Remainder of a Huge Number

We are given of the following number: ${ 2 }^{ 1000 } + { 2 }^{ 1001 } + { 2 }^{ 1002 } + { 2 }^{ 1003 }$

Factoring that number, we get: ${ 2 }^{ 1000 }({ 2 }^{ 0 }+ { 2 }^{ 1 } +{ 2 }^{ 2 }+ { 2 }^{ 3 })\Rightarrow { 2 }^{ 1000 }(1+2+4+8)\Rightarrow { 2 }^{ 1000 }(15)$

That number is also equivalent to: ${ 2 }^{ 1000 }(14+1)\Rightarrow { 2 }^{ 1000 }(14)+{ 2 }^{ 1000 }$

Dividing the number by 7, it's quotient will be expressed as: ${ [2 }^{ 1000 }(14)+{ 2 }^{ 1000 }]/(7)$

$\Rightarrow \quad [{ 2 }^{ 1000 }(14)/7]+({ 2 }^{ 1000 }/7)$

Upon observation, you would see that the first term is already divisible by 7 (from dividing 14 by 7); thus, we can only get the remainder from the second term.

Organizing a table with the powers of 2 and the corresponding remainder when divided by 7, we obtain a pattern: ${ 2 }^{ 1 }\rightarrow 2\\ { 2 }^{ 2 }\rightarrow 4\\ { 2 }^{ 3 }\rightarrow 1\\ { 2 }^{ 4 }\rightarrow 2\\ { 2 }^{ 5 }\rightarrow 4\\ { 2 }^{ 6 }\rightarrow 1\\ \quad...$

wherein, for every set of 3 numbers, the remainders keep on repeating.

Notice that 3 and 6 are divisible by 3, and both have the same remainder. Since 999 is also divisible by 3, the remainder [i.e. 1] of ${ 2 }^{ 999 }$ will also be the same when divided by 7. Following that, the table is extended to: $\quad ...\\ { 2 }^{ 999 }\rightarrow 1\\ { 2 }^{ 1000 }\rightarrow 2$

Hence, the remainder of the number upon division by 7 is $\boxed{2}$

We can solve this question using $mod$ function.

$2^{1000}+2^{1001}+2^{1002}+2^{1003}$ $mod$ $7$

$2^{1000}(2^{0}+2^{1}+2^{2}+2^{3})$ $mod$ $7$

$2^{1000}(1+2+4+8)$ $mod$ $7$

$2^{1000}\times15$ $mod$ $7=2$

Thus, the answer is: $2^{1000}\times15$ $mod$ $7=\boxed{2}$

exp. = 2^{1000} (1 + 2 + 4 +8) = 2^{1000} 15 = 2^{1000} (14 + 1)
Since 2^{1000}
14 is divisible by 7, we need to test only 2^{1000}
2^{1000} ={ ( 2^10)^10 }^10
( 2^10)/7 gives remainder as 2 , repeating twice gives 2. This is only another method.
How ever the method Rafael Ticzon has given for the pattern of power of 2 is really better.
Hats off young Rafael Ticzon.

{ 2 }^{ 3 }\quad =\quad 1(mod\quad 7)\ { 2 }^{ 999 }\quad =\quad 1(mod\quad 7)\ { 2 }^{ 1000 }=\quad 2(mod\quad 7)\ { 2 }^{ 1001 }\quad =\quad 2\times 2(mod\quad 7)\quad =\quad 4(mod\quad 7)\ { 2 }^{ 1002 }\quad =\quad 8(mod\quad 7)\quad =\quad 1(mod\quad 7)\ { 2 }^{ 1003 }\quad =\quad 2(mod\quad 7)\ total\quad =\quad 2+4+1+2=9\ 9\quad =\quad 2(mod\quad 7) so the answer is 2.