What is the modulo 5 remainder of the sum 1+2+3+4+5+...+120+121+122+123??
Sum of 1st 120 integers = (120) (121) / 2 = 60 * 121
(60 * 121) mod 5 = 0 since 60 is evenly divisiblle by 5
121 + 122 + 123 =
366
366 mod 5 = 1
Remainder = 1
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