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Problem:

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.

For example:

 persistence(for: 39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                       // and 4 has only one digit

persistence(for: 999) === 4 // because 999 = 729, 729 = 126,
// 126 = 12, and finally 1*2 = 2


persistence(for: 4) === 0 // because 4 is already a one-digit number

 persistence(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                       // and 4 has only one digit

 persistence(999) === 4 // because 9*9*9 = 729, 7*2*9 = 126,
                        // 1*2*6 = 12, and finally 1*2 = 2

 persistence(4) === 0 // because 4 is already a one-digit number
persistence(39) === 3; // because 3 * 9 = 27, 2 * 7 = 14, 1 * 4 = 4 and 4 has only one digit
persistence(999) === 4; // because 9 * 9 * 9 = 729, 7 * 2 * 9 = 126, 1 * 2 * 6 = 12, and finally 1 * 2 = 2
persistence(4) === 0; // because 4 is already a one-digit number
 persistence(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                       // and 4 has only one digit

 persistence(999) === 4 // because 9*9*9 = 729, 7*2*9 = 126,
                        // 1*2*6 = 12, and finally 1*2 = 2

 persistence(4) === 0 // because 4 is already a one-digit number
 persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                      // and 4 has only one digit

 persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
                       // 1*2*6 = 12, and finally 1*2 = 2

 persistence(4) == 0 // because 4 is already a one-digit number
persistence 39 = 3  // because 3*9 = 27, 2*7 = 14, 1*4=4
                    // and 4 has only one digit

persistence 999 = 4 // because 9*9*9 = 729, 7*2*9 = 126,
                    // 1*2*6 = 12, and finally 1*2 = 2

persistence 4 = 0   // because 4 is already a one-digit number
 persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                      // and 4 has only one digit

 persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
                       // 1*2*6 = 12, and finally 1*2 = 2

 persistence(4) == 0 // because 4 is already a one-digit number
 (persistence 39) ; returns 3, because 3*9=27, 2*7=14, 1*4=4
                  ; and 4 has only one digit

 (persistence 999) ; returns 4, because 9*9*9=729, 7*2*9=126,
                   ; 1*2*6=12, and finally 1*2=2

 (persistence 4) ; returns 0, because 4 is already a one-digit number
 persistence(39) => 3  # Because 3*9 = 27, 2*7 = 14, 1*4=4
                       # and 4 has only one digit.

 persistence(999) => 4 # Because 9*9*9 = 729, 7*2*9 = 126,
                       # 1*2*6 = 12, and finally 1*2 = 2.

 persistence(4) => 0   # Because 4 is already a one-digit number.
 persistence 39 -- returns 3, because 3*9=27, 2*7=14, 1*4=4
                -- and 4 has only one digit

 persistence 999 -- returns 4, because 9*9*9=729, 7*2*9=126,
                 -- 1*2*6=12, and finally 1*2=2

 persistence 4 -- returns 0, because 4 is already a one-digit number
 persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit

 persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

 persistence(4) # returns 0, because 4 is already a one-digit number
 persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit

 persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

 persistence(4) # returns 0, because 4 is already a one-digit number
 persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit

 persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

 persistence(4) # returns 0, because 4 is already a one-digit number
 persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                 # and 4 has only one digit

 persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                  # 1*2*6=12, and finally 1*2=2

 persistence(4) # returns 0, because 4 is already a one-digit number
 persistence(39) == 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                       // and 4 has only one digit

 persistence(999) == 4 // because 9*9*9 = 729, 7*2*9 = 126,
                        // 1*2*6 = 12, and finally 1*2 = 2

 persistence(4) == 0 // because 4 is already a one-digit number
 // because 3*9 = 27, 2*7 = 14, 1*4=4 and 4 has only one digit:
 persistence(39) == 3

 // because 9*9*9 = 729, 7*2*9 = 126, 1*2*6 = 12, and finally 1*2 = 2
 persistence(999) == 4

 // because 4 is already a one-digit number:
 persistence(4) == 0
persistence(39) # returns 3, because 3*9=27, 2*7=14, 1*4=4
                # and 4 has only one digit

persistence(999) # returns 4, because 9*9*9=729, 7*2*9=126,
                 # 1*2*6=12, and finally 1*2=2

persistence(4) # returns 0, because 4 is already a one-digit number
persistence(39) // returns 3, because 3*9=27, 2*7=14, 1*4=4
                // and 4 has only one digit

persistence(999) // returns 4, because 9*9*9=729, 7*2*9=126,
                 // 1*2*6=12, and finally 1*2=2

persistence(4) // returns 0, because 4 is already a one-digit number

Solution