what are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 449 * 349 ) ^ 2
"( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 449 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 6,7 , 9,9 ; 5 & 9 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,46 and final three as 47 * 49 * 49 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 46 ) ( 47 * 01 ) ) ^ 2 = ( 44 * 47 ) ^ 2 = 68 ^ 2 = 24 hence answer is e"
multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4)