if n = ( 33 ) ^ 43 + ( 53 ) ^ 33 what is the units digit of n ?
"first of all , the units digit of ( 33 ) ^ 53 is the same as that of 3 ^ 53 and the units digit of ( 53 ) ^ 33 is the same as that of 3 ^ 33 . so , we need to find the units digit of 3 ^ 53 + 3 ^ 33 . next , the units digit of 3 in positive integer power repeats in blocks of four { 3 , 9 , 7 , 1 } : 3 ^ 1 = 3 ( the units digit is 3 ) 3 ^ 2 = 9 ( the units digit is 9 ) 3 ^ 3 = 27 ( the units digit is 7 ) 3 ^ 4 = 81 ( the units digit is 1 ) 3 ^ 5 = 243 ( the units digit is 3 again ! ) . . . thus : the units digit of 3 ^ 53 is the same as the units digit of 3 ^ 3 , so 7 ( 53 divided by the cyclicity of 4 gives the remainder of 3 ) . the units digit of 3 ^ 33 is the same as the units digit of 3 ^ 1 , so 3 ( 33 divided by the cyclicity of 4 gives the remainder of 1 ) . therefore the units digit of ( 33 ) ^ 53 + ( 53 ) ^ 33 is 7 + 7 = 4 . answer : c ."
subtract(reminder(add(power(reminder(33, const_10), reminder(43, const_4)), power(reminder(43, const_10), const_4)), const_100), multiply(const_2, const_3))