1235 1253 1325 . . . . . . . . . + 5321 the addition problem above shows 4 of the 24 different integers that can be formed by using each of the digits 12 , 35 exactly once in each integer . what is the sum of these 24 integers ?
in each of the units , tens , hundreds , and thousands place , each digit will appear 6 times . then the sum equals 6 ( 1111 ) + 6 ( 2222 ) + 6 ( 3333 ) + 6 ( 5555 ) = 73,326 the answer is a .
divide(add(add(add(multiply(add(add(add(floor(divide(12, const_10)), reminder(12, const_10)), floor(divide(35, const_10))), reminder(35, const_10)), divide(24, 4)), multiply(multiply(add(add(add(floor(divide(12, const_10)), reminder(12, const_10)), floor(divide(35, const_10))), reminder(35, const_10)), divide(24, 4)), const_1000)), multiply(multiply(add(add(add(floor(divide(12, const_10)), reminder(12, const_10)), floor(divide(35, const_10))), reminder(35, const_10)), divide(24, 4)), const_100)), multiply(multiply(add(add(add(floor(divide(12, const_10)), reminder(12, const_10)), floor(divide(35, const_10))), reminder(35, const_10)), divide(24, 4)), const_10)), const_1000)