Problem: a palindrome is a number that reads the same forward and backward , such as 120 . how many odd , 4 - digit numbers are palindromes ?
Rationale: "a palindrome is a number that reads the same forward and backward . examples of four digit palindromes are 1221 , 4334 , 2222 etc you basically get to choose the first two digits and you repeat them in opposite order . say , you choose 45 as your first two digits . the next two digits are 54 and the number is 4554 . also , you need only odd palindromes . this means that you need an odd digit at the end i . e . 1 / 3 / 5 / 7 / 9 . this means that you need to start the number with an odd digit . only then will it end with an odd digit . in how many ways can you pick two digits such that the first one is an odd digit ? the first digit can be selected in 5 ways . ( 1 / 3 / 5 / 7 / 9 ) the second digit can be selected in 10 ways . ( 0 / 1 / 2 / 3 . . . 8 / 9 ) total = 5 * 11 = 55 ways b"
annotated_formula: divide(power(const_10, divide(4, const_2)), const_2)