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By congruency, we know that

$2^{100}$ is congruent to 2 modulo 7

$3^{100}$ is congruent to 4 modulo 7

$2^{200}$ is congruent to 4 modulo 7

$5^{100}$ is congruent to 2 modulo 7

Adding all, we get, req. is congruent to 5 modulo 7. Hence, 5

Watch my video for the solution to this problem -> http://youtu.be/Oh-b3DT_xow

For the solution in text: ---Part 1. - Finding the remainder of $2^{ 100 }$ --- We can rewrite any number as a + b, where 'a' is divisible by 7 and 'b' is a remainder. When we square this number, we will get $(a+b)^{ 2 }$ = $a^{ 2 } + 2ab + b^{ 2 }$ , so the remainder will also be squared (the first two terms are divisible by 7 since they contain 'a'). The remainder when $2^{ 3 }$ is divided by 7 is 1, so when 8 is taken to any power of a power of 2 (as in $8^{ 2^{ n } })$ and divided by 7, the remainder will still be 1. When substituting n with 5, you get $2^{ 96 }$ . To get to $2^{ 100 }$ from $2^{ 96 }$ , one must multiply by $2^{ 4 }$ (16), in turn multiplying the remainder (1) by 16. Having a remainder of 16 is mathematically equivalent to having a remainder of 2 since 16 is still divisible by 7. The remainder when $2^{ 100 }$ is divided by 7 is therefore 2.

---Part 2. The remainder of $4^{ 100 }$ --- The remainder when $2^{ 100 } + 3^{ 100 } + 4^{ 100 } + 5^{ 100 }$ is divided by 7 is equivalent to the sum of the remainders when each of the 4 terms are divided by 7. We already know the remainder of $\frac{ 2^{ 100 } }{ 7 } = 2$ . From exponent laws we know $4^{ 100 } = 2^{ 200 } = (2^{ 100 })^{ 2 }$ , and as stated in part 1 of this solution, we showed how squaring a number will result in it's remainder becoming squared too. Therefore the remainder when $2^{ 200 }$ is divided by 7 is $2^{ 2 } = 4$ .

---Part 3. The remainder of $3^{ 100 }$ and $5^{ 100 }$ --- Note that 5 can be rewritten as 7-2, so $5^{ 100 } = (7-2)^{ 100 }$ . The binomial expansion looks something like $(7-2)^{ 100 } = 7^{ 100 } - 7^{ 99 } \times 2^{ 1 } + 7^{ 98 } \times 2^{ 2 } -.... +7^{ 2 } \times 2^{ 98 } - 7^{ 1} \times 2^{ 99 } +2^{ 100 }$ . As you can see all terms except the last one contain at least one 7 and are therefore divisible by 7. Since only the term $2^{ 100 }$ is not divisible by 7 in the expansion, the remainder of $5^{ 100 }$ is equivalent to that of $2^{ 100 }$ , which is equal to 2. We can do the same thing for $3^{ 100 }$ , as 3 can be rewritten as 7-4 and $(7-4)^{ 100 } = 7^{ 100 } - 7^{ 99 } \times 4^{ 1 } +.... -7^{ 1 } \times 4^{ 99 } + 4^{ 100 }$ . Again the only term not divisible by 7 is $4^{100}$ , which means the remainder of $3^{100}$ is the same as that of $4^{100}$ which is equal to 4.

Adding all the remainders of $2^{100}+3^{100}+4^{100}+5^{100}=2+4+4+2=12$ . Having a remainder of 12 is equivalent to having a remainder of 5 since 12 is still divisible by 7.

Therefore, the remainder when $2^{100}+3^{100}+4^{100}+5^{100}$ is divided by 7 = $\boxed{5}$

2+3+4+5=14 modulo 7 =0 2^2 + 3^2 + 4^2 + 5^2= 54 modulo 7 =5 generalizing 2^n + 3^n + 4^n + 5^n =x modulo 7 = 0 or 5 when n is odd or even...