\quad mod\quad n

The last 3 digits are 049 049 , so the sum of the digits are 13 \boxed{13}

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