Find the total number of 5-digit positive integers N such that:
The sum of the first digit and the third digit of N is equal to its second digit, and:
The sum of the third digit and the fifth digit of N is one more than its fourth digit.

*** Assume non leading zeroes for each of the values of N.

Let abcde be the number.

a+c = b   <=  9
c+e = d+1 <= 10

a > 0

b and d are determined by c, a and e.

c = 0: 9 choices for a * 9 choices for e  =  81 choices (1<=a<=9; 1<=e<=9)
c = 1: 8 choices for a * 10 choices for e =  80 choices (1<=a<=8; 0<=e<=9)
c = 2: 7 choices for a * 9 choices for e  =  63 choices (1<=a<=7; 0<=e<=8)   
c = 3: 6 choices for a * 8 choices for e  =  48 choices        etc.
c = 4: 5 choices for a * 7 choices for e  =  35 choices
c = 5: 4 choices for a * 6 choices for e  =  24 choices
c = 6: 3 choices for a * 5 choices for e  =  15 choices
c = 7: 2 choices for a * 4 choices for e  =   8 choices
c = 8: 1 choices for a * 3 choices for e  =   3 choices
c = 9: 0 choices for a * 2 choices for e  =   0 choices
                                            ---
                                            357 such numbers abcde.

Edited on October 12, 2015, 3:30 pm

Posted by Charlie on 2015-10-12 15:29:02