what are the last two digits of ( 301 * 402 * 503 * 604 * 645 * 547 * 448 * 349 ) ^ 2
( ( 301 * 402 * 503 * 604 * 645 ) * ( 547 * 448 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 5,7 , 8,9 ; 6 is 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 , 04,45 and final three as 47 * 48 * 49 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 45 ) ( 56 * 49 ) ) ^ 2 = ( 00 * 44 ) ^ 2 = 00 ^ 2 = 00 hence answer is e
multiply(add(divide(const_100, const_4), multiply(multiply(multiply(multiply(const_2, const_3), subtract(const_1, 2)), subtract(const_1, 2)), subtract(const_1, 2))), const_4)