Problem: how many 3 digit positive integers t exist that when divided by 7 leave a remainder of 5 ?
Rationale: "minimum three digit number is 100 and maximum three digit number is 999 . the first three digit number that leaves remainder 5 when divided by 7 is 103 . 14 * 7 = 98 + 5 = 103 the second three digit number that leaves remainder 5 when divided by 7 is 110 . 15 * 7 = 105 + 5 = 110 the third three digit number that leaves remainder 5 when divided by 7 is 117 and so on the last three digit number that leaves remainder 5 when divided by 7 is 999 142 * 7 = 994 + 5 = 999 therefore , we identify the sequence 103 , 110,117 . . . . . 999 use the formula of last term last term = first term + ( n - 1 ) * common difference you will get the answer 129 that is definitely e ."
annotated_formula: divide(subtract(subtract(multiply(const_100, const_10), const_1), add(multiply(add(const_10, const_4), 7), 5)), 7)