All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Integer Count Illation (Posted on 2015-10-12) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution Comment 1 of 1
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

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information