But how?

Little tedious calculation will get you the answer.

$2^{100}\equiv 2 mod 7$

$3^{100}\equiv 4 mod 7$

$4^{100}\equiv 4 mod 7$

$5^{100}\equiv 2 mod 7$

$2^{100}+3^{100}+4^{100}+5^{100} \equiv 12 mod 7$

And

$12 \equiv 5 mod 7$

Thus the remainder is $5$

(2^100+3^100+4^100+5^100)/7

= (2.2^99 + 3.3^99 + 2^200 + 5.5^99)/7 = [2 (2^3)^33 + 3 (3^3)^33 + 2^2 (2^3)^66 + 5 (5^3)^33]/7 = [2 8^33 + 3 27^33 + 2^2 8^66 + 5 125^33]/7 = [2 (7+1)^33 + 3 (28-1)^33 + 2^2 (7+1)^66 + 5 (126-1)^33]/7   Taking remainders, [2 1^33 + 3 (-1)^33 + 2^2 1^66 + 5 (-1)^33]/7

= [2+ 3 (7-1)+ 4 + 5 (7-1)]/7; For [(-1)^n]/m; if n is odd, remainder is m - n. = [2 + 3 6 + 4 + 5 6]/7 = 54/7   => Remainder is 5